Toán Lớp 8: Tìm x : a, (8x-3).(3x+2)-(4x+7).(x+4)=(2x+1).(5x-1) b, 2x^2 +3.(x-1).(x+1)= 5x.(x+1)

Question

Toán Lớp 8:  Tìm x : a, (8x-3).(3x+2)-(4x+7).(x+4)=(2x+1).(5x-1) b, 2x^2  +3.(x-1).(x+1)= 5x.(x+1)

dananhthumai · Answer

$ text {&rarr;T&igrave;m }$ x:   a,   (8x-3)(3x+2)-(4x+7)(x+4) = (2x+1)(5x-1)   &hArr;24x^2+16x-9x-6-4x^2-16x-7x-28 = 10x^2-2x+5x-1   &hArr;20x^2-16x-34 = 10x^2-2x+5x-1   &hArr;20x^2-10x^2-16x-5x = 34-1   &hArr;10x^2-21x = 35   &hArr;10x^2-21x-35 = 0   &hArr;10x^2-30x+11x-33 = 0   &hArr;(10x^2+11x)-(30x+33) = 0   &hArr;x(10x+11)-3(10x+11) = 0   &hArr;(x-3)(10x+11) = 0   &hArr; ( left[  begin{array}{l}x-3=0  10x+11=0 end{array}  right. )    &hArr; ( left[  begin{array}{l}x=3  10x=-11 end{array}  right. )    &hArr; ( left[  begin{array}{l}x=3  x= frac{-11}{10}  end{array}  right. )    $ text {Vậy}$ x=3 $ text {hoặc}$ x=(-11)/(10)   b,   2x^2+3(x-1)(x+1) = 5x(x+1)   &hArr;2x^2+3(x^2-1) = 5x^2+5x   &hArr;2x^2+3x^2-3 = 5x^2+5x   &hArr;5x^2-5x^2-5x = 3   &hArr;-5x = 3   &hArr;x = (-3)/5   $ text {Vậy}$ x=-3/5   $ text {@lamtung2007}$

minhtamnguyet88 · Answer

Giải đáp + Lời giải và giải thích chi tiết:   a,   (8x-3)(3x+2)-(4x+7)(x+4)=(2x+1)(5x-1)   -> 24x^2 + 7x - 6 - 4x^2 - 23x - 28 = 10x^2 - 2x + 5 - 1   -> 20x^2 - 16x - 34 = 10x^2 + 3x - 1   -> 20x^2 - 16x - 33 = 10x^2 + 3x   -> 20x^2 - 19x - 33 = 10x^2   -> 10x^2 - 19x - 33 = 0   -> (10x^2+11x)+(-30x-33) = 0   -> x(10x+11) - 3(10x+11) = 0   -> (10x+11)(x-3) = 0   ->  ( left[  begin{array}{l}10x+11=0  x-3=0 end{array}  right. )    ->  ( left[  begin{array}{l}x=- dfrac{11}{10}  x=3 end{array}  right. )    Vậy x  in {-11/10 , 3}   $  $   b,   2x^2 + 3(x-1)(x+1) = 5x(x+1)   -> 2x^2 + 3x^2 - 3 = 5x^2 + 5x   -> 5x^2 = 5x^2 + 5x + 3   -> 0 = 5x + 3   -> -5x = 3   -> x = -3/5   Vậy x  in {-3/5}