Câu hỏi của Vũ Phương Anh

Question

Tìm x

a,$\left(x+2\right)\left(x^2-2x+4\right)-x\left(x-1\right)\left(x+1\right)+3x=2$

b,$2x^3-50x=0$

c,$5x^2-4\left(x^2-2x+1\right)-5=0$

Ťɧε⚡₣lαsɧ · Accepted Answer

b)  $2x^3-50x=0$

$\Leftrightarrow2\left(x^2-25\right)=0$

$\Leftrightarrow x^2-25=0$

$\Leftrightarrow\left(x-5\right)\left(x+5\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}x-5=0\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\x=-5\end{matrix}\right.$

Vậy $x\in\left\{5;-5\right\}$Câu trả lời: b)  $2x^3-50x=0$

$\Leftrightarrow2\left(x^2-25\right)=0$

$\Leftrightarrow x^2-25=0$

$\Leftrightarrow\left(x-5\right)\left(x+5\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}x-5=0\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\x=-5\end{matrix}\right.$

Vậy $x\in\left\{5;-5\right\}$

Ťɧε⚡₣lαsɧ · Answer

a) $\left(x+2\right)\left(x^2-2x+4\right)-x\left(x-1\right)\left(x+1\right)+3x=2$

$\Leftrightarrow x^3+2^3-x\left(x^2-1\right)+3x=2$

$\Leftrightarrow x^3+8-x^3+x+3x=2$

$\Leftrightarrow4x+8=2$

$\Leftrightarrow4x=-6$

$\Leftrightarrow x=-\frac{3}{2}$

Vậy $x=-\frac{3}{2}$Câu trả lời: a) $\left(x+2\right)\left(x^2-2x+4\right)-x\left(x-1\right)\left(x+1\right)+3x=2$

$\Leftrightarrow x^3+2^3-x\left(x^2-1\right)+3x=2$

$\Leftrightarrow x^3+8-x^3+x+3x=2$

$\Leftrightarrow4x+8=2$

$\Leftrightarrow4x=-6$

$\Leftrightarrow x=-\frac{3}{2}$

Vậy $x=-\frac{3}{2}$

Ťɧε⚡₣lαsɧ · Answer

c)  $5x^2-4\left(x^2-2x+1\right)-5=0$

$\Leftrightarrow5x^2-4x^2+8x-4-5=0$

$\Leftrightarrow x^2+8x-9=0$

$\Leftrightarrow x^2+9x-\left(x+9\right)=0$

$\Leftrightarrow x\left(x+9\right)-\left(x+9\right)=0$

$\Leftrightarrow\left(x-1\right)\left(x+9\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}x-1=0\x+9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\x=-9\end{matrix}\right.$

Vậy $x\in\left\{1;-9\right\}$

Chúc bạn học tốt  ✔✔✔Câu trả lời: c)  $5x^2-4\left(x^2-2x+1\right)-5=0$