\(\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}=\sqrt{2}\) giai phuong trinh
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NV
20 tháng 5 2017
ĐKXĐ: \(x\ge\frac{1}{2}\)
Đề \(\Rightarrow\sqrt{\frac{x+7}{x+1}}-\sqrt{3}+8-2x^2-\left(\sqrt{2x-1}-\sqrt{3}\right)=0\)
Nhân liên hợp ta được:
\(\frac{\left(\sqrt{\frac{x+7}{x+1}}-\sqrt{3}\right)\left(\sqrt{\frac{x+7}{x+1}}+\sqrt{3}\right)}{\sqrt{\frac{x+7}{x+1}}+\sqrt{3}}+2\left(4-x^2\right)-\frac{\left(\sqrt{2x-1}-\sqrt{3}\right)\left(\sqrt{2x+1}+\sqrt{3}\right)}{\sqrt{2x+1}+\sqrt{3}}=0\)
\(\Rightarrow\frac{\frac{x+7}{x+1}-3}{\sqrt{\frac{x+7}{x+1}}+\sqrt{3}}+2\left(4-x^2\right)-\frac{2x-1-3}{\sqrt{2x+1}+\sqrt{3}}=0\)
\(\Rightarrow\frac{\frac{-2x+4}{x+1}}{\sqrt{\frac{x+7}{x+1}}+\sqrt{3}}+2\left(2-x\right)\left(2+x\right)-\frac{2x-4}{\sqrt{2x+1}+\sqrt{3}}=0\)
\(\Rightarrow\left(x-2\right)\left[\frac{-2}{\left(x+1\right)\left(\sqrt{\frac{x+7}{x+1}}+\sqrt{3}\right)}-2\left(2+x\right)-\frac{2}{\sqrt{2x+1}+\sqrt{3}}\right]=0\)
mà \(-\frac{2}{\left(x+1\right)\left(\sqrt{\frac{x+7}{x+1}}+\sqrt{3}\right)}-2\left(2+x\right)-\frac{2}{\sqrt{2x+1}+\sqrt{3}}< 0\)
=> x - 2 = 0 => x = 2
Vậy x = 2
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Nguyễn Việt Lâm
Giáo viên
12 tháng 11 2018
TXĐ: \(x\ge0\)
Phương trình đã cho tương đương:
\(\dfrac{\left(\sqrt{2x+1}-\sqrt{3x}\right)\left(\sqrt{2x+1}+\sqrt{3x}\right)}{\sqrt{2x+1}+\sqrt{3x}}=x-1\)
\(\Leftrightarrow\dfrac{2x+1-3x}{\sqrt{2x+1}+\sqrt{3x}}=x-1\Leftrightarrow\dfrac{-\left(x-1\right)}{\sqrt{2x+1}+\sqrt{3x}}=x-1\)
\(\Leftrightarrow\left(x-1\right)\left(1+\dfrac{1}{\sqrt{2x+1}+\sqrt{3x}}\right)=0\)
\(\Leftrightarrow x-1=0\) (do \(1+\dfrac{1}{\sqrt{2x+1}+\sqrt{3x}}>0\) \(\forall x\ge0\))
\(\Leftrightarrow x=1\)
12 tháng 11 2018
\(\sqrt{2x+1}-\sqrt{3x}=x-1\)
Điều kiện : x\(\ge0\)
\(\Leftrightarrow\sqrt{2x+1}=x-1+\sqrt{3x}\)
\(\Leftrightarrow\left(\sqrt{2x+1}\right)^2=\left(x-1+\sqrt{3x}\right)^2\)
\(\Leftrightarrow2x+1=\left(x-1\right)^2+2\left(x-1\right)\sqrt{3x}+3x\)
\(\Leftrightarrow2x+1=x^2-2x+1+2\left(x-1\right)\sqrt{3x}+3x\)
\(\Leftrightarrow2x+1-x^2-x-x-2\left(x-1\right)\sqrt{3x}=0\)
\(\Leftrightarrow-x^2+x-2\left(x-1\right)\sqrt{3x}=0\)
\(\Leftrightarrow-x\left(x-1\right)-2\left(x-1\right)\sqrt{3x}=0\)
\(\Leftrightarrow\left(x-1\right)\left(-x-2\sqrt{3x}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\-x-2\sqrt[]{3x}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\-\sqrt{x}\left(\sqrt{x}+2\sqrt{3}\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\-\sqrt{x}=0\\\sqrt{x}+2\sqrt{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=0\\\sqrt{x}=-2\sqrt{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=0\\x\in\varnothing\end{matrix}\right.\) Vậy pt tập nghiệm S={1;0}
NV
Nguyễn Việt Lâm
Giáo viên
20 tháng 11 2018
\(\sqrt{1+2005^2+\dfrac{2005^2}{2006^2}}=\dfrac{1}{2006}\sqrt{2006^2+2005^2+\left(2005.2006\right)^2}\)
\(=\dfrac{1}{2006}\sqrt{\left(2006-2005\right)^2+2.2005.2006+\left(2005.2006\right)^2}\)
\(=\dfrac{1}{2006}\sqrt{1+2.2005.2006+\left(2005.2006\right)^2}\)
\(=\dfrac{1}{2006}\sqrt{\left(2005.2006+1\right)^2}=\dfrac{2005.2006+1}{2006}=2005+\dfrac{1}{2006}\)
Phương trình tương đương:
\(\sqrt{\left(x-1\right)^2}+\sqrt{\left(x-2\right)^2}=2005+\dfrac{1}{2006}+\dfrac{2005}{2006}\)
\(\Leftrightarrow\left|x-1\right|+\left|x-2\right|=2006\)
TH1: \(x\ge2\): \(x-1+x-2=2006\Rightarrow2x=2009\Rightarrow x=\dfrac{2009}{2}\)
TH2: \(x\le1\) : \(1-x+2-x=2006\Rightarrow-2x=2003\Rightarrow x=\dfrac{-2003}{2}\)
TH3: \(1< x< 2:\) \(x-1+2-x=2006\Rightarrow3=2006\) (vô nghiệm)
Vậy \(\left[{}\begin{matrix}x=\dfrac{2009}{2}\\x=\dfrac{-2003}{2}\end{matrix}\right.\)
\(\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}=\sqrt{2}\)
\(\Leftrightarrow x+\sqrt{2x-1}+x-\sqrt{2x-1}+2\sqrt{\left(x+\sqrt{2x-1}\right)\cdot\left(x-\sqrt{2x-1}\right)}=2\)
\(\Leftrightarrow2x+2\sqrt{x^2-2x+1}=2\Leftrightarrow2\left(x+\sqrt{\left(x-1\right)^2}\right)=2\Leftrightarrow x+|^{ }_{ }x-1|=1\)
\(\Leftrightarrow|^{ }_{ }x-1|^{ }_{ }=1-x\Leftrightarrow x-1< 0\Leftrightarrow x< 1\)
vậy x<1
mik hỏi bạn mik bảo x<10