Câu hỏi của Ẩn danh

Question

tìm x:   a)x^4-2x^3+5x^2-10x=0            b)(3x+5)^2=(2x-2)^2.           c)x^3–2x^2+x=0.           d)x^2(x-1)-4x^2+8x-4=0

Nguyễn Hoàng Minh · Accepted Answer

$a,x^4-2x^3+5x^2-10x=0\
\Leftrightarrow x^3\left(x-2\right)+5x\left(x-2\right)=0\
\Leftrightarrow x\left(x^2+5\right)\left(x-2\right)=0\
\Leftrightarrow\left[{}\begin{matrix}x=0\x=2\x^2+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\x=2\x\in\varnothing\left(x^2+5>0\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\x=2\end{matrix}\right.$$b,\left(3x+5\right)^2=\left(2x-2\right)^2\
\Leftrightarrow\left(3x+5\right)^2-\left(2x-2\right)^2=0\
\Leftrightarrow\left(3x+5+2x-2\right)\left(3x+5-2x+2\right)=0\
\Leftrightarrow\left(5x+3\right)\left(x+7\right)=0\
\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{5}\x=-7\end{matrix}\right.$$c,x^3-2x^2+x=0\
\Leftrightarrow x\left(x-1\right)^2=0\
\Leftrightarrow\left[{}\begin{matrix}x=0\x=1\end{matrix}\right.$$d,x^2\left(x-1\right)-4x^2+8x-4=0\
\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\
\Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\
\Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\
\Leftrightarrow\left[{}\begin{matrix}x=1\x=2\end{matrix}\right.$Câu trả lời: $a,x^4-2x^3+5x^2-10x=0\
\Leftrightarrow x^3\left(x-2\right)+5x\left(x-2\right)=0\
\Leftrightarrow x\left(x^2+5\right)\left(x-2\right)=0\
\Leftrightarrow\left[{}\begin{matrix}x=0\x=2\x^2+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\x=2\x\in\varnothing\left(x^2+5>0\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\x=2\end{matrix}\right.$$b,\left(3x+5\right)^2=\left(2x-2\right)^2\
\Leftrightarrow\left(3x+5\right)^2-\left(2x-2\right)^2=0\
\Leftrightarrow\left(3x+5+2x-2\right)\left(3x+5-2x+2\right)=0\
\Leftrightarrow\left(5x+3\right)\left(x+7\right)=0\
\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{5}\x=-7\end{matrix}\right.$$c,x^3-2x^2+x=0\
\Leftrightarrow x\left(x-1\right)^2=0\
\Leftrightarrow\left[{}\begin{matrix}x=0\x=1\end{matrix}\right.$$d,x^2\left(x-1\right)-4x^2+8x-4=0\
\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\
\Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\
\Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\
\Leftrightarrow\left[{}\begin{matrix}x=1\x=2\end{matrix}\right.$

