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a)
\(\sqrt{x^2-2x+1}=x^2-1\)
\(\Rightarrow\left|x-1\right|=x^2-1\)
Với \(x< 1\Rightarrow\left(x-1\right)< 0\)
\(\Rightarrow1-x=x^2-1\)
\(\Leftrightarrow x^2+x-2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(l\right)\\x=-2\left(tm\right)\end{matrix}\right.\)
Với \(x\ge1\Rightarrow\left(x-1\right)\ge0\)
\(\Rightarrow x-1=x^2-1\)
\(\Leftrightarrow x^2-x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(l\right)\\x=1\left(tm\right)\end{matrix}\right.\)
Vậy phương trình đã cho có tập nghiệm \(S=\left\{-2;1\right\}\)
Gõ hết thì mệt lắm , từ câu a --> e làm tương tự còn câu f) bạn bình phương 2 vế lên là được.
2: \(\Leftrightarrow\left|x-1\right|=x^2-1\)
\(\Leftrightarrow\left(x-1\right)^2=\left(x-1\right)^2\left(x+1\right)^2\)
\(\Leftrightarrow\left(x-1\right)^2\cdot x\cdot\left(x+2\right)=0\)
hay \(x\in\left\{1;0;-2\right\}\)
3: \(\Leftrightarrow\left\{{}\begin{matrix}x>=1\\\left(2x-1\right)^2-\left(x-1\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=1\\\left(2x-1-x+1\right)\left(2x-1+x-1\right)=0\end{matrix}\right.\)
hay \(x\in\varnothing\)
\(\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1\) ( SỬA ĐỀ)
\(\sqrt{x-1-2.2.\sqrt{x-1}+4}+\sqrt{x-1-2.3.\sqrt{x-1}+9}=1\)
\(|x-1-2|+|x-1-3|=1\)
\(|x-3|+|x-4|=1\)
Với \(x\le3\)thì PT thành \(3-x+4-x=1\) \(\Rightarrow-2x=-6\Rightarrow x=3\)(thõa mãn)
Với \(3\le x< 4\)thì PT thành \(x-3+4-x=1\Leftrightarrow0x=0\Rightarrow\)Đúng với mọi x từ \(3\le x< 4\)
Với \(x\ge4\)thì PT thành \(x-3+x-4=1\Leftrightarrow2x=8\Leftrightarrow x=4\)(thõa mãn)
Vậy \(3\le x\le4\)
b: \(=\dfrac{\left|x\right|+\left|x-2\right|+1}{2x-1}=\dfrac{x+x-2+1}{2x-1}=\dfrac{2x-1}{2x-1}=1\)
c: \(=\left|x-4\right|+\left|x-6\right|\)
=x-4+6-x=2
\(a.\sqrt{1-4a+4a^2}-2a=\sqrt{\left(1-2a\right)^2}-2a=\left|1-2a\right|-2a\)
*\(a>\dfrac{1}{2}\Rightarrow\left|1-2a\right|-2a=2a-1-2a=4a-1\)
* \(a\le\dfrac{1}{2}\Rightarrow\left|1-2a\right|-2a=1-2a-2a=1-4a\)
\(b.x-2y-\sqrt{x^2-4xy+4y^2}=x-2y-\sqrt{\left(x-2y\right)^2}=x-2y-\left|x-2y\right|\)
* \(x\ge2y\Rightarrow x-2y-\left|x-2y\right|=x-2y-x+2y=2x\)
* \(x< 2y\Rightarrow x-2y-\left|x-2y\right|=x-2y-2y+x=2x-4y\)
\(c.x^2+\sqrt{x^4-8x^2+16}=x^2+\sqrt{\left(x^2-4\right)^2}=x^2+\left|x^2-4\right|\)
* \(x^2-4\ge0\Rightarrow x^2+\left|x^2-4\right|=x^2+x^2-4=2x^2-4\)
* \(x^2-4< 0\Rightarrow x^2+\left|x^2-4\right|=x^2+4-x^2=4\)
\(d.2x-1-\dfrac{\sqrt{x^2-10x+25}}{x-5}=2x-1-\dfrac{\sqrt{\left(x-5\right)^2}}{x-5}=2x-1-\dfrac{\left|x-5\right|}{x-5}\)
* \(x\ge5\Rightarrow2x-1-\dfrac{\left|x-5\right|}{x-5}=2x-1-1=2x-2\)
* \(x< 5\Rightarrow2x-1-\dfrac{\left|x-5\right|}{x-5}=2x-1+1=2x\)
\(e.\dfrac{\sqrt{x^4-4x^2+4}}{x^2-2}=\dfrac{\sqrt{\left(x^2-2\right)^2}}{x^2-2}=\dfrac{\left|x^2-2\right|}{x^2-2}\)
