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BÀI TẬP LÀM TRONG DỊP NGHỈ TẾT NĂM 2017  Yêu cầu: Các em làm tập vào giấy men, buổi học sau Tết nộp cho thầy chúc em làm hết tập! Bài Giaûi phương trình sau: a) 3x – = 2x – c) – 2x = 22 – 3x e) x – 12 + 4x = 25 + 2x – g) 11 + 8x – = 5x – + x b) d) f) h) – 4y + 24 + 6y = y + 27 + 3y 8x – = 5x + 12 x + 2x + 3x – 19 = 3x + – 2x + 15 = 9x + – 2x a) c) e) g) i) b) d) f) h) j) 2x(x + 2)2 – 8x2 = 2(x – 2)(x2 + 2x + 4) (x – 2)3 + (3x – 1)(3x + 1) = (x + 1)3 (x – 1)3 – x(x + 1)2 = 5x(2 – x) – 11(x + 2) (x – 3)(x + 4) – 2(3x – 2) = (x – 4)2 (x + 1)(x2 – x + 1) – 2x = x(x + 1)(x – 1) – (x – 6) = 4(3 – 2x) – (2x + 4) = – (x + 4) (x + 1)(2x – 3) = (2x – 1)(x + 5) (x – 1) – (2x – 1) = – x x(x + 3)2 – 3x = (x + 2)3 + a) 1,2 – (x – 0,8) = –2(0,9 + x) c) 2,3x – 2(0,7 + 2x) = 3,6 – 1,7x e) + 2,25x +2,6 = 2x + + 0,4x a) c) e) g) i) m) p) r) t) v) b) 3,6 – 0,5(2x + 1) = x – 0,25(2 – 4x) d) 0,1 – 2(0,5t – 0,1) = 2(t – 2,5) – 0,7 f) 5x + 3,48 – 2,35x = 5,38 – 2,9x + 10,42 5x   3x  3    13  2 x       x  5    7x  16  x  2x  3x  3x     2x 4x  6x  5x    3 2x  x  x    15 x 2x  x   x 6 3x  11 x 3x  5x     11 2x  3x  9x  3x     12 b) d) f) h) k) n) q) s) u) 5x  2x  x  x    10 15 30 w) 5(x  1)  7x  2(2x  1)   5 2(x  3) 3x 2(x  7) c) 14    3(2x  1) 3x  2(3x  2) e)  1  10 3(x  3) 4x  10,5 3(x  1) g)   6 10 a) Bài Giải phương trình sau: 10x   8x  1 12 20x  1,5 x  5(x  9)  5x  4(0,5  1,5x)   x4 x x2 x4  5x  8x  4x    5 1 (x  3)   (x  1)  (x  2) 2x  2x  0,5x   0,25 9x  0,7 5x  1,5 7x  1,1 5(0,4  2x)    6 x  2x  6x  2x     3 12  3x x3 2x  7x    x 1 15 3(x  30) 7x 2(10x  2)  24   15 10 x  3(2x  1) 2x  3(x  1)  12x    d) 12 10x  f) x  (2x  1)  (1  2x)  17 34 2(3x  1)  2(3x  1) 3x  h) 5  10 b) x ThuVienDeThi.com (2x  1)2 (x  1)2 7x  14x  (x  10)(x  4) (x  4)(2  x) (x  10)(x  2)   b)   15 12 2 ( x  2) (2x  3)(2x  3) (x  4) c)   0 a) Bài Giải phương trình sau: a) x x 1  2x 3x   1 3 2x   2x 3x  x 1 2x  6  2  3x   b) Bài Giải phương trình sau: x  23 x  23 x  23 x  23 x2  x3  x4  x5     b)   1    1    1    1 96 97 98 24 25 26 27        95  x 1 x  x  x  201  x 203  x 205  x c) d)   3    99 97 95 2004 2003 2002 2001 x 1 x  x  x  x  45 x  47 x  55 x  53    e) f)    55 53 45 47 x2 x4 