ILoveMath · Answer

a) $x^4-2x^3+5x^2-10x=0\
\Rightarrow\left(x^4-2x^3\right)+\left(5x^2-10x\right)=0\
\Rightarrow x^3\left(x-2\right)+5x\left(x-2\right)=0\
\Rightarrow\left(x^3+5x\right)\left(x-2\right)=0\
\Rightarrow x\left(x^2+5\right)\left(x-2\right)=0\
\Rightarrow\left[{}\begin{matrix}x=0\x^2+5=0\x-2=0\end{matrix}\right.\
\Rightarrow\left[{}\begin{matrix}x=0\x=\pm\sqrt{5}\x=2\end{matrix}\right.$Vậy $x=\left\{-\sqrt{5};0;\sqrt{5};2\right\}$b) $\left(3x+5\right)^2=\left(2x-2\right)^2\
\Rightarrow\left[{}\begin{matrix}3x+5=2x-2\3x+5=-2x+2\end{matrix}\right.\ \Rightarrow\left[{}\begin{matrix}x=-7\x=-\dfrac{3}{5}\end{matrix}\right.$c) $x^3-2x^2+x=0\
\Rightarrow x\left(x^2-2x+1\right)=0\
\Rightarrow x\left(x-1\right)^2=0\
\Rightarrow\left[{}\begin{matrix}x=0\\left(x-1\right)^2=0\end{matrix}\right.\
\Rightarrow\left[{}\begin{matrix}x=0\x=1\end{matrix}\right.$vậy ... d) $x^2\left(x-1\right)-4x^2+8x-4=0\ x^2\left(x-1\right)-\left(4x^2-8x+4\right)=0\ x^2\left(x-1\right)-\left(2x-2\right)^2=0\ \Rightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\ \Rightarrow\left(x-1\right)\left[x^2-4\left(x-1\right)\right]=0\
\Rightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\
\Rightarrow\left(x-1\right)\left(x-2\right)^2=0$    $\Rightarrow\left[{}\begin{matrix}x-1=0\\left(x-2\right)^2=0\end{matrix}\right.\
\Rightarrow\left[{}\begin{matrix}x=1\x=2\end{matrix}\right.$Câu trả lời: a) $x^4-2x^3+5x^2-10x=0\
\Rightarrow\left(x^4-2x^3\right)+\left(5x^2-10x\right)=0\
\Rightarrow x^3\left(x-2\right)+5x\left(x-2\right)=0\
\Rightarrow\left(x^3+5x\right)\left(x-2\right)=0\
\Rightarrow x\left(x^2+5\right)\left(x-2\right)=0\
\Rightarrow\left[{}\begin{matrix}x=0\x^2+5=0\x-2=0\end{matrix}\right.\
\Rightarrow\left[{}\begin{matrix}x=0\x=\pm\sqrt{5}\x=2\end{matrix}\right.$Vậy $x=\left\{-\sqrt{5};0;\sqrt{5};2\right\}$b) $\left(3x+5\right)^2=\left(2x-2\right)^2\
\Rightarrow\left[{}\begin{matrix}3x+5=2x-2\3x+5=-2x+2\end{matrix}\right.\ \Rightarrow\left[{}\begin{matrix}x=-7\x=-\dfrac{3}{5}\end{matrix}\right.$c) $x^3-2x^2+x=0\
\Rightarrow x\left(x^2-2x+1\right)=0\
\Rightarrow x\left(x-1\right)^2=0\
\Rightarrow\left[{}\begin{matrix}x=0\\left(x-1\right)^2=0\end{matrix}\right.\
\Rightarrow\left[{}\begin{matrix}x=0\x=1\end{matrix}\right.$vậy ... d) $x^2\left(x-1\right)-4x^2+8x-4=0\ x^2\left(x-1\right)-\left(4x^2-8x+4\right)=0\ x^2\left(x-1\right)-\left(2x-2\right)^2=0\ \Rightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\ \Rightarrow\left(x-1\right)\left[x^2-4\left(x-1\right)\right]=0\
\Rightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\
\Rightarrow\left(x-1\right)\left(x-2\right)^2=0$    $\Rightarrow\left[{}\begin{matrix}x-1=0\\left(x-2\right)^2=0\end{matrix}\right.\
\Rightarrow\left[{}\begin{matrix}x=1\x=2\end{matrix}\right.$

Nguyễn Lê Phước Thịnh · Answer

a: Ta có: $x^4-2x^3+5x^2-10x=0$$\Leftrightarrow x\left(x^3-2x^2+5x-10\right)=0$$\Leftrightarrow x\left(x-2\right)=0$$\Leftrightarrow\left[{}\begin{matrix}x=0\x=2\end{matrix}\right.$b:Ta có: $\left(3x+5\right)^2=\left(2x-2\right)^2$$\Leftrightarrow\left(3x+5\right)^2-\left(2x-2\right)^2=0$$\Leftrightarrow\left(3x+5-2x+2\right)\left(3x+5+2x-2\right)=0$$\Leftrightarrow\left(x+7\right)\left(5x+3\right)=0$$\Leftrightarrow\left[{}\begin{matrix}x=-7\x=-\dfrac{3}{5}\end{matrix}\right.$Câu trả lời: a: Ta có: $x^4-2x^3+5x^2-10x=0$$\Leftrightarrow x\left(x^3-2x^2+5x-10\right)=0$$\Leftrightarrow x\left(x-2\right)=0$$\Leftrightarrow\left[{}\begin{matrix}x=0\x=2\end{matrix}\right.$b:Ta có: $\left(3x+5\right)^2=\left(2x-2\right)^2$$\Leftrightarrow\left(3x+5\right)^2-\left(2x-2\right)^2=0$$\Leftrightarrow\left(3x+5-2x+2\right)\left(3x+5+2x-2\right)=0$$\Leftrightarrow\left(x+7\right)\left(5x+3\right)=0$$\Leftrightarrow\left[{}\begin{matrix}x=-7\x=-\dfrac{3}{5}\end{matrix}\right.$

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