* \(x^2\ge2\Rightarrow\dfrac{\left|x^2-2\right|}{x^2-2}=1\)
* \(x^2< 2\Rightarrow\dfrac{\left|x^2-2\right|}{x^2-2}=-1\)
\(f.\sqrt{\left(x-4\right)^2}+\dfrac{x-4}{\sqrt{x^2-8x+16}}=\left|x-4\right|+\dfrac{x-4}{\sqrt{\left(x-4\right)^2}}=\left|x-4\right|+\dfrac{x-4}{\left|x-4\right|}\)
* \(x\ge4\Rightarrow\left|x-4\right|+\dfrac{x-4}{\left|x-4\right|}=x-4+\dfrac{x-4}{x-4}=x-5\)
* \(x< 4\Rightarrow\left|x-4\right|+\dfrac{x-4}{\left|x-4\right|}=4-x-1=5-x\)
b)\(\frac{4}{x}+\sqrt{x-\frac{1}{x}}=x+\sqrt{2x-\frac{5}{x}}\)
\(pt\Leftrightarrow\frac{4}{x}+\sqrt{x-\frac{1}{x}}-\sqrt{\frac{3}{2}}=x+\sqrt{2x-\frac{5}{x}}-\sqrt{\frac{3}{2}}\)
\(\Leftrightarrow\left(\frac{4}{x}-x\right)+\frac{x-\frac{1}{x}-\frac{3}{2}}{\sqrt{x-\frac{1}{x}}+\sqrt{\frac{3}{2}}}=\frac{2x-\frac{5}{x}-\frac{3}{2}}{\sqrt{2x-\frac{5}{x}}+\sqrt{\frac{3}{2}}}\)
\(\Leftrightarrow\frac{-\left(x-2\right)\left(x+2\right)}{x}+\frac{\frac{\left(x-2\right)\left(2x+1\right)}{2x}}{\sqrt{x-\frac{1}{x}}+\sqrt{\frac{3}{2}}}-\frac{\frac{\left(x-2\right)\left(4x+5\right)}{2x}}{\sqrt{2x-\frac{5}{x}}+\sqrt{\frac{3}{2}}}=0\)
\(\Leftrightarrow\left(x-2\right)\left(\frac{-\left(x+2\right)}{x}+\frac{\frac{\left(2x+1\right)}{2x}}{\sqrt{x-\frac{1}{x}}+\sqrt{\frac{3}{2}}}-\frac{\frac{\left(4x+5\right)}{2x}}{\sqrt{2x-\frac{5}{x}}+\sqrt{\frac{3}{2}}}\right)=0\)
Pt trong ngoặc VN suy ra x=2
a)\(x^2+3\sqrt{x^2-1}=\sqrt{x^4-x^2+1}\)
\(\Leftrightarrow x^2+3\sqrt{x^2-1}-1=\sqrt{x^4-x^2+1}-1\)
\(\Leftrightarrow\frac{x^2\left(3\sqrt{x^2-1}+1\right)}{3\sqrt{x^2-1}+1}+\frac{9\left(x^2-1\right)-1}{3\sqrt{x^2-1}+1}=\frac{x^4-x^2+1-1}{\sqrt{x^4-x^2+1}+1}\)
\(\Leftrightarrow\frac{9x^2-10+3x^2\sqrt{x^2-1}+x^2}{3\sqrt{x^2-1}+1}=\frac{x^4-x^2}{\sqrt{x^4-x^2+1}+1}\)
\(\Leftrightarrow\frac{\sqrt{x^2-1}\left(3x^2+10\sqrt{x^2-1}\right)}{3\sqrt{x^2-1}+1}=\frac{x^2\left(x-1\right)\left(x+1\right)}{\sqrt{x^4-x^2+1}+1}\)
\(\Leftrightarrow\frac{\sqrt{\left(x-1\right)\left(x+1\right)}\left(3x^2+10\sqrt{x^2-1}\right)}{3\sqrt{x^2-1}+1}-\frac{x^2\left(x-1\right)\left(x+1\right)}{\sqrt{x^4-x^2+1}+1}=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(\frac{\frac{1}{\sqrt{x^2-1}}\left(3x^2+10\sqrt{x^2-1}\right)}{3\sqrt{x^2-1}+1}-\frac{x^2}{\sqrt{x^4-x^2+1}+1}\right)=0\)
pt trong căn vô nghiệm
suy ra x=1; x=-1
\(1.\sqrt{1-2x^2}=x-1\)
\(\Leftrightarrow1-2x^2=x^2-2x+1\)
\(\Leftrightarrow-2x^2-x^2+2x=0\)
\(\Leftrightarrow x\left(2-3x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\2-3x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{3}\end{matrix}\right.\)
Vậy phương trình có nghiệm x = 0 hoặc x = \(\dfrac{2}{3}\) .
\(2.\sqrt{x^4-2x^2+1}=x-1\)
\(\Leftrightarrow x^2-1=x-1\)
\(\Leftrightarrow x\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
Vậy phương trình có nghiệm x = 0 hoặc x = -1 .