Toán Lớp 7: Tìm x biết: 1. (2x-1)^3=(2x-1)^5 2. x^2+8x-9=0 3. x^2+11x-12=0

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Toán Lớp 7:  Tìm x biết: 1. (2x-1)^3=(2x-1)^5 2. x^2+8x-9=0 3. x^2+11x-12=0

hienthucthu97 · Answer

Giải đáp + Lời giải và giải thích chi tiết:    1 , (2x - 1)^3 = (2x - 1)^5   (2x - 1)^5 - (2x - 1)^3 = 0   (2x - 1)^3 . [ (2x - 1)^2 - 1] = 0   &rArr; ( left[  begin{array}{l}(2x - 1)^3 = 0  (2x - 1)^2- 1= 0 end{array}  right. )    &rArr; ( left[  begin{array}{l}2x - 1 = 0  2x - 1 = 1   2x - 1 = -1 end{array}  right. )    &rArr; ( left[  begin{array}{l}2x = 1  2x =2   2x =0 end{array}  right. )    &rArr; ( left[  begin{array}{l}x =  frac{1}{2}  x=1   x =0 end{array}  right. )    Vậy x  in {1/2 ; 1 ; 0}   2 , x^2 + 8x - 9 = 0    x^2 + 9x - x - 9 = 0    x(x + 9) - (x + 9) = 0   (x - 1)(x + 9) = 0   &rArr; ( left[  begin{array}{l}x-1=0  x+9=0 end{array}  right. )    &rArr; ( left[  begin{array}{l}x=1  x=-9 end{array}  right. )    Vậy x  in {1 ; -9}   3 , x^2 + 11x - 12 = 0    x^2 + 12x - x - 12 = 0    x(x + 12) - (x + 12) = 0    (x - 1)(x + 12) = 0   &rArr; ( left[  begin{array}{l}x-1=0  x+12=0 end{array}  right. )    &rArr; ( left[  begin{array}{l}x=1  x=-12 end{array}  right. )    Vậy x  in {1 ; -12}