Cho \(\sqrt{x}+\sqrt{y}-\sqrt{z}=0\) 0.
CMR: \(\frac{1}{x+y-z}-\frac{1}{y+z-x}+\frac{1}{z+x-y}=\)0.
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\(\frac{1}{\sqrt{k-1}+\sqrt{k}}>\frac{1}{\sqrt{k}+\sqrt{k}}=\frac{1}{2\sqrt{k}}\)
\(\Leftrightarrow2x+1+x-3+2\sqrt{\left(2x+1\right)\left(x-3\right)}=4\)
\(\Leftrightarrow2\sqrt{2x^2-5x-3}=6-3x\)
\(\Leftrightarrow8x^2-20x-12=9x^2-36x+36\)
\(\Leftrightarrow x^2-16x+48=0\)
\(\Leftrightarrow x=4;12\)
\(\sqrt{2x+1}+\sqrt{x-3}=4\)
ĐK x >= 3
\(\Leftrightarrow2x+1+x-3+2\sqrt{\left(2x+1\right)\left(x-3\right)}=16\)
\(\Leftrightarrow2\sqrt{\left(2x+1\right)\left(x-3\right)}=18-3x\)
ĐK \(x\le6\)
\(\Leftrightarrow4\left(2x^2-5x-3\right)=9\left(36-12x+x^2\right)\)
\(\Leftrightarrow8x^2-20x-12=324-108x+9x^2\)
\(\Leftrightarrow x^2-88x+336=0\)
\(\Leftrightarrow x^2-4x-84x+336=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-84\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=4\left(n\right)\\x=84\left(l\right)\end{cases}}\)vậy \(S=\left\{4\right\}\)
\(2x^2+x+\sqrt{x^2+3}+2x\sqrt{x^2+3}=9\)
\(\Leftrightarrow2x^2+x-3+\left(\sqrt{x^2+3}-2\right)+\left(2x\sqrt{x^2+3}-4\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x+3\right)+\frac{x^2+3-4}{\sqrt{x^2+3}+2}+\frac{4x\left(x^2+3\right)-16}{2x\sqrt{x^2+3}+4}=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x+3\right)+\frac{x^2-1}{\sqrt{x^2+3}+2}+\frac{4x^3+12x-16}{2x\sqrt{x^2+3}+4}=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x+3\right)+\frac{\left(x-1\right)\left(x+1\right)}{\sqrt{x^2+3}+2}+\frac{4\left(x-1\right)\left(x^2+x+4\right)}{2x\sqrt{x^2+3}+4}=0\)
\(\Leftrightarrow\left(x-1\right)\left(\left(2x+3\right)+\frac{\left(x+1\right)}{\sqrt{x^2+3}+2}+\frac{4\left(x^2+x+4\right)}{2x\sqrt{x^2+3}+4}\right)=0\)
Dễ thấy: \(\left(2x+3\right)+\frac{\left(x+1\right)}{\sqrt{x^2+3}+2}+\frac{4\left(x^2+x+4\right)}{2x\sqrt{x^2+3}+4}>0\)
Nên x-1=0 suy ra x=1
\(\sqrt{2x^2+x+9}+\sqrt{2x^2-x+1}=x+4\)
\(\Leftrightarrow\sqrt{2x^2+x+9}-\left(\frac{1}{2}x+3\right)+\sqrt{2x^2-x+1}-\left(\frac{1}{2}x+1\right)=0\)
\(\Leftrightarrow\frac{2x^2+x+9-\left(\frac{1}{2}x+3\right)^2}{\sqrt{2x^2+x+9}+\frac{1}{2}x+3}+\frac{2x^2-x+1-\left(\frac{1}{2}x+1\right)^2}{\sqrt{2x^2-x+1}+\frac{1}{2}x+1}=0\)
\(\Leftrightarrow\frac{\frac{1}{4}x\left(7x-8\right)}{\sqrt{2x^2+x+9}+\frac{1}{2}x+3}+\frac{\frac{1}{4}x\left(7x-8\right)}{\sqrt{2x^2-x+1}+\frac{1}{2}x+1}=0\)
\(\Leftrightarrow\frac{1}{4}x\left(7x-8\right)\left(\frac{1}{\sqrt{2x^2+x+9}+\frac{1}{2}x+3}+\frac{1}{\sqrt{2x^2-x+1}+\frac{1}{2}x+1}\right)=0\)
Dễ thấy: \(\frac{1}{\sqrt{2x^2+x+9}+\frac{1}{2}x+3}+\frac{1}{\sqrt{2x^2-x+1}+\frac{1}{2}x+1}>0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\7x-8=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{8}{7}\end{cases}}\)
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