phương trình, hệ phương trình không chứa căn

1/ Gii h phng trỡnh : (2) Gii: (2) ( t Khi ú (2) ( ( hoc ( ; ; ; x x + y y + = 2 x y + x + y 22 = 2 ( x 2) + ( y 3) = u 2+xv2 u2=2 420= u ( x + 4)( yx3xx=+==3) 22+ x 20 = v 3u==+02vv2) = u.v+y4( yyy===5533 2) Gii phng trỡnh: (1) x x 5.3 t =3 Gii: t (1) ( ( 5t 7t + 3t 31 = x3xy 3=+log 273= 18 y Gii: pt (*) ( x2 y + x3 =3y t a = 2x; b = (*) ( (2xx)3a=++3blog =ữ33 = 18 ( H ó cho cú nghim: (x:y) = = ab y, y13+ ; ữ x; + 13+ yữ 4/ Gii h phng trỡnh: (x, y ( y + x ) = y ữ 22x+ 2ữ x+ 4ữ = 3 ) (*) (x + y1)( y + x y 2) = y 3/ Gii h phng trỡnh: (*) xx==12 x2 + Gii: (*) ( hoc + y + x = x2 + =1 5/ Gii phng trỡnh: 3x.2x = 3x y y = 52 y + 2x + x + ( y + x 2) = y + x =1 y x (2 x 1) Gii: PT ( (1) Ta thy = x +1 x= khụng phi l nghim ca (1) Vi , ta cú: (1) ( ( x11+ x x x2+ 3 x= = t +x12x21 1x x 2x f (x) = =3 Ta cú: 2x x 11 x > 0, x Do ú f(x) ng bin trờn cỏc f ( x ) = ln + 11(2 x111) ; ữữ; + ữ ; ữ;, + khong v ( Phng 22 22 trỡnh f(x) = cú nhiu nht nghim trờn tng khong x = x = x = 1, x = x + + y (R x + y) = y ( x + 1)( x + y 2) = y x2 + + x+ y2=2 y 2x + u=x + 12( x, + v= x+ y2 y 2) = u +yvyx= + = u1 = v = y (R x + y) = y xuv+=11+y + xy + y2 = 2) =y ( x +x1)( x +1 + x+ y2=2 y u 2+x2vx+=21+ 21 u=x + ( x, + v=y u1 x= +2)vy==112 uvy=y1 y x + y = x + x + y = 2 y3 x=39+xx2yy=+592x xyx+ x52x = 18 Ta thy l cỏc nghim ca f(x) x + x x5=x 18 x + 18 = = Vy PT cú nghim 6/ Gii h phng trỡnh: (x, y x = x = ) Gii: y = khụng phi l nghim H PT ( + t Ta cú h ( 7/ Gii h phng trỡnh: (x, y ) Gii: y = khụng phi l nghim H PT ( t Ta cú h ( ==> Nghim ca hpt ó cho l (1; 2), (2; 5) 8/ Gii h phng trỡnh: Gii: H PT ( ( 9/ Gii h phng trỡnh: x y3 + 27 = y (1) Gii: T (1) ( y ( Khi ú H 2t8=x 3xy 3 y + 272 = y xy +3 26 x2 = y (2) PT ( ( 4t =+y6t 84tx +y27+=6 xy ( t = xy 31 131 ( Vi : T (1) ( y = t x== t t;==t = =ữ ; y =; t (loi) ( Vi : T (1) ( 223 222 ( Vi : T (1) ( x = t =2; y = ữ 10/ Gii h phng trỡnh: y 4x 2= 3 2 Gii: Ta cú: x y = ( y x ) ( y3 x )3 x + x y + xy y = x yy = =0 y x Khi thỡ h VN 3 32 x xyy 00 x c: ữ + ữ + ữ = ty3y+=2xt + y2t 5x=0y t =1 t , ta cú : t = x = y = 1, x = y = 11/ Gii h phng trỡnh: 31 2x y 3y = xy y = Gii: Ta cú : x y2 =2 xy = x y = 93 ( Khi: , ta cú: v y 3y=) 3== 