tìm x bt |x-3|-1=x
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\(x+\dfrac{1}{x}=3\Leftrightarrow\left(x+\dfrac{1}{x}\right)^3=27\\ \Leftrightarrow x^3+\left(\dfrac{1}{x}\right)^3+3x\cdot\dfrac{1}{x}\left(x+\dfrac{1}{x}\right)=27\\ \Leftrightarrow x^3+\dfrac{1}{x^3}+3\cdot3=27\\ \Leftrightarrow x^3+\dfrac{1}{x^3}=18\)
![](https://rs.olm.vn/images/avt/0.png?1311)
ĐKXĐ: \(x>0;x\ne1\)
\(A=\left(\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\dfrac{2\left(\sqrt{x}+1\right)}{x\left(\sqrt{x}+1\right)}-\dfrac{2-x}{x\left(\sqrt{x}+1\right)}\right)\)
\(=\left(\dfrac{x+2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\dfrac{x+2\sqrt{x}}{x\left(\sqrt{x}+1\right)}\right)\)
\(=\dfrac{\left(x+2\sqrt{x}\right).x.\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\left(x+2\sqrt{x}\right)}=\dfrac{x}{\sqrt{x}-1}\)
b.
\(x=4+2\sqrt{3}=\left(\sqrt{3}+1\right)^2\Rightarrow\sqrt{x}=\sqrt{3}+1\)
\(\Rightarrow A=\dfrac{4+2\sqrt{3}}{\sqrt{3}+1-1}=\dfrac{4+2\sqrt{3}}{\sqrt{3}}=\dfrac{6+4\sqrt{3}}{3}\)
c.
Để \(\sqrt{A}\) xác định \(\Rightarrow\sqrt{x}-1>0\Rightarrow x>1\)
Ta có:
\(\sqrt{A}=\sqrt{\dfrac{x}{\sqrt{x}-1}}=\sqrt{\dfrac{x}{\sqrt{x}-1}-4+4}=\sqrt{\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}-1}+4}\ge\sqrt{4}=2\)
Dấu "=" xảy ra khi \(\sqrt{x}-2=0\Rightarrow x=4\)
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\(\left(x+1\right)^3+\left(x+2\right)^3=\left(2x+3\right)^3\)
\(\Leftrightarrow\left(2x+3\right)\left(x^2+2x+1-x^2-3x-2+x^2+4x+4\right)=\left(2x+3\right)^3\)
\(\Leftrightarrow\left(2x+3\right)\left(x^2+3x+3\right)-\left(2x+3\right)^3=0\)
\(\Leftrightarrow\left(2x+3\right)\left(4x^2+12x+9-x^2-3x-3\right)=0\)
\(\Leftrightarrow\left(2x+3\right)\left(3x^2+9x+6\right)=0\)
\(\Leftrightarrow\left(2x+3\right)\left(x+1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=-1\\x=-2\end{matrix}\right.\)
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1) \(A=\dfrac{\left(\sqrt{x}+1\right)^2+\left(\sqrt{x}-1\right)^2-3\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x+2\sqrt{x}+1+x-2\sqrt{x}+1-3\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{2x-3\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\left(2x-2\sqrt{x}\right)-\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}\)
b) \(A=\dfrac{2\sqrt{9}-1}{\sqrt{9}+1}=\dfrac{5}{4}\)
c) \(A=\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}< 1\Rightarrow2\sqrt{x}-1< \sqrt{x}+1\Rightarrow\sqrt{x}< 2\Rightarrow x< 4\)
\(1,A=\dfrac{x+2\sqrt{x}+1+x-2\sqrt{x}+1-3\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ A=\dfrac{2x-3\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\left(\sqrt{x}-1\right)\left(2\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}\\ 2,x=9\Leftrightarrow A=\dfrac{6-1}{3+1}=\dfrac{5}{4}\\ 3,A< 1\Leftrightarrow\dfrac{2\sqrt{x}-1-\sqrt{x}-1}{\sqrt{x}+1}< 0\\ \Leftrightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}+1}< 0\Leftrightarrow\sqrt{x}-2< 0\left(\sqrt{x}+1>0\right)\\ \Leftrightarrow x< 4\Leftrightarrow0\le x< 4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left(x+3\right)^2-x\left(x-1\right)=2\)
\(\Leftrightarrow x^2+6x+9-x^2+x=2\)
\(\Leftrightarrow7x+9=2\)
\(\Leftrightarrow7x=2-9\)
\(\Leftrightarrow7x=-7\)
\(\Leftrightarrow x=\dfrac{-7}{7}=-1\)
b) \(\left(2x+3\right)^2-\left(x+1\right)\left(4x-3\right)=-1\)
\(\Leftrightarrow4x^2+12x+9-\left(4x^2-3x+4x-3\right)=-1\)
\(\Leftrightarrow4x^2+12x+9-4x^2+3x-4x+3=-1\)
\(\Leftrightarrow11x+12=-1\)
\(\Leftrightarrow11x=-13\)
\(\Leftrightarrow x=\dfrac{-13}{11}\)
* Nếu \(x< 3\)
\(\Rightarrow-\left(x-3\right)-1=x\Rightarrow2-x=x\Rightarrow x=1\)
* Nếu \(x\ge3\)
\(\Rightarrow\left(x-3\right)-1=x\Rightarrow x-4=x\Rightarrow0=-4\)( vô lí )
Vậy x=1