Toán Lớp 8: Tìm x a) x^3+x^2=36 b) 10x^2-7=33x

Question

Toán Lớp 8:  Tìm x a) x^3+x^2=36 b) 10x^2-7=33x

aivilanlan · Answer

Giải đáp:    a) x=3    b)x=$ frac{7}{2}$, x=$ frac{-1}{5}$    Lời giải và giải thích chi tiết:    a) x&sup3;+x&sup2;=36     &hArr;x&sup3;+x&sup2;-36=0     &hArr;x&sup3;-3x&sup2;+4x&sup2;-36=0     &hArr;x&sup2;(x-3)+4(x&sup2;-9)=0     &hArr;x&sup2;(x-3)+4(x-3)(x+3)=0     &hArr;(x-3)(x&sup2;+4x+12)=0     &hArr; ( left[  begin{array}{l}x-3=0  x^2+4x+12=0 end{array}  right. )       &hArr; ( left[  begin{array}{l}x=3  x^2+4x+4=-8 end{array}  right. )      &hArr; ( left[  begin{array}{l}x=3  (x+2)^2=-8(loai) end{array}  right. )   Vậy x=3   b) 10x&sup2;-7=33x     &hArr;10x&sup2;-33x-7=0     &hArr;10x&sup2;-35x+2x-7=0     &hArr;5x(2x-7)+(2x-7)=0     &hArr;(5x+1)(2x-7)=0     &hArr; ( left[  begin{array}{l}5x+1=0  2x-7=0 end{array}  right. )      &hArr; ( left[  begin{array}{l} frac{7}{2}   frac{-1}{5} end{array}  right. )     Vậy x=$ frac{7}{2}$; x=$ frac{-1}{5}$   #truongtuan

maingoctruc · Answer

a) x^3+x^2=36 ->x^3+x^2-36=0   ->x^3-3x^2+4x^2-12x+12x-36=0   ->x^2(x-3)+4x(x-3)+12(x-3)=0   ->(x-3)(x^2+4x+12)=0   -> ( left[  begin{array}{l}x-3=0  x^2+4x+12=0 end{array}  right. )   -> ( left[  begin{array}{l}x=3  x^2+4x+4=-8 end{array}  right. )   -> ( left[  begin{array}{l}x=3  (x+2)^2=-8(KTM) end{array}  right. )   Vậy S={3}   b)10x^2-7=33x ->10x^2-33x-7=0   ->10x^2+2x-35x-7=0   ->2x(5x+1)-7(5x+1)=0   ->(5x+1)(2x-7)=0   -> ( left[  begin{array}{l}5x=-1  2x=7 end{array}  right. )   -> ( left[  begin{array}{l}x= frac{-1}{5}  x= frac{7}{2} end{array}  right. )    Vậy S={-1/5;7/2}