. Dovay dÚcJng tron (K) tiep xuc trong vdi dtfdng
HI;6NG DAN GIA
Bai 1: a) a + b + c = 0 => a + b = - c => (a + b)^ = (-c)' => â + b^ + 3ab(a + b) = c' => á + b^ + c' = -3ab(a + b) ii> â + b'' + c^ = 3abc
Do vay, ta co: 3abc = 0 => a, b, c c6 it nhát mot so bkng 0.
X + y + z = 3 (l)
b) ixy + yz + zx = -1 (2) x-^ + y ^ + z ^ + 6 = 3(x^+y^+z2) (3)
Tuf (1), (2) co: x^ + y^ + z^ = (X + y + z)^ - 2(xy + yz + zx) = 11 Nen tir (3) co: x ' + y ' + z'+ 6 = 33 o x^ + y^ + z'= 27
Do do: (x + y + z)' - (x' + y^ + z^) = 0 o 3(x + y)(y + z)(z + x) = 0 o X + y, y + z, X + z CO it nhát mot so b^ng 0.
Cty TNHH MTV DWH Khang Vi§t
, Xetx + y = 0, kethdp(l)c6z = 3.
i Nen tilf (2) co: -x^ - 3x + 3x = -1 x^ = 1 <=> x = ±1
Vay co: (X ; y ; z) = (1 ; - 1 ; 3) ; (X ; y ; z) = (-1 ; 1 ; 3)
• Xet y + z = 0, lap luan nhiT tren, ta c6: ^ (X ; y ; z) = (3 ; 1 ; - 1) ; (X ; y ; z) = (3 ; - 1 ; 1) OA * MA,
0 Xet z + X = 0, lap luan nhiT tren, ta co: s,i,'(:,:; , (X ; y ; z) = (1 ; 3 ; -1) ; (X ; y ; z) = (-1 ; 3 ; 1)
Vay nghiem (x ; y ; z) cua he phiTdng trinh la:
(1 ; - 1 ; 3), (-1 ; 1 ; 3), (3 ; 1 ; -1), (3 ; - 1 ; 1), (1 ; 3 ; -1), (-1 ; 3 ; 1)
Nhanxet: i -
a) Tuf a + b + c = 0, ta co: á + bS c' = 3abc (bai toan quen thupc d Idp 8).
Do vay CO abc = 0. ,s > „ ,
b) Tijr X + y + z = 3 va xy + yz + zx = -1 giup cd x^ + y^ + z^ = 11.
Tuf do, cho ta (x + y + z)^ - (x' + y^ + z^) = 0 hay 3(x + y)(y + z)(z + x) = 0.
Bai 2: a) ( 2 x - 1)^ = 12Vx^ - X - 2 + 1 c:>4x^-4x + 1 = 12Vx^ - x - 2 + 1 ox^ - X - 3Vx^ - x - 2 = 0 Dat y = Vx^ - X - 2 (y > 0), ta cd: x^ - X - 2 = y^ o x' - X = ý + 2 PhiTdng tnnh trd thanh y^ - 3y + 2 = 0 Ta cd: y, = 1, y2 = 2 (vi a + b + c = 0) yi = 1, ta cd: Vx^ - x - 2 = 1 <=> -i I o x ^ - x - 3 = 0 o x = i ^ 1 > 0 x^ - X - 2 - 1 yi = 2, ta cd: Vx^ - x - 2 = 2 • 2 > 0 - X - 2 = 4 o x - x - 6 = 0 o ( x - 3)(x + 2) = 0 o x = 3 hoSc x = -2. 1 v« y S = | ^ ; ^ ; 3 ; - 2
b) AABC vuong tai A cd dien tich b^ng 1 => ^ AB.AC = 1 AB^ + AC^ = BC^ (dinh li Py-ta-go)
LuyQn giii M truOc kl thi viio lOp 10 ba mign BJc. Trung, Nam mOn Toan _ Nguyjn Pijfc Tin
B C = VBC^ = VAB^ + AC^ > V2AB.AC = 2 ( 1 )
^ (AB + AC) = V2(AB + AC)^ = yJliAB^ + AC^ + 2AB.AC)
= V2(BC2 + 4 ) = V(BC + if + ( B C - if > V(BC + if = B C + 2
=;> V2(AB + A C ) - 2 > B C = > >/2(AB + A C - > / 2 ) > B C (2)
>,, Tir (1) va (2) ta c6: 2 < B C < V2 (AB + AC - y/l) ^ x
Nhfinxet: ' = '
a) De nhan ra, dat an phu Vx^ - x - 2 = y se di/a phiTdng trlnh can giSi ve giai
phiTdng trinh bac hai mot an.
b) AABC vuong tai A AB^ + AC^ = BC^ (dinh li Py-ta-go)
Bai 3 : a) Bp 4 so nguyen dÚcfng phan biet tdng ba so baft ki trong chung la
mot so nguyen to c6 the la (1 ; 5 ; 7 ; 11).
That vay: 1 + 5 + 7 = 13 ; 1 + 5 + 11 = 17 ; 1 + 7 + 11 = 1 9 ; 5 + 7 + 11 = 23.
^ C^c s6' 13 ; 17 ; 19 ; 23 deu la so nguyen tọ