427 x3 x( xy Suy ra: l cỏc X X 27 x3=; ( y ) X = 31 nghim ca phng trỡnh: Vy nghim ca H PT l: hoc x = + 31, y = + 31 327 Khi: , ta cú: v y=3 )== xx3 (xy Suy ra: l nghim X + X + x27 ( PTVN ) ; =0y ca phng trỡnh: 12/ Gii h phng trỡnh: y + 22 = 2 Gii: iu kin: x 0, x y+y0, x + y x t H PT tr thnh: 2 2 3x x + =u1x= x+ y+ y+4 1; == (1) uy+=vv22 u v y u + + 4v = 22 u = 21 4v (2) Thay (2) vo (1) ta c: v = 3 2 + = 2v 13v + 21 = v = 21 v ( Nu v = thỡ u = 9, x + y = 9v x + y = 10 x = x = ta cú H PT: x x = 3y y = y = ( Nu thỡ u = 7, = y v = ta cú H PT: 2 2 2 y = x + y = y = x + y = So sỏnh iu 53 53 kin ta c x = x = y x = 14 x = 14 nghim ca H y 53 53 PT 13/ Gii h phng trỡnh: x + y + xy + = y 2y Gii: T h PT ( y ( x + y ) = x + 7x y + 12+ x + y = Khi ú ta cú: x + y + xy + = y y t 2 x + x2 + x + y y ( x + y ) = x u+=7 y + , v = Ta cú h: u+v = uy = v ( x + y ) v2= 3, u ==17 y v = 3, u =1 ( Vi v 2u = v + 2v 15 = v = 5, u = Ta cú h: x2 + = y x2 + = y x2 + x = x = 1, y = 5, u = ( Vi ta cú h: x + y=x 23 + =9yy= xv2 =+ 1= x + x + x46==02, y = 9y = x , x + y = y = x y = x H ny vụ nghim (1; 2), ( 2; 5) Kt lun: H ó cho cú hai nghim: 14/ Gii h phng log1 x ( xy x + y + 2) + log + y ( x x + 1) = trỡnh: =1 log1 x ( y + 5) log 2+ y ( x + 4) Khi , chia v cho ta ( ) ( ) Gii: iu kin: xy x + y + > 0, x x + > 0, y + > 0, x + > (*) 2log1 x [(1 0x)( log ( 4) log ( x + x (1) x + ) (1 +;5) (1;+) k: ( x + 4x x>0(7;x5 x 5) > log2 ( T (1) log ( x + + x 15 > x + 14 x + 49 7x2 log x +7 log > 02 ( x 2+ 4x x+7)x5)> > Kt hp iu 27 x+7 27 x (7; ) kin: Vy BPT 10 x > 54 x < 5 cú nghim: 19/ Gii h phng trỡnh : x + y = Gii: 2 x y + 23xy +3 y = 3 x + y = x + y = (1) y Ta cú: 3 x +3 y = ( ) x+ y x y xy = (2) t : (4) cú dng : x y + xy + y 3= 2 x1, = t 2t3 t2 2t + = t = t = x x y22 x + = ( 4) a) Nu t = ta cú h yx + y = y x = y = b) Nu t = -1 ta cú h h vụ x + y =1 x = y x = y 3.log 25 33 x + y = x = , y = x y= 23 3 nghim c) Nu t = ta cú h 20/ Gii phng 3.25 x + ( 3x 10) x2 = x trỡnh: 3.5 x x + x = x2 3.5 x2 11 + x 3x.52x2 13 3.5 x = Gii: = x + (1) 5x23=(.52)x x1==20+ log(15 ) = log5 3 x32 + x = ( 2) V trỏi l hm ng bin v phi l hm nghch bin m (2) cú nghim x = nờn l nghim nht log Vy Pt cú nghim l: x = v x = log x (cos x sin x) + log (cos x + cos x) = 21/ Gii phng trỡnh: < x x Gii: iu kin: cos xk sin x > x = + k sin x = x + + Khi ú Pt cos x = x cos k 22 x = cos x + 2cos x = + x + cos x > Kt hp vi iu kin ta c: x = x + k x = + k 22/ Gii phng trỡnh: x > Gii: iu kin: Bin i theo x < x logarit c s thnh phng x > trỡnh ( ( ) ( )( ) () ) x = ( loaùi ) log ( x + ) ( x 1) = log ( x ) x x = x = x = 23/ Gii h phng trỡnh x + xy y0 = Gii: K : h a h v dng t y22xy2 2+xux=2xy12 =2=2 ==> u =v2u +u v1y = 7v = hoc 3+0u =7v = u = v 2 v + v u = y u = +7 ux2= =u0= v = T ú ta cú nghim ca h 3y2+ (x:y) = (-1 ;-1),(1 ;1), (), () 2v + v u2 y2,;= 07 + 1 + 24/ Gii h phng log ( x +yv2 = v log (2 x ) + = log ) 4 ( x + y) 2 trỡnh: Gii : x x=2 x = +1) log (4 y + y + 4) = log xvaứ vụự i >0 tuyứ yự 4 25/ Gii phng trỡnh: log ( xylog ( x + 2) + log ( x 5) + log 1y=1 = y ữ 2y = { Gii: iu kin: x > v x ( (*) Vi iu kin ú, ta cú phng trỡnh ó cho tng ng vi phng trỡnh: log (x + 2) x = log (x + 2) x = (x 3x 18)(x 3x 2) = x 3x 18 = 17 So iu kin (*), ta c x 3x = x3 =x6=173; x = 6; x = x= cỏc nghim ca phng trỡnh ó cho l: v 26/ Gii h phng trỡnh log ( y x ) log = Gii: iu kin: y x > y H phng yx 12 + y y= >25 yx log ( y x ) + log4 x = 0log = = trỡnh y y y ( loi) x = 3y 152 52 x + y = x25= y x + y = 25 x; y ) = x = 3x y ;+ y =ữ25 ( Vy h 210 10 25 phng trỡnh ó cho vụ x + y= 25 y + y = 25 y = 10 15 nghim ; ( x; y ) = ữ 10 10 27/ Gii hpt : xy + 4(( x + y ) xy )) + =7 log1 x ( 3xy x + y +2) + log 22+(yx(+x 2y ) 2 x +31) = xy + 4(( x + y ) xy )) + ( x + y ) =1 4( x + y ) xy + ( x + y ) = x + 4) = log ( y + ) log ( ++y ( x x x + y + y) = x + y x + y + x + y + + ( x y) = + ( x y) = x+ y x+ y 3 2 2 3( x + y ) + (( x + y ) xy ) + ( x + y ) = 3( x + y ) + ( x + y xy ) + ( x + y ) = x + y + + ( x y) = x + y + + ( x y) = x+ y x+ y 2 + ( x y)2 = ( x + y ) + 3( x + y ) + ( x + y ) + ( x y ) = ( x + y ) x + y + + ( x y) = x + y + + ( x y) = x+ y x+ y Gii: 28/ Gii h phng trỡnh: Gii: K < x < 1, x a phng trỡnh th nht ca h log1 x (2 y+ >y )+2;log y 2+y 1(1 x ) = v dng t (=x;log y ) 1= x((22+;1y) ) t , Tỡm c T=1, kt hp vi phng trỡnh th hai ca h, i chiu vi iu kin trờn, tỡm c nghim 29/ Gii h phng trỡnh: x y + 27 = 18y (1) Gii: (1) ( y ( x y + x = y (2) H ( x + 27 (2 x )3 + 3 = 