Rút gọn biểu thức:
\(A=\frac{\sqrt{x+2\sqrt{x-1}+\sqrt{x-2\sqrt{x-1}}}}{\sqrt{x+\sqrt{2x-1}+\sqrt{x-\sqrt{2x-1}}}}.\sqrt{2x-1}\)
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29 tháng 7 2017
\(\frac{A}{\sqrt{2}}\)=\(\frac{\sqrt{x-1+2\sqrt{x-1}+1}+\sqrt{x-1-2\sqrt{x-1}+1}}{\sqrt{2x-1+2\sqrt{2x-1}+1}-\sqrt{2x-1-2\sqrt{2x-1}+1}}\) (DK \(x\ge1\)
\(=\frac{\sqrt{x-1}+1+\left|\sqrt{x-1}-1\right|}{\sqrt{2x-1}+1-\left|\sqrt{2x-1}-1\right|}\)
vs \(x\ge2\) \(\frac{\sqrt{x-1}+1+\sqrt{x-1}-1}{\sqrt{2x-1}+1-\sqrt{2x-1}+1}=\frac{2\sqrt{x-1}}{2}=\sqrt{x-1}\) \(\Rightarrow A=\sqrt{2x-2}\)
vs \(1\le x< 2\) \(\frac{\sqrt{x-1}+1+1-\sqrt{x-1}}{\sqrt{2x-1}+1-1+\sqrt{2x-1}}=\frac{1}{\sqrt{2x-1}}\) \(\Rightarrow A=\frac{\sqrt{2}}{\sqrt{2x-1}}\)
29 tháng 7 2017
\(\sqrt{2X-1}\ge1\Leftrightarrow X\ge1\)NEN SUY RA THEO CACH LAM CUA TO
THOI U AM BUSY SEE YOU AGAIN
8 tháng 11 2021
\(A=\left(a-1\right)\sqrt{\frac{a}{a-1}}+\sqrt{a\left(a-1\right)}-a\sqrt{\frac{a-1}{a}}\)
\(A=\sqrt{\left(a-1\right)^2.\frac{a}{a-1}}+\sqrt{a\left(a-1\right)}-\sqrt{a^2.\frac{a-1}{a}}\)
\(A=\sqrt{\left(a-1\right)a}+\sqrt{a\left(a-1\right)}-\sqrt{a\left(a-1\right)}\)
\(A=\sqrt{a\left(a-1\right)}\)
25 tháng 8 2020
ĐKXĐ: \(x\ge1\); x khác 2; 3
Ta có:
\(\frac{1}{\sqrt{x}-\sqrt{x-1}}=\frac{\sqrt{x}+\sqrt{x-1}}{x-\left(x-1\right)}=\sqrt{x}+\sqrt{x-1}\)
\(\frac{x-3}{\sqrt{x-1}-\sqrt{2}}=\frac{\left(x-3\right)\left(\sqrt{x-1}+\sqrt{2}\right)}{x-1-2}=\sqrt{x-1}+\sqrt{2}\)
=> \(\frac{1}{\sqrt{x}-\sqrt{x-1}}-\frac{x-3}{\sqrt{x-1}-\sqrt{2}}=\sqrt{x}+\sqrt{x-1}-\left(\sqrt{x-1}+\sqrt{2}\right)=\sqrt{x}-\sqrt{2}\)
\(\frac{2}{\sqrt{2}-\sqrt{x}}-\frac{\sqrt{x}+\sqrt{2}}{\sqrt{2x}-x}=\frac{2\sqrt{x}-\sqrt{x}-2}{\sqrt{x}\left(\sqrt{2}-\sqrt{x}\right)}=\frac{\sqrt{x}-2}{\sqrt{x}\left(\sqrt{2}-\sqrt{x}\right)}\)
=> \(P=\left(\sqrt{x}-\sqrt{2}\right).\frac{\sqrt{x}-2}{\sqrt{x}\left(\sqrt{2}-\sqrt{x}\right)}=\frac{2-\sqrt{x}}{\sqrt{x}}\)