x6 x8 2x 1 x x g) h)    1   98 96 94 92 2002 2003 2004 2 2 x  10x  29 x  10x  27 x  10x  1971 x  10x  1973 i)    1971 1973 29 27 a) Bài Giải phương trình sau: a) c) e) g) i) (3x – 2)(4x + 5) = (4x + 2)(x2 + 1) = (x – 1)(2x + 7)(x2 + 2) = (3,5 – 7x)(0,1x + 2,3) = 15(x + 9)(x – 3) (x + 21) = b) d) f) h) j) (2,3x – 6,9)(0,1x + 2) = (2x + 7)(x – 5)(5x + 1) = (4x – 10)(24 + 5x) = (5x + 2)(x – 7) = (x2 + 1)(x2 – 4x + 4) = l) (3,3 – 11x)  (3x + 2)(x2 – 1) = (9x2 – 4)(x + 1) 2x(x – 3) + 5(x – 3) = (x + 2)(3 – 4x) = x2 + 4x + 3x – 15 = 2x(x – 5) 0,5x(x – 3) = (x – 3)(1,5x – 1) x(2x – 9) = 3x(x – 5) 2x(x – 1) = x2 - b) d) f) h) j) l) x(x + 3)(x – 3) – (x + 2)(x2 – 2x + 4) = (3x – 1)(x2 + 2) = (3x – 1)(7x – 10) x(2x – 7) – 4x + 14 = (2x + 1)(3x – 2) = (5x – 8)(2x + 1) (2x2 + 1)(4x – 3) = (x – 12)(2x2 + 1) (x – 1)(5x + 3) = (3x – 8)(x – 1) x   x(3x  7) 7 1      (x  1) x x  3   1  p)  x     x   x    4   2   3x    3x   r) (2x  3)  1  (x  5)  1   7x    7x   2(x  3) 4x    =   k) (3x – 2)  a) c) e) g) i) k) m) o) q)  x  2(1  x)   =   n) (2 – 3x)(x + 11) = (3x – 2)(2 – 5x) s) (x + 2)(x – 3)(17x2 – 17x + 8) = (x + 2)(x – 3)(x2 – 17x +33) a) c) e) g) i) k) m) (2x – 5)2 – (x + 2)2 = (x2 – 2x + 1) – = (x + 1)2 = 4(x2 – 2x + 1)2 9(x – 3)2 = 4(x + 2)2 (2x – 1)2 = 49 (2x + 7)2 = 9(x + 2)2 (x2 – 16)2 – (x – 4)2 = b) d) f) h) j) l) n) o) x  32  x  52  25  3x  x 2 p)        3  3 (3x2 + 10x – 8)2 = (5x2 – 2x + 10)2 4x2 + 4x + = x2 (x2 – 9)2 – 9(x – 3)2 = (4x2 – 3x – 18)2 = (4x2 + 3x)2 (5x – 3)2 – (4x – 7)2 = 4(2x + 7)2 = 9(x + 3)2 (5x2 – 2x + 10)2 = (3x2 + 10x – 8)2 ThuVienDeThi.com GIÁO VIÊN: ĐINH BÁ SANG ... 53 45 47 x2 x4 x6 x? ?8 2x 1 x x g) h)    1   98 96 94 92 2002 2003 2004 2 2 x  10x  29 x  10x  27 x  10x  1971 x  10x  1973 i)    1971 1973 29 27 a) Bài Giải phương trình... (3x2 + 10x – 8) 2 = (5x2 – 2x + 10)2 4x2 + 4x + = x2 (x2 – 9)2 – 9(x – 3)2 = (4x2 – 3x – 18) 2 = (4x2 + 3x)2 (5x – 3)2 – (4x – 7)2 = 4(2x + 7)2 = 9(x + 3)2 (5x2 – 2x + 10)2 = (3x2 + 10x – 8) 2 ThuVienDeThi.com... 1)(7x – 10) x(2x – 7) – 4x + 14 = (2x + 1)(3x – 2) = (5x – 8) (2x + 1) (2x2 + 1)(4x – 3) = (x – 12)(2x2 + 1) (x – 1)(5x + 3) = (3x – 8) (x – 1) x   x(3x  7) 7 1      (x  1) x x  3

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