18 33 = 18 3 t a = 2x; b = Ta cú h: ay + b = 18 a + b y=ữ ab(a + b) y ab = = x + x2 = x.3 x + = ữ y y y y ( H ó cho cú nghim 30/ Gii phng trỡnh: ; , + ; ữ ữ (33x+ 1)5+1= log (23x+ 15) log Gii: K : (*) x> Vi iu kin trờn phng trỡnh log (3x 1) + 13= log (2 x + 1) ó cho log 5(3 x 1) = log (2 x + 1) = (3x 2) (8 x 1) = so k ta c nghim x x 33 x52(3+x36 1x) =4 (20x + 1) ca phng trỡnh ó cho x = l x = x x + y y + = 31/ Gii h phng trỡnh: x y + x + y 22 = (2x+2 (y2) 23)+2 (=y 3) = ( x 2) Gii: H phuong trỡnh ó 2 cho tng ng vi x +x22 =20 20 = ( x 2(+x4)(+y2) y3 + 3) 22 * Thay vo h phng u +xv =2 4= u trỡnh ta cú 3u =+ v ) = u.v+y 4( hoc u = 02 Th vo cỏch t ta c cỏc nghim ca xxxvx====02222 yy==33 h l:;;;; yy==55 32/ Gii h phng trỡnh: 1 x + x + (1 + ) = yy y Gii: K 1 x + x + y (1 + y ) = 42 x + y + x + y = 1 t x +x + x+3 = x3 a = x x =y4 xy3 y x 3y + + x2 ( + x ) = Ta cú a a + a 2b y=2 4+ y +a 2y + = 2b y a y + ya = 2b a = x Khi ú 3 x =2 yba= 4) = a a + = b = y = a 2ab = a a(a + y KL x= 2 + 9x 33/ Gii phng trỡnh log (x + 5x + 6)x ++ log=3 (x + 120) = + log x Gii: (*) log (x + 5x + 6) + log (x + 9x + 20) = + log + log = log 24 + iu kin : , v x < x + 5x + > x < x > cú : < x < x < x > 9x 2++20 x+ + log (x + 5x 6)(x 9x>+020) = log 24 (x + 5x + 6)(x x > + 9x + 20) = 24 (x < 5) (4 < x < 3) (x > 2) (x < 5) (4 < x < 3) (x > 2) + PT (*) (x + 2)(x + 3)(x + 4)(x + 5) = 24 (*) + t , PT (*) tr t = (x + 3)(x 12 2)(x 5) = t < 4) 5)=x(+47x < x+< 3) (x(x+ > 2)+(**) (x + thnh : t(t-2) = 24 (t 1) = 25 t = t = t = : ( tha kin x = x + 7x + 12 = x + 7x + = x = (**)) t = - : : vụ nghim x + 7x + 12 = x + 7x + 16 = + Kt lun : PT cú hai nghim l x = -1 v x = - 34/ Cho h phng trỡnh Gii x + xy + y = m + 2 phng trỡnh vi m=3 x y + xy = m + Gii: t K S 2x + 4yP=S0 , xy = P Vit li h phng trỡnh di ( x + y ) + xy = m +( I2) S + P = m + dng SP = m + ( x + y ) xy = m + Khi ú S,P l nghim ca phng trỡnh bc Vi m=-3 ta cú t = t ( m + 2) t + m + = t = m +1 f ( u ) =ux2+yu=+1m + 1( 1) xy = m + g u ( ) =u ( m + 1) u + 1( ) x + y = m + u = x = 1; y = 2 ( 1) u u = =u1= x = 2; y = xy ( ) u + u + = u = x = y = Vy vi m=3, h phng trỡnh ó cho ( 1; ) , ( 2; 1) , ( 1; 1) cú nghim l 35/ Gii h phng trỡnh sau: xy + 4( x + y2 ) + =7 Gii K: x + y ( x + y)2 Ta cú h 21 3( x + y ) + ( x y )2 + =7 x + = 2 t u = x + y + ( ) ; v = x u ( x + y ) u + v = 13 x+ y x+ y y ta c h : x + y + 1u ++v x=3y = Gii h ta c u = 2, v = x+ y u ( ) T ú gii h x+ y+ =2 x + y = x = 36/ Gii h phng trỡnh: ( x yx)+ xy2 + y 2=13 x y (=x,1 y Ă y) = x y = 2 Gii: ( x(+xxy3) y+x)xyx22y+xy22=y 25 =y13 = 13 ( 1) ( 1' ) Ly (2) - (1) ta c: x2 y x y xy = ( ) y xy 22 + x 22y x3 = 25 ( ' ) xy2 = (3) ( x + y ) x y = 25 ( ) Kt hp vi (1) ta cú : ( ) ( ( ) ) ( ) t y = - z ta cú: ( x y ) ( x + y ) = 13 ( x + z ) ( x( +I ) z ) x + z ) ( x + z ) = 13 t S = x +z ( I ) ( ( x y ) xy = 2 V P = x.z ta cú : Ta cú: h ny cú ( x + z ) xz = ( ) ( x + z ) xz = 2xz = 13 S S 2P = 13 S 2SP = 13 S = 3=21 xx+=z SP = P = SP = z ==326 x.z nghim hoc Vy h phng trỡnh ó cho cú nghim l: ( ; 2) v ( -2 ; -3 ) log ( x + x + 1) log x = x x 37/ Gii phng trỡnh: Gii: (1) x2 + x + 1 log = x ( 2x( 3x ( x ) = x + + ) xx) t:f(x)= ; g(x)= x x x3+ + (x0) x Dựng pp kshs =>max f(x)=3; g(x)=3=>PT f(x)= g(x) max f(x)= g(x)=3 ti x=1 =>PT cú nghim x= 38/ Gii h phng trỡnh: ( x 1)( y 1)( x + y 2) = ( x 1)( y 1)( x + yx12) += y62 u2=xuvx( 2u y1+v )3 == 60 uv(u + v) = 2 2 1v = v =u y + ( x 1) + ( y 1) = (u + v ) 2uv = Gii: H vi t: c P.S = 6S = u + v S = x = u, v l nghim ca PP XS =2 x5==u1.0v= P =2 phng trỡnh: X2 3X + X = y = y = 2=0 Vy nghim ca h: (3 ; 2), (2 ; 3) 39 / Gii phng 1 log ( x + 3) + log ( x 1) = log (4 x ) trỡnh: Gii: iu kin: x > Bin i theo logarit c s x < x thnh phng trỡnh x > x = ( loaùi ) log ( x + ) ( x 1) = log ( x ) x x = x = x = 40/ Gii h phng trỡnh sau: y2 + 3y = Gii: iu kin x>0, y>0 Khi ú h x y = y 2 x+ 3xy = x22 + tng ng 3x = x + Tr v theo v hai phng trỡnh ta x = y2 y c: (x-y)(3xy+x+y) = thay li phng trỡnh Gii tỡm c nghim ca h l: (1;1) 41/ gii h phng trỡnh : log (x + y ) = + log (xy) x2 xy + y2 (x, y ( R) = 81 Gii: iu kin x, y > ((( ( hay log (x + y ) = log 2 + log (xy) = log (2xy) 2 x xy + y = 42 == y222xy 2= x(x x2=y) +xy yxy y=4=+=2y422 = = xy x xy ( 42/ Gii h phng log y log x = ( y x ) x xy + y 3 trỡnh : 2 Gii: iu kin : x > ; 2 x + y = y>0 x, y y 2VT(*) Ta cú : >0 ; Xột x > y 2> 2 log x < log y x xy + y = x + y >0 (*) vụ nghim nờn h vụ 3 VP(*) log y nờn h vụ nghim VP(*) > 2 Khi x = y h cho ta x = ( x;0y=) 0=2 2; y = ( x, y > 0) Vy h cú 2 x = y = nghim nht ( ) ) 43/ Gii h phng trỡnh: xy 18 = 12 x Gii h: xy 18 = 12 xx=2 3121 2xy x2 = 18 x xy = + y xy = + y x y y x 3 , tng ng x 23;33;23 {{ } } y Th li, tho h ó cho Vy, ( x; y ) {( )( 3;3 , 3;3 )} 44/ Gii h phng trỡnh log ( y x ) log y = ( x, y Ă ) Gii: iu kin: ; y x > x + y 1= 25 y > 15 yx x = yy x x y ( x=; y)1= x =0log y ;4 ữ = log y x = ( ) =+3log y y y 25 10 10 2 2 y =102 x + y = x25+ y= 25 y x+2 y+ y=2 25 =525 15 x + y = 25 ; ( x; y ) = ữ H phng trỡnh 10 10 (ko tha k) Vy h phng trỡnh ó cho vụ nghim 45/ Gii h phng trỡnh : x + y + x + y = 2 22 Gii: (I) 4===402 y) x+y y++1) y++x++ x( xx+(x yx(xyy++=+yy1) x 2xy +xyy==2022hay x + y = x + y +x + y + xy = hay V V V y= =1=2221y x 2+xx= xy= x y =2 12 xy+x=x =2222= 149/ 1) Gii h phng x x+, yy 2+R1 =2 trỡnh: () x+ y+2 ( xy x y + 1)( x + y 2) = [...]... x, y > 0) Vậy hệ có 2 2 x = 2 y = 4 nghiệm duy nhất ( ) ) 43/ Giải hệ phương trình: ﾧ  xy − 18 = 12 − x 2  Giải hệ: ﾧ  xy − 18 = 12 ⇒−xx=2 2⇒ 3121⇒ − 2xy x2 = ≥ 18 0⇒ x ≤2 3  xy = 9 + y  ﾧ  3 1 2  xy = 9 + y ⇒ x y ≥ 2 3 y ⇒ x ≥ 2 3 3  ﾧ, tương ứng ⇒∈x ∈ − 3− 23;33;23 3 {{ } } yﾧ Thử lại, thoả mãn hệ đã cho Vậy, ﾧ ( x; y ) ∈ {( − 2 )( 3;−3 3 , 2 3;3 3 )} 44/ Giải hệ phương trình  1 log... ⇔ ⇔ 25  2   10 10     2 2  2 2  y =102 2  x 2 + y 2 = x25+ y⇔= 25 9 y x+2 y+ y=2 25 =525   15   x + y = 25 ;− ( x; y ) =  − ÷ Hệ phương trình ﾧ 10    10 ﾧﾧ(ko thỏa đk) Vậy hệ phương trình đã cho vô nghiệm 45/ Giải hệ phương trình : ﾧ  x 2 + y 2 + x + y = 4   2 2 2 22 Giải: (I) ﾧﾧﾧ 4===402 y) x+y y++1) y++x++ x( xx+(x yx(xyy++=+yy1) ⇔ ⇔ ⇔    ﾧﾧ x 2xy +xyy==2−−022hay... 42/ Giải hệ phương log y − log x = ( y − x ) x 2 − xy + y 2 3 3  trình : 2 2  Giải: Điều kiện : x > 0 ;  2 2 x + y = 4 y>0 ∀ x, y y 2VT(*) Ta có : >0 ; Xét x > y 3 2> 0 ⇒  2 2 ⇒ log x < log y ⇒ x − xy + y = x − + y >0    (*) vô nghiệm nên hệ vô 3 3 2 VP(*) 4 log 3 y ⇒  ⇒ nên hệ vô nghiệm VP(*) > 0  2 2 Khi x = y hệ cho ta... 2xy +xyy==2−−022hay x + y = − 1 ⇔ x + y +⇔x + y + xy = 2 ﾧ hay ﾧﾧ ﾧ V ﾧ V ﾧ V y=− =1=−2−2−21y x 2+xx= xy= x−  ⇔  2 y =2 1−2 ﾧ xy+x=x −=2222= 0 149/ 1) Giải hệ phương  x 2 x+, yy 2∈+R1 =2  trình: ﾧ (ﾧ) x+ y+2  ( xy − x − y + 1)( x + y − 2) = 6 