\(\sqrt{x-\sqrt{x-\sqrt{x-\sqrt{x-5}}}}=5\)
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9 tháng 9 2021
\(\dfrac{x+\sqrt{5}}{\sqrt{x}+\sqrt{x+\sqrt{5}}}+\dfrac{x-\sqrt{5}}{\sqrt{x}-\sqrt{x-\sqrt{5}}}\)
\(=\dfrac{\left(x+\sqrt{5}\right)\cdot\left(\sqrt{x}-\sqrt{x+\sqrt{5}}\right)}{x-x-\sqrt{5}}+\dfrac{\left(x-\sqrt{5}\right)\left(\sqrt{x}+\sqrt{x-\sqrt{5}}\right)}{x-x+\sqrt{5}}\)
\(=\dfrac{\left(x+\sqrt{5}\right)\left(\sqrt{x}-\sqrt{x+\sqrt{5}}\right)+\left(-x+\sqrt{5}\right)\left(\sqrt{x}+\sqrt{x-\sqrt{5}}\right)}{\sqrt{5}}\)
\(=\dfrac{\left(3+\sqrt{5}\right)\left(\sqrt{3}-\sqrt{3+\sqrt{5}}\right)-\left(3-\sqrt{5}\right)\left(\sqrt{3}+\sqrt{3-\sqrt{5}}\right)}{\sqrt{5}}\)
\(=\dfrac{\left(6+2\sqrt{5}\right)\left(\sqrt{6}-\sqrt{6+2\sqrt{5}}\right)-\left(6-2\sqrt{5}\right)\left(\sqrt{6}+\sqrt{6-2\sqrt{5}}\right)}{\sqrt{5}}\)
\(=\dfrac{\left(6+2\sqrt{5}\right)\left(\sqrt{6}-\sqrt{5}-1\right)-\left(6-2\sqrt{5}\right)\left(\sqrt{6}+\sqrt{5}-1\right)}{\sqrt{5}}\)
\(=\dfrac{-12\sqrt{5}+4\sqrt{30}}{\sqrt{5}}\)
\(=-12+4\sqrt{6}\)
BC
9 tháng 9 2021
Nguyễn Lê Phước Thịnh CTV, sai rồi bn ơi. Mk thay vào không bằng nhau
26 tháng 8 2021
\(A=\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}=x-\sqrt{xy}+y\)
\(B=\dfrac{\sqrt{x}-\sqrt{y}}{x\sqrt{x}-y\sqrt{y}}=\dfrac{1}{x+\sqrt{xy}+y}\)
\(C=\dfrac{3\sqrt{3}+x\sqrt{x}}{3-\sqrt{3x}+x}=\sqrt{x}+\sqrt{3}\)
\(D=\dfrac{x+\sqrt{5x}+5}{x\sqrt{x}-5\sqrt{5}}=\dfrac{1}{\sqrt{x}-\sqrt{5}}\)
9 tháng 12 2023
a: \(\dfrac{6}{2-\sqrt{10}}-\dfrac{2\sqrt{5}-5\sqrt{2}}{\sqrt{2}-\sqrt{5}}+\sqrt{49+4\sqrt{10}}\)
\(=\dfrac{6\left(2+\sqrt{10}\right)}{4-10}-\dfrac{\sqrt{10}\left(\sqrt{2}-\sqrt{5}\right)}{\sqrt{2}-\sqrt{5}}+\sqrt{49+2\cdot2\sqrt{10}}\)
\(=\dfrac{6\left(2+\sqrt{10}\right)}{-6}-\sqrt{10}+\sqrt{49+2\cdot\sqrt{40}}\)
\(=-2-\sqrt{10}-\sqrt{10}+\sqrt{49+4\sqrt{10}}\)
\(=-2-2\sqrt{10}+\sqrt{49+4\sqrt{10}}\)
b: ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x< >1\end{matrix}\right.\)
\(\left(\dfrac{x-\sqrt{x}}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{x+\sqrt{x}}\right):\dfrac{\sqrt{x}+1}{x}\)
\(=\left(\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\cdot\dfrac{x}{\sqrt{x}+1}\)
\(=\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right)\cdot\dfrac{x}{\sqrt{x}+1}\)
\(=\dfrac{x-1}{\sqrt{x}}\cdot\dfrac{x}{\sqrt{x}+1}=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\cdot\sqrt{x}\cdot\sqrt{x}}{\sqrt{x}\cdot\left(\sqrt{x}+1\right)}\)
\(=\sqrt{x}\left(\sqrt{x}-1\right)\)
G
10 tháng 8 2023
a) Đk: \(x>0;x\ne9;x\ne25\)
Đặt \(A=\left(\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{2x}{9-x}\right):\left(\dfrac{\sqrt{x}-1}{x-3\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\)
\(=\left[\dfrac{\sqrt{x}\left(3-\sqrt{x}\right)}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}+\dfrac{2x}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\right]\)\(:\left[\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\dfrac{2\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-3\right)}\right]\)
