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1) đặt đk rùi bình phương 2 vế là ok
2) \(pt\Leftrightarrow\frac{\sqrt{x}-\sqrt{x+2}}{x-x-2}+\frac{\sqrt{x+2}-\sqrt{x+4}}{x+2-x-4}+\frac{\sqrt{x+4}-\sqrt{x+6}}{x+4-x-6}=\frac{\sqrt{10}}{2}-1\)(ĐKXĐ : \(x\ge0\))
<=> \(\frac{\sqrt{x}-\sqrt{x+6}}{-2}=\frac{\sqrt{10}}{2}-1\)
<=> \(\frac{\sqrt{x+6}-\sqrt{x}}{2}=\frac{\sqrt{10}-2}{2}\)
<=> \(\sqrt{x+6}-\sqrt{x}=\sqrt{10}-2\)
<=> \(\sqrt{x+6}+2=\sqrt{10}+\sqrt{x}\)
đến đây bình phương 2 vế rùi giải bình thường nhé
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
\(\frac{4}{x}+\sqrt{x-\frac{1}{x}}=x+\sqrt{2x-\frac{5}{x}}\)
\(\Leftrightarrow\sqrt{x-\frac{1}{x}}-\sqrt{2x-\frac{5}{x}}=x-\frac{4}{x}\)
\(\Leftrightarrow\frac{\frac{4}{x}-x}{\sqrt{x-\frac{1}{x}}+\sqrt{2x-\frac{5}{x}}}=x-\frac{4}{x}\)
\(\Leftrightarrow\left(\frac{4}{x}-x\right).\left(\frac{1}{\sqrt{x-\frac{1}{x}}+\sqrt{2x-\frac{5}{x}}}+1\right)=0\)
\(\frac{1}{\sqrt{x-\frac{1}{x}}+\sqrt{2x-\frac{5}{x}}}+1>0\Rightarrow\frac{4}{x}-x=0\Rightarrow x=2;x=-2\)
Thử lại, ta có nghiệm \(x=2\) thỏa mãn.
Vậy, \(x=2\).
a: \(\Leftrightarrow\left\{{}\begin{matrix}x^2-4x+4=x^2-1\\x>=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-4x=-5\\x>=2\end{matrix}\right.\Leftrightarrow x=\dfrac{5}{4}\left(loại\right)\)
b: \(\Leftrightarrow\sqrt{2x^2+1}=5\)
\(\Leftrightarrow2x^2+1=25\)
\(\Leftrightarrow2x^2=24\)
hay \(x\in\left\{2\sqrt{3};-2\sqrt{3}\right\}\)
c: \(\Leftrightarrow\left|x\right|+\left|x-1\right|=2\)
Trường hợp 1: x<0
Pt trở thành -x-x+1=2
=>-2x=1
hay x=-1/2(nhận)
TRường hợp 2:0<=x<1
Pt trở thành x+1-x=2
=>1=2(loại)
Trường hợp 3: x>=1
Pt trở thành x+x-1=2
=>2x-1=2
hay x=3/2(nhận)
\(x-\frac{1}{x}+\sqrt{x-\frac{1}{x}}+\frac{1}{4}=2x-\frac{5}{x}+\sqrt{2x-\frac{5}{x}}+\frac{1}{4}\)
=>\(\left(\sqrt{x-\frac{1}{x}}+\frac{1}{2}\right)^2=\left(\sqrt{2x-\frac{5}{x}}+\frac{1}{2}\right)^2\)
dễ suy ra đc:\(\sqrt{x-\frac{1}{x}}=\sqrt{2x-\frac{5}{x}}\)
từ đây=>x=?