Câu hỏi của tràn thị trúc oanh

Question

Câu 1:Tìm x:

a) x3-16x=0                          b) x4-2x3+10x2-20x=0

c) (2x-1)2=(x+3)2                 d) x2(x-2)-2x2+8x-8=0

T.Thùy Ninh · Accepted Answer

$a,x^3-16x=0$

$\Leftrightarrow x\left(x^2-16\right)=0$

$\Leftrightarrow x\left(x-4\right)\left(x+4\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}x=0\x-4=0\x+4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\x=4\x=-4\end{matrix}\right.$

$b,x^4-2x^3+10x^2-20x=0$

$\Leftrightarrow x^3\left(x-2\right)+10x\left(x-2\right)=0$

$\Leftrightarrow\left(x-2\right)\left(x^3+10x\right)=0$

$\Leftrightarrow\left(x-2\right)x\left(x^2+10\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}x-2=0\x=0\x^2+10=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\x=0\\left[{}\begin{matrix}x^2=10\x^2=-10\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\x=0\x=\sqrt{10}\x=-\sqrt{10}\end{matrix}\right.$$c,\left(2x-1\right)^2=\left(x+3\right)^2$

$\Leftrightarrow4x^2-4x+1=x^2+6x+9$

$\Leftrightarrow4x^2-4x+1-x^2-6x-9=0$

$\Leftrightarrow3x^2-10x-8=0$

$\Leftrightarrow3x^2-12x+2x-8=0$

$\Leftrightarrow3x\left(x-4\right)+2\left(x-4\right)=0$

$\Leftrightarrow\left(3x-2\right)\left(x-4\right)=0$

$\Rightarrow\left[{}\begin{matrix}3x-2=0\x-4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\x=4\end{matrix}\right.$

Phần d tương tựCâu trả lời: $a,x^3-16x=0$

$\Leftrightarrow x\left(x^2-16\right)=0$

$\Leftrightarrow x\left(x-4\right)\left(x+4\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}x=0\x-4=0\x+4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\x=4\x=-4\end{matrix}\right.$

$b,x^4-2x^3+10x^2-20x=0$

$\Leftrightarrow x^3\left(x-2\right)+10x\left(x-2\right)=0$

$\Leftrightarrow\left(x-2\right)\left(x^3+10x\right)=0$

$\Leftrightarrow\left(x-2\right)x\left(x^2+10\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}x-2=0\x=0\x^2+10=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\x=0\\left[{}\begin{matrix}x^2=10\x^2=-10\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\x=0\x=\sqrt{10}\x=-\sqrt{10}\end{matrix}\right.$$c,\left(2x-1\right)^2=\left(x+3\right)^2$

$\Leftrightarrow4x^2-4x+1=x^2+6x+9$

$\Leftrightarrow4x^2-4x+1-x^2-6x-9=0$

$\Leftrightarrow3x^2-10x-8=0$

$\Leftrightarrow3x^2-12x+2x-8=0$

$\Leftrightarrow3x\left(x-4\right)+2\left(x-4\right)=0$

$\Leftrightarrow\left(3x-2\right)\left(x-4\right)=0$

$\Rightarrow\left[{}\begin{matrix}3x-2=0\x-4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\x=4\end{matrix}\right.$

Phần d tương tự

DƯƠNG PHAN KHÁNH DƯƠNG · Answer

Câu a :

$x^3-16x=0$

$\Leftrightarrow x\left(x^2-4^2\right)=0$

$\Leftrightarrow x\left[\left(x+4\right)\left(x-4\right)\right]=0$

$\Rightarrow$      $x=0$

$\Rightarrow$     $x+4=0\Rightarrow x=-4$

$\Rightarrow x-4=0\Rightarrow x=4$

Câu b :

$x^4-2x^3+10x^2-20x=0$

$\Leftrightarrow x^3\left(x-2\right)+10x\left(x-2\right)$ $=0$

$\Leftrightarrow\left(x-2\right)\left(x^3+10x\right)=0$

$\Leftrightarrow x\left(x-2\right)\left(x^2+10\right)=0$

$\Rightarrow x=0$

$\left(x-2\right)=0\Rightarrow x=2$

$x^2+10=0$ $\Rightarrow$ x ( loại )Câu trả lời: Câu a :

$x^3-16x=0$

$\Leftrightarrow x\left(x^2-4^2\right)=0$

$\Leftrightarrow x\left[\left(x+4\right)\left(x-4\right)\right]=0$

$\Rightarrow$      $x=0$

$\Rightarrow$     $x+4=0\Rightarrow x=-4$

$\Rightarrow x-4=0\Rightarrow x=4$

Câu b :

$x^4-2x^3+10x^2-20x=0$

$\Leftrightarrow x^3\left(x-2\right)+10x\left(x-2\right)$ $=0$

$\Leftrightarrow\left(x-2\right)\left(x^3+10x\right)=0$

$\Leftrightarrow x\left(x-2\right)\left(x^2+10\right)=0$

$\Rightarrow x=0$

$\left(x-2\right)=0\Rightarrow x=2$

$x^2+10=0$ $\Rightarrow$ x ( loại )

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