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1) Ta có: \(\sqrt{21-x}+1=x\)
\(\Leftrightarrow21-x=\left(x-1\right)^2\)
\(\Leftrightarrow x^2-2x+1-21+x=0\)
\(\Leftrightarrow x^2-3x-20=0\)
\(\text{Δ}=\left(-3\right)^2-4\cdot1\cdot\left(-20\right)=9+80=89\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{3+\sqrt{89}}{2}\\x_2=\dfrac{3-\sqrt{89}}{2}\end{matrix}\right.\)
1)\(\sqrt{21-x}+1=x\)
\(\Leftrightarrow21-x=\left(x-1\right)^2\)
\(\Leftrightarrow21-x=x^2-2x+1\)
\(\Leftrightarrow x^2-x-20=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-4\end{matrix}\right.\)
2)\(\sqrt{8-x}+2=x\)
\(\Leftrightarrow8-x=\left(x-2\right)^2\)
\(\Leftrightarrow8-x=x^2-4x+4\)
\(\Leftrightarrow x^2-3x-4=0\Leftrightarrow\left(x-4\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-1\end{matrix}\right.\)
\(P=\frac{3x+3\sqrt{x}-3-\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{x+\sqrt{x}-2}\)
\(P=\frac{3x+3\sqrt{x}-3-x+1-x+4}{x+\sqrt{x}-2}\)
\(P=\frac{x+3\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(P=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
Bài 2
b, `\sqrt{3x^2}=x+2` ĐKXĐ : `x>=0`
`=>(\sqrt{3x^2})^2=(x+2)^2`
`=>3x^2=x^2+4x+4`
`=>3x^2-x^2-4x-4=0`
`=>2x^2-4x-4=0`
`=>x^2-2x-2=0`
`=>(x^2-2x+1)-3=0`
`=>(x-1)^2=3`
`=>(x-1)^2=(\pm \sqrt{3})^2`
`=>` $\left[\begin{matrix} x-1=\sqrt{3}\\ x-1=-\sqrt{3}\end{matrix}\right.$
`=>` $\left[\begin{matrix} x=1+\sqrt{3}\\ x=1-\sqrt{3}\end{matrix}\right.$
Vậy `S={1+\sqrt{3};1-\sqrt{3}}`
\(\sqrt{3x^2-5x+1}-\sqrt{x^2-2}=\sqrt{3\left(x^2-x-1\right)}-\sqrt{x^2-3x+4}\)
\(\Leftrightarrow\left(\sqrt{3x^2-5x+1}-\sqrt{3}\right)-\left(\sqrt{x^2-2}-\sqrt{2}\right)=\left(\sqrt{3\left(x^2-x-1\right)}-\sqrt{3}\right)-\left(\sqrt{x^2-3x+4}-\sqrt{2}\right)\)
\(\Leftrightarrow\frac{3x^2-5x+1-3}{\sqrt{3x^2-5x+1}+\sqrt{3}}-\frac{x^2-2-2}{\sqrt{x^2-2}+\sqrt{2}}=\frac{3\left(x^2-x-1\right)-3}{\sqrt{3\left(x^2-x-1\right)}+\sqrt{3}}-\frac{x^2-3x+4-2}{\sqrt{x^2-3x+4}+\sqrt{2}}\)
\(\Leftrightarrow\frac{3x^2-5x-2}{\sqrt{3x^2-5x+1}+\sqrt{3}}-\frac{x^2-4}{\sqrt{x^2-2}+\sqrt{2}}-\frac{3x^2-3x-6}{\sqrt{3\left(x^2-x-1\right)}+\sqrt{3}}+\frac{x^2-3x+2}{\sqrt{x^2-3x+4}+\sqrt{2}}=0\)
\(\Leftrightarrow\frac{\left(x-2\right)\left(3x+1\right)}{\sqrt{3x^2-5x+1}+\sqrt{3}}-\frac{\left(x-2\right)\left(x+2\right)}{\sqrt{x^2-2}+\sqrt{2}}-\frac{3\left(x-2\right)\left(x+1\right)}{\sqrt{3\left(x^2-x-1\right)}+\sqrt{3}}+\frac{\left(x-1\right)\left(x-2\right)}{\sqrt{x^2-3x+4}+\sqrt{2}}=0\)
\(\Leftrightarrow\left(x-2\right)\left(\frac{3x+1}{\sqrt{3x^2-5x+1}+\sqrt{3}}-\frac{x+2}{\sqrt{x^2-2}+\sqrt{2}}-\frac{3\left(x+1\right)}{\sqrt{3\left(x^2-x-1\right)}+\sqrt{3}}+\frac{x-1}{\sqrt{x^2-3x+4}+\sqrt{2}}\right)=0\)
Dễ thấy: \(\frac{3x+1}{\sqrt{3x^2-5x+1}+\sqrt{3}}-\frac{x+2}{\sqrt{x^2-2}+\sqrt{2}}-\frac{3\left(x+1\right)}{\sqrt{3\left(x^2-x-1\right)}+\sqrt{3}}+\frac{x-1}{\sqrt{x^2-3x+4}+\sqrt{2}}=0\) vô nghiệm
\(\Rightarrow x-2=0\Rightarrow x=2\)
1) \(\sqrt[3]{x+1}=5\)
\(\Rightarrow x+1=125\)
\(\Rightarrow x=124\)
2) \(\sqrt[3]{1-3x^3}=-2\)
\(\Rightarrow1-3x^3=-8\)
\(\Rightarrow3x^3=9\)
\(\Rightarrow x=\sqrt[3]{3}\)
a: \(C=\dfrac{3x+3\sqrt{x}-3-x+1}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}-2}{\sqrt{x}}\cdot\dfrac{1-1+\sqrt{x}}{1-\sqrt{x}}\)
\(=\dfrac{2x+3\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}-2}{1-\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}-1-\sqrt{x}+2}{\sqrt{x}-1}=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
b: Để \(C=\sqrt{x}\) thì \(x-\sqrt{x}=\sqrt{x}+1\)
=>\(x-2\sqrt{x}-1=0\)
=>\(\Leftrightarrow x=3+2\sqrt{2}\)
c: |2x-5|=3
=>2x-5=3 hoặc 2x-5=-3
=>2x=2 hoặc 2x=8
=>x=4(nhận) hoặc x=1(loại)
Khi x=4 thì \(C=\dfrac{2+1}{2-1}=3\)