(x+4)(x^2 - 4x + 16) - (x-2)(x^2+2x+4)
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a) \(\sqrt[]{x^2-2x+4}=2x-2\)
\(\Leftrightarrow\sqrt[]{x^2-2x+4}=2\left(x-1\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}2\left(x-1\right)\ge0\\x^2-2x+4=4\left(x-1\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1\ge0\\x^2-2x+4=4x^2-8x+4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\3x^2-6x=0\end{matrix}\right.\) \(\left(1\right)\)
Giải pt \(3x^2-6x=0\)
\(\Leftrightarrow3x\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=2\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow x=2\)
c) \(\sqrt{x^2-3x+2}=\sqrt[]{x-1}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1\ge0\\x^2-3x+2=x-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\x^2-4x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\x=1\cup x=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)
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Câu d đề có đúng ko bn
mk thấy hơi sai Nguyen Thi tuong Vi
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a) \(\dfrac{16+x}{x^2-2x}+\dfrac{18}{2x-x^2}\)
\(=\dfrac{16+x}{x^2-2x}-\dfrac{18}{x^2-2x}\)
\(=\dfrac{16+x-18}{x^2-2x}\)
\(=\dfrac{x-2}{x\left(x-2\right)}\)
\(=\dfrac{1}{x}\)
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2:
=>x^3-1-2x^3-4x^6+4x^6+4x=6
=>-x^3+4x-7=0
=>x=-2,59
4: =>8x-24x^2+2-6x+24x^2-60x-4x+10=-50
=>-62x+12=-50
=>x=1
\(\left(x+4\right)\left(x^2-4x+16\right)-\left(x-2\right)\left(x^2+2x+4\right)\)
\(=x^3-4x^2+16x+4x^2-16x+64-\left(x^3+2x^2+4x-2x^2-4x-8\right)\)
\(=x^3+\left(-4x^2+4x^2\right)+\left(16x-16x\right)+64-\left(x^3-8\right)\)
\(=x^3+0+0+64-x^3+8\)
\(=\left(x^3-x^3\right)+\left(64+8\right)\)
\(=0+72=72\)
(x + 4)(x² - 4x + 16) - (x - 2)(x² + 2x + 4)
= x³ + 4³ - (x³ - 2³)
= x³ + 64 - x³ + 8
= 72