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a) \(\frac{1}{x-1+\sqrt{x^2-2x+3}}+\frac{1}{x-1-\sqrt{x^2-2x+3}}=1\)
ĐKXĐ : \(x\inℝ\)
\(\Leftrightarrow\frac{x-1-\sqrt{x^2-2x+3}}{\left(x-1+\sqrt{x^2-2x+3}\right)\left(x-1-\sqrt{x^2-2x+3}\right)}+\frac{x-1+\sqrt{x^2-2x+3}}{\left(x-1+\sqrt{x^2-2x+3}\right)\left(x-1-\sqrt{x^2-2x+3}\right)}=\frac{\left(x-1+\sqrt{x^2-2x+3}\right)\left(x-1-\sqrt{x^2-2x+3}\right)}{\left(x-1+\sqrt{x^2-2x+3}\right)\left(x-1-\sqrt{x^2-2x+3}\right)}\)
\(\Rightarrow2x-2=\left[\left(x-1\right)+\left(\sqrt{x^2-2x+3}\right)\right]\left[\left(x-1\right)-\left(\sqrt{x^2-2x+3}\right)\right]\)
\(\Leftrightarrow2x-2=\left(x-1\right)^2-\left(\sqrt{x^2-2x+3}\right)^2\)
\(\Leftrightarrow2x-2=x^2-2x+1-\left(x^2-2x+3\right)\)
\(\Leftrightarrow2x-2=x^2-2x+1-x^2+2x-3\)
\(\Leftrightarrow2x-2=-2\)
\(\Leftrightarrow2x=0\)
\(\Leftrightarrow x=0\)
Vậy phương trình có nghiệm duy nhất x = 0
ĐKXĐ: \(x\ge-\dfrac{1}{2};x\ne0\)
\(\dfrac{1}{x^2}-\dfrac{1}{x}=\sqrt{2x+1}-\sqrt{x+2}\)
\(\Leftrightarrow-\dfrac{x-1}{x^2}=\dfrac{x-1}{\sqrt{2x+1}+\sqrt{x+2}}\)
\(\Leftrightarrow\left(x-1\right)\left(\dfrac{1}{\sqrt{2x+1}+\sqrt{x+2}}+\dfrac{1}{x^2}\right)=0\)
\(\Leftrightarrow x-1=0\) (do \(\dfrac{1}{\sqrt{2x+1}+\sqrt{x+2}}+\dfrac{1}{x^2}\) luôn dương)
\(\Leftrightarrow x=1\)
Đk: \(x\ge-\dfrac{1}{2},x\ne0\)
pt \(\Leftrightarrow\dfrac{1}{x^2}-\dfrac{1}{x}=\sqrt{2x+1}-\sqrt{x+2}\)
\(\Leftrightarrow\dfrac{1-x}{x^2}=\dfrac{2x+1-\left(x+2\right)}{\sqrt{2x+1}+\sqrt{x+2}}\)
\(\Leftrightarrow\dfrac{1-x}{x^2}=\dfrac{x-1}{\sqrt{2x+1}+\sqrt{x+2}}\)
\(\Leftrightarrow\left(x-1\right)\left(\dfrac{1}{\sqrt{2x+1}+\sqrt{x+2}}+\dfrac{1}{x^2}\right)=0\)
\(\Leftrightarrow x=1\) (vì \(\dfrac{1}{\sqrt{2x+1}+\sqrt{x+2}}+\dfrac{1}{x^2}>0\))
Vậy \(S=\left\{1\right\}\)