Giải pt sau:
\(2\sqrt{2+x-x^2}=1+\frac{1}{x}\)
Những câu hỏi liên quan
1.Giải pt sau:(sqrt{2} +2)(xsqrt{2} -1)2xsqrt{2} -sqrt{2}
2.Cho pt: 2(a-1).x-a(x-1)2a+3
3.Giải pt sau:
a) frac{2}{x+frac{text{1}}{text{1}+frac{x+text{1}}{x-2}}}frac{6}{3x-text{1}}
b) frac{frac{x+text{1}}{x-text{1}}-frac{x-text{1}}{x+text{1}}}{text{1}+frac{x+text{1}}{x-text{1}}}frac{x-text{1}}{2left(x+text{1}right)}
Đọc tiếp
1.Giải pt sau:(\(\sqrt{2}\) +2)(x\(\sqrt{2}\) -1)=2x\(\sqrt{2}\) -\(\sqrt{2}\)
2.Cho pt: 2(a-1).x-a(x-1)=2a+3
3.Giải pt sau:
a) \(\frac{2}{x+\frac{\text{1}}{\text{1}+\frac{x+\text{1}}{x-2}}}=\frac{6}{3x-\text{1}}\)
b) \(\frac{\frac{x+\text{1}}{x-\text{1}}-\frac{x-\text{1}}{x+\text{1}}}{\text{1}+\frac{x+\text{1}}{x-\text{1}}}=\frac{x-\text{1}}{2\left(x+\text{1}\right)}\)
1) Nhìn cái pt hết ham, nhưng bấm nghiệm đẹp v~`~
\(\left(\sqrt{2}+2\right)\left(x\sqrt{2}-1\right)=2x\sqrt{2}-\sqrt{2}\)
\(\Leftrightarrow\left(\sqrt{2}+2\right)\left(x\sqrt{2}-1\right)-2x\sqrt{2}+\sqrt{2}=0\)
\(\Leftrightarrow2x-\sqrt{2}+2x\sqrt{2}-2-2x\sqrt{2}+\sqrt{2}=0\)
\(\Leftrightarrow2x-2=0\Leftrightarrow2x=2\Rightarrow x=1\)
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Mấy bài kia sao cái phương trình dài thê,s giải sao nổi
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giải bất pt sau:
\(\frac{\sqrt{x^{2^{ }}-x-2}}{\sqrt{x-1}}+\sqrt{x-1}< \frac{2x+1}{\sqrt{x-1}}\)
Dk 1<x<2
√x^2 -x -2<x+2
5x+6>0
X > -6/5
Bpt vô nghiệm
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Giải PT sau: \(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=\frac{x+3}{2}\)
Giải PT sau: \(1+\frac{2}{3}\sqrt{x-x^2}=\sqrt{x}+\sqrt{1-x}\)
ĐK: \(0\le x\le1\)
Đặt \(t=\sqrt{x}+\sqrt{1-x}\) ( \(t>0\) )
\(\Leftrightarrow t^2=x+1-x+2\sqrt{x\left(1-x\right)}\)
\(\Leftrightarrow t^2-1=2\sqrt{x-x^2}\)
\(\Leftrightarrow\frac{t^2-1}{2}=\sqrt{x-x^2}\)
Ta có \(pt\Leftrightarrow1+\frac{2}{3}\cdot\frac{t^2-1}{2}=t\)
\(\Leftrightarrow1+\frac{t^2-1}{3}-t=0\)
\(\Leftrightarrow t^2-1-3t+3=0\)
\(\Leftrightarrow t^2-3t+2=0\)
\(\Leftrightarrow\left(t-1\right)\left(t-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=1\\t=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}+\sqrt{1-x}=1\\\sqrt{x}+\sqrt{1-x}=2\end{matrix}\right.\)
TH1: \(\sqrt{x}+\sqrt{1-x}=1\)
\(\Leftrightarrow x+1-x+2\sqrt{x\left(1-x\right)}=1\)
\(\Leftrightarrow\sqrt{x\left(1-x\right)}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)( thỏa (
TH2: \(\sqrt{x}+\sqrt{1-x}=2\)
\(\Leftrightarrow x+1-x+2\sqrt{x\left(1-x\right)}=4\)
\(\Leftrightarrow\sqrt{x\left(1-x\right)}=\frac{3}{2}\)
\(\Leftrightarrow x\left(1-x\right)=\frac{9}{4}\)
\(\Leftrightarrow4x\left(1-x\right)=9\)
\(\Leftrightarrow4x^2-4x+9=0\)
\(\Leftrightarrow\left(2x+1\right)^2+8=0\)( vô lý )
Vậy \(x\in\left\{0;1\right\}\)
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Giải PT sau : \(\sqrt[3]{\frac{1}{2}+x}+\sqrt{\frac{1}{2}-x}=6\)
Giải pt sau :
\(x^2-1+\sqrt{143}=\frac{1}{x^2-1}-\sqrt{143}\)
\(x^2-1+\sqrt{143}=\frac{1}{x^2-1}-\sqrt{143}\)(đk: \(x\ne1\))
Đặt \(x^2-1=a\left(a\ge-1,a\ne0\right)\)
Có \(a+\sqrt{143}=\frac{1}{a}-\sqrt{143}\)
<=> \(a-\frac{1}{a}+2\sqrt{143}=0\)
<=> \(\frac{a^2-1+2\sqrt{143}a}{a}=0\)
<=> \(a^2+2\sqrt{143}a+143=144\)
<=> \(\left(a+\sqrt{143}\right)^2=144\)
=> \(\left[{}\begin{matrix}a+\sqrt{143}=12\\a+\sqrt{143}=-12\left(ktm\right)\end{matrix}\right.\) <=> \(a=12-\sqrt{143}\)
<=> \(x^2-1=12+\sqrt{143}\)
Làm nốt nha :))
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Giải pt: \(\frac{1}{\sqrt{x+3}+\sqrt{x+2}}+\frac{1}{\sqrt{x+2}+\sqrt{x+1}}+\frac{1}{\sqrt{x+1}+\sqrt{x}}=1\)
giải các pt sau
\(\frac{3}{\sqrt{x}+15}=\frac{\sqrt{x}}{5}\)
\(\frac{x+2\sqrt{x}+1}{\sqrt{x}}=\frac{9}{2}\)
Giải pt \(\frac{1}{\sqrt{x-1}+\sqrt{x-2}}+\frac{1}{\sqrt{x-2}+\sqrt{x-3}}+...+\frac{1}{\sqrt{x-9}+\sqrt{x-10}}=1\)