\(=\dfrac{\sqrt{x}\left(3-\sqrt{x}\right)+2x}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}:\dfrac{\sqrt{x}-1-2\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\dfrac{3\sqrt{x}+x}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}:\dfrac{-\sqrt{x}+5}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\dfrac{\sqrt{x}\left(3+\sqrt{x}\right)}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}.\dfrac{\sqrt{x}\left(3-\sqrt{x}\right)}{\sqrt{x}-5}\)
\(=\dfrac{x}{\sqrt{x}-5}\)
G
10 tháng 8 2023
b) Đk: \(x\ge0;x\ne1;x\ne25\)
Biểu thức
\(=\left[\dfrac{\sqrt{x}-2}{\sqrt{x}+5}+\dfrac{\sqrt{x}}{\sqrt{x}-5}-\dfrac{x+9}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\right]:\dfrac{1-\sqrt{x}}{5+\sqrt{x}}\)
\(=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-5\right)+\sqrt{x}\left(\sqrt{x}+5\right)-x-9}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}.\dfrac{\sqrt{x}+5}{1-\sqrt{x}}\)
\(=\dfrac{x-7\sqrt{x}+10+x+5\sqrt{x}-x-9}{\left(\sqrt{x}-5\right)\left(1-\sqrt{x}\right)}\)
\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-5\right)\left(1-\sqrt{x}\right)}\)\(=\dfrac{\left(1-\sqrt{x}\right)^2}{\left(\sqrt{x}-5\right)\left(1-\sqrt{x}\right)}=\dfrac{1-\sqrt{x}}{\sqrt{x}-5}\)
NV
Nguyễn Việt Lâm
Giáo viên
26 tháng 9 2019
\(g\left(3\right)=\frac{3+\sqrt{5}}{\sqrt{3}+\sqrt{3+\sqrt{5}}}+\frac{3-\sqrt{5}}{\sqrt{3}-\sqrt{3-\sqrt{5}}}\)
\(=\frac{\left(3+\sqrt{5}\right)\left(\sqrt{3}+\sqrt{3+\sqrt{5}}\right)}{-\sqrt{5}}+\frac{\left(3-\sqrt{5}\right)\left(\sqrt{3}-\sqrt{3-\sqrt{5}}\right)}{\sqrt{5}}\)
\(=\frac{\left(3+\sqrt{5}\right)\left(\sqrt{6}+\sqrt{6+2\sqrt{5}}\right)}{-\sqrt{10}}+\frac{\left(3-\sqrt{5}\right)\left(\sqrt{6}-\sqrt{6-2\sqrt{5}}\right)}{\sqrt{10}}\)
\(=\frac{\left(3+\sqrt{5}\right)\left(\sqrt{6}+\sqrt{5}+1\right)}{-\sqrt{10}}+\frac{\left(3-\sqrt{5}\right)\left(\sqrt{6}-\sqrt{5}+1\right)}{\sqrt{10}}\)
\(=\frac{3\sqrt{6}-4\sqrt{5}-\sqrt{30}+8}{\sqrt{10}}-\frac{3\sqrt{6}+4\sqrt{5}+\sqrt{30}+8}{\sqrt{10}}\)
\(=\frac{-8\sqrt{5}-2\sqrt{30}}{\sqrt{10}}=\frac{-8-2\sqrt{6}}{\sqrt{2}}=-4\sqrt{2}-2\sqrt{3}\)
Bạn kiểm tra lại
16 tháng 9 2021
\(=\dfrac{\sqrt{x}\left(x+2\sqrt{x}\right)+2\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)+50-5\sqrt{x}}{2\sqrt{x}\left(\sqrt{x}+5\right)}\\ =\dfrac{x\sqrt{x}+2x+2x-50+50-5\sqrt{x}}{2\sqrt{x}\left(\sqrt{x}+5\right)}\\ =\dfrac{x\sqrt{x}-5\sqrt{x}+4x}{2\sqrt{x}\left(\sqrt{x}+5\right)}=\dfrac{\sqrt{x}\left(x+4\sqrt{x}-5\right)}{2\sqrt{x}\left(\sqrt{x}+5\right)}\\ =\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+5\right)}{2\sqrt{x}\left(\sqrt{x}+5\right)}=\dfrac{\sqrt{x}-1}{2}\)
16 tháng 9 2021
\(\dfrac{x+2\sqrt{x}}{2\sqrt{x}+10}+\dfrac{\sqrt{x}-5}{\sqrt{x}}+\dfrac{50-5\sqrt{x}}{2\sqrt{x}\left(\sqrt{x}+5\right)}\left(đk:x>0\right)\)
\(=\dfrac{\sqrt{x}\left(x+2\sqrt{x}\right)+2\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)+50-5\sqrt{x}}{2\sqrt{x}\left(\sqrt{x}+5\right)}\)
\(=\dfrac{x\sqrt{x}+2x+2x-50+50-5\sqrt{x}}{2\sqrt{x}\left(\sqrt{x}+5\right)}=\dfrac{x\sqrt{x}+4x-5\sqrt{x}}{2\sqrt{x}\left(\sqrt{x}+5\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+5\right)}{2\sqrt{x}\left(\sqrt{x}+5\right)}=\dfrac{\sqrt{x}-1}{2}\)