Toán Lớp 8: B1 gải pt a,2x-1/x^2-1=6/3x+1 b,1/x+3+1/x-1=10/(x+3)(x-1) B2 giải pt a,1/x+1-4/x^2-x+1=2x^2+1/x^3+1 b,3x+1/x+1-2x-5/x-3=1-4/(x+1)(x-3)

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Toán Lớp 8:  B1 gải pt a,2x-1/x^2-1=6/3x+1 b,1/x+3+1/x-1=10/(x+3)(x-1) B2 giải pt a,1/x+1-4/x^2-x+1=2x^2+1/x^3+1 b,3x+1/x+1-2x-5/x-3=1-4/(x+1)(x-3)

maitrucloan · Answer

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quynhthanhnghi · Answer

Giải đáp:       Lời giải và giải thích chi tiết:   a, (2x-1)/(x^2-1) = 6/(3x+1)   ĐKXĐ :  ( left { begin{array}{l}x ne1  x ne-1  x ne- dfrac{1}{3} end{array}  right. )    -> (2x-1)(3x+1) = (x^2-1)*6   -> 6x^2 - x - 1 = 6x^2 - 6   -> -x - 1 = -6   -> -x = -5   -> x = 5(TM)   Vậy x  in {5}   b, 1/(x+3) + 1/(x-1) = 10/((x+3)(x-1))   -> (1(x-1))/((x+3)(x-1)) + (1(x+3))/((x+3)(x-1)) = 10/((x+3)(x-1))   -> (x-1) + (x+3) = 10   -> (x+x) - 1 + 3 = 10   -> 2x + 2 = 10   -> 2x = 8   -> x = 4(TM)   Vậy x  in {4}   $  $   a, 1/(x+1) - 4/(x^2-x+1) = (2x^2+1)/(x^3+1)   ĐKXĐ : x  ne -1   -> (1(x^2-x+1))/((x+1)(x^2-x+1)) - (4(x+1))/((x+1)(x^2-x+1)) = (2x^2+1)/((x+1)(x^2-x+1))   -> x^2 - x + 1 - 4(x+1) = 2x^2 + 1   -> x^2 - 5x - 3 = 2x^2 + 1   -> x^2 - 5x - 4 = 2x^2   -> -x^2 - 5x - 4 = 0   -> -(x^2+5x+4) = 0   -> -(x+1)(x+4) = 0   -> ( left {  begin{array}{l}x+1=0  x+4=0 end{array}  right. )    -> ( left[  begin{array}{l}x=-1(KTM)  x=-4(TM) end{array}  right. )    Vậy x  in {-4}   b, (3x+1)/(x+1) - (2x-5)/(x-3) = 1 - 4/((x+1)(x-3))   ĐKXĐ :  ( left {  begin{array}{l}x ne-1  x ne3 end{array}  right. )    -> ((3x+1)(x-3))/((x+1)(x-3)) - ((2x-5)(x+1))/((x+1)(x-3)) = 1(x+1)(x-3) - 4/((x+1)(x-3))   -> (3x+1)(x-3) - (2x-5)(x+1) = (x+1)(x-3) - 4   -> x^2 - 5x + 2 = x^2 - 2x - 7   -> x^2 - 5x = x^2 - 2x - 9   -> -5x = -2x - 9   -> -3x = -9   -> x = 3(KTM)   Vậy x  in &empty;