Câu hỏi của Thủy Hồ

Question

Tìm x biết:
(x+2)^3-x^2(x-6)-4=0
6x^2-(2x-3)(3x+2)=1

ILoveMath · Accepted Answer

$\left(x+2\right)^3-x^2\left(x-6\right)-4=0\
\Leftrightarrow x^3-6x^2+12x-8-x^3+6x^2-4=0\
\Leftrightarrow12x-12=0\
\Leftrightarrow12x=12\
\Leftrightarrow x=1$$6x^2-\left(2x-3\right)\left(3x+2\right)=1\
\Leftrightarrow6x^2-\left[3x.\left(2x-3\right)+2.\left(2x-3\right)\right]=1\
\Leftrightarrow6x^2-\left(6x^2-9x+4x-6\right)=1\
\Leftrightarrow6x^2-\left(6x^2-5x-6\right)=1\
\Leftrightarrow6x^2-6x^2+5x+6=1\
\Leftrightarrow5x=-5\
\Leftrightarrow x=-1$Câu trả lời: $\left(x+2\right)^3-x^2\left(x-6\right)-4=0\
\Leftrightarrow x^3-6x^2+12x-8-x^3+6x^2-4=0\
\Leftrightarrow12x-12=0\
\Leftrightarrow12x=12\
\Leftrightarrow x=1$$6x^2-\left(2x-3\right)\left(3x+2\right)=1\
\Leftrightarrow6x^2-\left[3x.\left(2x-3\right)+2.\left(2x-3\right)\right]=1\
\Leftrightarrow6x^2-\left(6x^2-9x+4x-6\right)=1\
\Leftrightarrow6x^2-\left(6x^2-5x-6\right)=1\
\Leftrightarrow6x^2-6x^2+5x+6=1\
\Leftrightarrow5x=-5\
\Leftrightarrow x=-1$

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