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31 tháng 8 2021

\(C=\sqrt{9x^2}-2x=\left|3x\right|-2x=-3x-2x=-5x\left(x< 0\right)\)

\(D=x-4+\sqrt{16-8x+x^2}\left(x>4\right)\)

\(=x-4+\sqrt{\left(x-4\right)^2}\)

\(=x-4+\left|x-4\right|\)

\(=x-4+x-4\)

\(=2x-8\)

31 tháng 8 2021

 

21 tháng 9 2023

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) \(\sqrt{x^2-3x+2}=\sqrt[]{x-1}\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-1\ge0\\x^2-3x+2=x-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\x^2-4x+3=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\x=1\cup x=3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)

14 tháng 7 2017

\(a,\sqrt{1-4a+4a^2}-2a\)

\(=\sqrt{\left(1-2a\right)^2}-2a\)

\(=1-2a-2a\)

\(=1-4a\)

\(b,x-2y-\sqrt{x^2-4xy+4y^2}\)

\(=x-2y-\sqrt{\left(x-2y\right)^2}\)

\(=x-2y-\left(x-2y\right)\)

\(=x-2y-x+2y\)

\(=0\)

\(c,x^2+\sqrt{x^4-8x^2+16}\)

\(=x^2+\sqrt{\left(x^2-4\right)^2}\)

\(=x^2+x^2-4\)

\(=2x^2-4\)

Các câu còn lại tương tự nha

14 tháng 7 2017

\(a,\sqrt{1-4a+4a^2}-2a\)

\(=\sqrt{\left(1-2a\right)^2}-2a\)

\(=\left(1-2a\right)-2a\)

\(=1-4a\)

\(b,x-2y-\sqrt{x^2-4xy+4y^2}\)

\(=x-2y-\sqrt{\left(x-2y\right)^2}\)

\(=x-2y-\left(x-2y\right)\)

\(=x-2y-x+2y\)

\(=0\)

\(c,x^2+\sqrt{x^4-8x^2+16}\)

\(=x^2+\sqrt{\left(x^2-2^2\right)^2}\)

\(=x^2+\left(x^2-4\right)\)

\(=x^2+x^2-4\)

\(=2x^2-4\)

\(d,2x-1-\frac{\sqrt{x^2-10x+25}}{x-5}\)

\(=2x-1-\frac{\sqrt{\left(x-5\right)^2}}{x-5}\)

\(=2x-1-\frac{x-5}{x-5}\)

\(=2x-1-1\)

\(=2x-2\)

\(=2\left(x-1\right)\)

11 tháng 8 2018

\(a.\sqrt{1-4a+4a^2}-2a=\sqrt{\left(1-2a\right)^2}-2a=\left|1-2a\right|-2a\)

*\(a>\dfrac{1}{2}\Rightarrow\left|1-2a\right|-2a=2a-1-2a=4a-1\)

* \(a\le\dfrac{1}{2}\Rightarrow\left|1-2a\right|-2a=1-2a-2a=1-4a\)

\(b.x-2y-\sqrt{x^2-4xy+4y^2}=x-2y-\sqrt{\left(x-2y\right)^2}=x-2y-\left|x-2y\right|\)

* \(x\ge2y\Rightarrow x-2y-\left|x-2y\right|=x-2y-x+2y=2x\)

* \(x< 2y\Rightarrow x-2y-\left|x-2y\right|=x-2y-2y+x=2x-4y\)

\(c.x^2+\sqrt{x^4-8x^2+16}=x^2+\sqrt{\left(x^2-4\right)^2}=x^2+\left|x^2-4\right|\)

* \(x^2-4\ge0\Rightarrow x^2+\left|x^2-4\right|=x^2+x^2-4=2x^2-4\)

* \(x^2-4< 0\Rightarrow x^2+\left|x^2-4\right|=x^2+4-x^2=4\)

\(d.2x-1-\dfrac{\sqrt{x^2-10x+25}}{x-5}=2x-1-\dfrac{\sqrt{\left(x-5\right)^2}}{x-5}=2x-1-\dfrac{\left|x-5\right|}{x-5}\)

* \(x\ge5\Rightarrow2x-1-\dfrac{\left|x-5\right|}{x-5}=2x-1-1=2x-2\)

* \(x< 5\Rightarrow2x-1-\dfrac{\left|x-5\right|}{x-5}=2x-1+1=2x\)

\(e.\dfrac{\sqrt{x^4-4x^2+4}}{x^2-2}=\dfrac{\sqrt{\left(x^2-2\right)^2}}{x^2-2}=\dfrac{\left|x^2-2\right|}{x^2-2}\)

* \(x^2\ge2\Rightarrow\dfrac{\left|x^2-2\right|}{x^2-2}=1\)

* \(x^2< 2\Rightarrow\dfrac{\left|x^2-2\right|}{x^2-2}=-1\)

\(f.\sqrt{\left(x-4\right)^2}+\dfrac{x-4}{\sqrt{x^2-8x+16}}=\left|x-4\right|+\dfrac{x-4}{\sqrt{\left(x-4\right)^2}}=\left|x-4\right|+\dfrac{x-4}{\left|x-4\right|}\)

* \(x\ge4\Rightarrow\left|x-4\right|+\dfrac{x-4}{\left|x-4\right|}=x-4+\dfrac{x-4}{x-4}=x-5\)

* \(x< 4\Rightarrow\left|x-4\right|+\dfrac{x-4}{\left|x-4\right|}=4-x-1=5-x\)

24 tháng 9 2023

a) \(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\) (ĐK: \(x\ge1\)

\(\Leftrightarrow\sqrt{x-1}+\sqrt{4\left(x-1\right)}-\sqrt{25\left(x-1\right)}+2=0\)

\(\Leftrightarrow\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}+2=0\)

\(\Leftrightarrow-2\sqrt{x-1}=-2\)

\(\Leftrightarrow\sqrt{x-1}=\dfrac{2}{2}\)

\(\Leftrightarrow\sqrt{x-1}=1\)

\(\Leftrightarrow x-1=1\)

\(\Leftrightarrow x=2\left(tm\right)\)

b) \(\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}=16\) (ĐK: \(x\ge-1\))

\(\Leftrightarrow\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}+\sqrt{4\left(x+1\right)}+\sqrt{x+1}=16\)

\(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)

\(\Leftrightarrow4\sqrt{x+1}=16\)

\(\Leftrightarrow\sqrt{x+1}=4\)

\(\Leftrightarrow x+1=16\)

\(\Leftrightarrow x=15\left(tm\right)\)

16 tháng 10 2023

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