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\(\sqrt{\frac{x+1}{x^2-x+1}}+\sqrt{\frac{x^2-x+1}{x+1}}=2\)
\(\Leftrightarrow\left(\sqrt{\frac{x+1}{x^2-x+1}}+\sqrt{\frac{x^2-x+1}{x+1}}\right)^2=2^2\)
\(\Leftrightarrow\frac{x+1}{x^2-x+1}+2.\sqrt{\frac{x+1}{x^2-x+1}}.\sqrt{\frac{x^2-x+1}{x+1}}+\frac{x^2-x+1}{x+1}=4\)
\(\Leftrightarrow\frac{x+1}{x^2-x+1}+\frac{x^2-x+1}{x+1}=4-2\)
\(\Leftrightarrow\frac{\left(x+1\right)^2}{\left(x^2-x+1\right)\left(x+1\right)}+\frac{\left(x^2-x+1\right)^2}{\left(x+1\right)\left(x^2-x+1\right)}=2\)
\(\Leftrightarrow\frac{\left(x+1\right)^2+\left(x^2+1-x\right)^2}{x^3+1}=2\)
\(\Leftrightarrow\frac{x^2+2x+1+x^4+1+x^2+2x^2-2x-2x^3}{x^3+1}=2\)
\(\Leftrightarrow x^2+2x+1+x^4+1+x^2+2x^2-2x-2x^3=2\left(x^3+1\right)\)
\(\Leftrightarrow4x^2+2+x^4-2x^3=2x^3+2\)
\(\Leftrightarrow x^4-2x^3-2x^3+4x^2=2-2\)
\(\Leftrightarrow x^4-4x^3+4x^2=0\)
\(\Leftrightarrow x^2\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2=0\\x^2-4x+4=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\\left(x-2\right)^2=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
Bạn tự tìm ĐKXĐ rồi so sánh kết quả nhé!
\(\sqrt{a}+2\sqrt{a}-1+\sqrt{a}-2\sqrt{a}-1\)
\(=2\sqrt{a}-2\)
vậy thôi à??
\(\frac{2}{\sqrt{7}-\sqrt{5}}+\frac{6}{\sqrt{11}+\sqrt{5}}\)
\(=\frac{2\left(\sqrt{7}+\sqrt{5}\right)}{\left(\sqrt{7}-\sqrt{5}\right)\left(\sqrt{7}+\sqrt{5}\right)}+\frac{6\left(\sqrt{11}-\sqrt{5}\right)}{\left(\sqrt{11}+\sqrt{5}\right)\left(\sqrt{11}-\sqrt{5}\right)}\)
\(=\frac{2\left(\sqrt{7}+\sqrt{5}\right)}{7-5}+\frac{6\left(\sqrt{11}-\sqrt{5}\right)}{11-5}\)
\(=\frac{2\left(\sqrt{7}+\sqrt{5}\right)}{2}+\frac{6\left(\sqrt{11}-\sqrt{5}\right)}{6}\)
\(=\sqrt{7}+\sqrt{5}+\sqrt{11}-\sqrt{5}\)
\(=\sqrt{7}+\sqrt{11}\)
\(A=\frac{1}{\sqrt{3}+1}+\frac{1}{\sqrt{3}-1}\)
\(=\frac{\sqrt{3}-1}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}+\frac{\sqrt{3}+1}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)
\(=\frac{\sqrt{3}-1+\sqrt{3}+1}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\)
\(=\frac{2\sqrt{3}}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\)
\(=\frac{2\sqrt{3}}{3-1}\)
\(=\frac{2\sqrt{3}}{2}\)
\(=\sqrt{3}\)
\(B=\frac{5+\sqrt{5}}{5-\sqrt{5}}+\frac{5-\sqrt{5}}{5+\sqrt{5}}\)
\(=\frac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{5}\left(\sqrt{5}-1\right)}+\frac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}\left(\sqrt{5}+1\right)}\)
\(=\frac{\left(\sqrt{5}+1\right)}{\left(\sqrt{5}-1\right)}+\frac{\left(\sqrt{5}-1\right)}{\left(\sqrt{5}+1\right)}\)
\(=\frac{\left(\sqrt{5}+1\right)^2}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}+\frac{\left(\sqrt{5}-1\right)^2}{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}\)
\(=\frac{5+2\sqrt{5}+1+5-2\sqrt{5}+1}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}\)
\(=\frac{12}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}\)
\(=\frac{12}{5-1}\)
\(=\frac{12}{4}\)
\(=3\)
\(\text{a)}x\sqrt{x}+\sqrt{x}-x-1\)
\(=\left(x\sqrt{x}+\sqrt{x}\right)-\left(x+1\right)\)
\(=\sqrt{x}\left(x+1\right)-\left(x+1\right)\)
\(=\left(x+1\right)\left(\sqrt{x}-1\right)\)
\(\text{b)}\sqrt{ab}+2\sqrt{a}+3\sqrt{b}+6\)
\(=\left(\sqrt{ab}+2\sqrt{a}\right)+\left(3\sqrt{b}+6\right)\)
\(=\sqrt{a}\left(\sqrt{b}+2\right)+3\left(\sqrt{b}+2\right)\)
\(=\left(\sqrt{b}+2\right)\left(\sqrt{a}+3\right)\)
\(\text{c)}\left(1+\sqrt{x}\right)^2-4\sqrt{x}\)
\(=\left(1+\sqrt{x}\right)^2-\left(2\sqrt{\sqrt{x}}\right)^2\)
\(=\left(1+\sqrt{x}+2\sqrt{\sqrt{x}}\right)\left(1+\sqrt{x}-2\sqrt{\sqrt{x}}\right)\)
\(\text{d)}\sqrt{ab}-\sqrt{a}-\sqrt{b}+1\)
\(=\left(\sqrt{ab}-\sqrt{a}\right)-\left(\sqrt{b}-1\right)\)
\(=\sqrt{a}\left(\sqrt{b}-1\right)-\left(\sqrt{b}-1\right)\)
\(=\left(\sqrt{b}-1\right)\left(\sqrt{a}-1\right)\)
\(\text{e)}a+\sqrt{a}+2\sqrt{ab}+2\sqrt{b}\)
\(=\left(a+\sqrt{a}\right)+\left(2\sqrt{ab}+2\sqrt{b}\right)\)
\(=\left[\left(\sqrt{a}\right)^2+\sqrt{a}\right]+\left(2\sqrt{ab}+2\sqrt{b}\right)\)
\(=\sqrt{a}\left(\sqrt{a}+1\right)+2\sqrt{b}\left(\sqrt{a}+1\right)\)
\(=\left(\sqrt{a}+1\right)\left(\sqrt{a}+2\sqrt{b}\right)\)
\(\text{f)}x-2\sqrt{x-1}-a^2\)
\(=\left(\sqrt{x-2}\right)^2\left(\sqrt{\sqrt{x-1}}\right)^2-a^2\)
\(=\left(\sqrt{x-2}\sqrt{\sqrt{x-1}}\right)^2-a^2\)
\(=\left(\sqrt{x-2\sqrt{x-1}}\right)^2-a^2\)
\(=\left(\sqrt{x-2\sqrt{x-1}}+a\right)\left(\sqrt{x-2\sqrt{x-1}}-a\right)\)