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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
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
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: \(\left\{{}\begin{matrix}x>0\\x< >1\end{matrix}\right.\)
\(\left(\dfrac{x-\sqrt{x}}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{x+\sqrt{x}}\right):\dfrac{\sqrt{x}+1}{x}\)
\(=\left(\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\cdot\dfrac{x}{\sqrt{x}+1}\)
\(=\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right)\cdot\dfrac{x}{\sqrt{x}+1}\)
\(=\dfrac{x-1}{\sqrt{x}}\cdot\dfrac{x}{\sqrt{x}+1}=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\cdot\sqrt{x}\cdot\sqrt{x}}{\sqrt{x}\cdot\left(\sqrt{x}+1\right)}\)
\(=\sqrt{x}\left(\sqrt{x}-1\right)\)
26 tháng 8 2021
\(A=\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}=x-\sqrt{xy}+y\)
\(B=\dfrac{\sqrt{x}-\sqrt{y}}{x\sqrt{x}-y\sqrt{y}}=\dfrac{1}{x+\sqrt{xy}+y}\)
\(C=\dfrac{3\sqrt{3}+x\sqrt{x}}{3-\sqrt{3x}+x}=\sqrt{x}+\sqrt{3}\)
\(D=\dfrac{x+\sqrt{5x}+5}{x\sqrt{x}-5\sqrt{5}}=\dfrac{1}{\sqrt{x}-\sqrt{5}}\)
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
\(\dfrac{x+2\sqrt{x}}{2\sqrt{x}+10}+\dfrac{\sqrt{x}-5}{\sqrt{x}}+\dfrac{50-5\sqrt{x}}{2\sqrt{x}\left(\sqrt{x}+5\right)}\left(đk:x>0\right)\)
\(=\dfrac{\sqrt{x}\left(x+2\sqrt{x}\right)+2\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)+50-5\sqrt{x}}{2\sqrt{x}\left(\sqrt{x}+5\right)}\)
\(=\dfrac{x\sqrt{x}+2x+2x-50+50-5\sqrt{x}}{2\sqrt{x}\left(\sqrt{x}+5\right)}=\dfrac{x\sqrt{x}+4x-5\sqrt{x}}{2\sqrt{x}\left(\sqrt{x}+5\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+5\right)}{2\sqrt{x}\left(\sqrt{x}+5\right)}=\dfrac{\sqrt{x}-1}{2}\)
27 tháng 5 2021
1,\(K=\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{x}}\)
\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{6-2\sqrt{5}}+\sqrt{6+2\sqrt{5}}\right)\)\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{\left(\sqrt{5}-1\right)^2}+\sqrt{\left(\sqrt{5}+1\right)^2}\right)\)
\(=\dfrac{1}{\sqrt{2}}\left(\left|\sqrt{5}-1\right|+\sqrt{5}+1\right)\)\(=\dfrac{1}{\sqrt{2}}\left|\sqrt{5}-1+\sqrt{5}+1\right|=\dfrac{1}{\sqrt{2}}.2\sqrt{5}\)\(=\sqrt{10}\)
2, \(\sqrt{x-3}-2\sqrt{x^2-3x}=0\left(đk:x\ge3\right)\)
\(\Leftrightarrow\sqrt{x-3}\left(1-2\sqrt{x}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-3}=0\\1-2\sqrt{x}=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=\left(\dfrac{1}{2}\right)^2=\dfrac{1}{4}\left(ktm\right)\end{matrix}\right.\)
Vậy pt có nghiệm x=3
3, \(\dfrac{9x-7}{\sqrt{7x+5}}=\sqrt{7x+5}\left(đk:x>-\dfrac{5}{7}\right)\)
\(\Leftrightarrow9x-7=7x+5\)
\(\Leftrightarrow x=6\left(tm\right)\)
4, \(x-5\sqrt{x}+4=0\)(đk: \(x\ge0\))
\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(\sqrt{x}-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=1\\\sqrt{x}=4\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=16\end{matrix}\right.\) (tm)
Vậy...
27 tháng 5 2021
1) Bạn tự làm
2) ĐK: \(x\ge3\)
PT \(\Leftrightarrow\sqrt{x-3}\left(1-2\sqrt{x}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-3}=0\\2\sqrt{x}=1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{4}\left(loại\right)\end{matrix}\right.\)
Vậy ...
3) ĐK: \(x>-\dfrac{5}{7}\)
PT \(\Rightarrow9x-7=7x+5\) \(\Leftrightarrow x=6\)
Vậy ...
4) ĐK: \(x\ge0\)
PT \(\Leftrightarrow x-4\sqrt{x}-\sqrt{x}+4=0\)
\(\Leftrightarrow\left(\sqrt{x}-4\right)\left(\sqrt{x}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=4\\\sqrt{x}=1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=16\\x=1\end{matrix}\right.\)
Vậy ...
vì x+x=0 => x = 2,689994048
sao lại tek