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a) \(\sqrt{17}-4\) b) \(\sqrt{3}\) c) \(\frac{\sqrt{2}}{2}\) d)\(\frac{\sqrt{x}+1}{\sqrt{x}-1}\) e) \(x-\sqrt{5}\)
f) \(4+2\sqrt{3}\) g) \(3+2\sqrt{2}\) h) \(x+\sqrt{x}+1\) i) \(\frac{3\sqrt{5}-\sqrt{15}}{10}\)
k) \(\sqrt{5}+\sqrt{6}\) i) 5 h) 0 l) \(\sqrt{5}+\sqrt{3}\) m) \(\frac{20\sqrt{3}}{3}\) d) 0
a, \(\frac{1}{\left(\sqrt{3}+\sqrt{2}\right)^2}\) +\(\frac{1}{\left(\sqrt{3}-\sqrt{2}\right)^2}\) =\(\frac{\left(\sqrt{3}+\sqrt{2}\right)^2+\left(\sqrt{3}-\sqrt{2}\right)^2}{\left(\sqrt{3}+\sqrt{2}\right)^2\left(\sqrt{3}-\sqrt{2}\right)^2}\)
\(=\frac{10}{1}=10\)
mấy câu còn lại bạn tự làm nốt nhé mk ban rồi
c. \(\frac{2}{\sqrt{5}+\sqrt{3}}-\frac{3-\sqrt{15}}{\sqrt{5}-\sqrt{3}}\)
= \(\frac{2\left(\sqrt{5}-\sqrt{3}\right)}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}-\frac{\left(3-\sqrt{15}\right)\left(\sqrt{5}+\sqrt{3}\right)}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}\)
= \(\frac{2\sqrt{5}-2\sqrt{3}}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}-\frac{3\sqrt{5}+3\sqrt{3}-5\sqrt{3}+3\sqrt{5}}{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}\)
= \(\frac{2\sqrt{5}-2\sqrt{3}-3\sqrt{5}+3\sqrt{3}-5\sqrt{3}+3\sqrt{5}}{5-3}\)
= \(\frac{2\sqrt{5}-2\sqrt{3}-2\sqrt{3}}{2}\)
= \(\frac{2\sqrt{5}-4\sqrt{3}}{2}\)
mk chỉ bik cách lm như z thoy còn kết quả thì mk chưa chắc đã đúng đâu nên pn xem lại nhá
\(\frac{1}{\sqrt{5}-1}+\frac{1}{1+\sqrt{5}}\)
= \(\frac{1}{\sqrt{5}-1}-\frac{1}{\sqrt{5}+1}\)
= \(\frac{\sqrt{5}+1}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}-\frac{\sqrt{5}-1}{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}\)
= \(\frac{\sqrt{5}+1}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}-\frac{\sqrt{5}+1}{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}\)
= \(\frac{2}{5-1}\)
= \(\frac{2}{4}\)
= \(\frac{1}{2}\)
những ai thích xem minecraft và blockman go thì hãy xem kênh youtube của mik kênh mik là M.ichibi các bn nhớ sud và chia sẻ cho nhiều người khác nhé
a) Ta có: \(\left(\sqrt{8}-3\sqrt{2}+\sqrt{10}\right)\sqrt{2}-\sqrt{5}\)
\(=\left(-\sqrt{2}+\sqrt{10}\right)\sqrt{2}-\sqrt{5}\)
\(=-2+2\sqrt{5}-\sqrt{5}\)
\(=-2+\sqrt{5}\)
b) \(\left(\frac{1}{2}\sqrt{\frac{1}{2}}-\frac{3}{2}\sqrt{2}+\frac{4}{5}\sqrt{200}\right)\div\frac{1}{8}\)
\(=\left(\frac{\sqrt{2}}{4}-\frac{3\sqrt{2}}{2}+8\sqrt{2}\right)\cdot8\)
\(=\frac{27\sqrt{2}}{4}\cdot8\)
\(=54\sqrt{2}\)
5/
Đặt \(\left\{{}\begin{matrix}\sqrt{2x-\frac{3}{x}}=a\ge0\\\sqrt{\frac{6}{x}-2x}=b\ge0\end{matrix}\right.\) \(\Rightarrow a^2+b^2=\frac{3}{x}\)
Pt trở thành:
\(a-1=\frac{a^2+b^2}{2}-b\)
\(\Leftrightarrow a^2+b^2-2a-2b+2=0\)
\(\Leftrightarrow\left(a^2-2a+1\right)+\left(b^2-2b+1\right)=0\)
\(\Leftrightarrow\left(a-1\right)^2+\left(b-1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{2x-\frac{3}{x}}=1\\\sqrt{\frac{6}{x}-2x}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x^2-x-3=0\\2x^2+x-6=0\end{matrix}\right.\) \(\Rightarrow x=\frac{3}{2}\)
4/
ĐKXĐ: \(x\ge\frac{1}{5}\)
\(\Leftrightarrow\frac{4x-3}{\sqrt{5x-1}+\sqrt{x+2}}=\frac{4x-3}{5}\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-3=0\Rightarrow x=\frac{3}{4}\\\sqrt{5x-1}+\sqrt{x+2}=5\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\sqrt{5x-1}-3+\sqrt{x+2}-2=0\)
\(\Leftrightarrow\frac{5\left(x-2\right)}{\sqrt{5x-1}+3}+\frac{x-2}{\sqrt{x+2}+2}=0\)
\(\Leftrightarrow\left(x-2\right)\left(\frac{5}{\sqrt{5x-1}+3}+\frac{1}{\sqrt{x+2}+2}\right)=0\)
\(\Leftrightarrow x=2\)
\(\left(\frac{2}{\sqrt{3}-1}+\frac{3}{\sqrt{3}-2}+\frac{15}{3-\sqrt{3}}\right).\frac{1}{\sqrt{3}+5}\)
\(=\frac{1}{5+\sqrt{3}}.\left(\frac{2}{\sqrt{3}-1}+\frac{3}{\sqrt{3}-2}+\frac{5\sqrt{3}}{3-\sqrt{3}}\right)\)
\(=\frac{1}{5+\sqrt{3}}.\left(\frac{3}{\sqrt{3}-2}+\frac{7\sqrt{3}+17}{2}\right)\)
\(=\frac{1}{5+\sqrt{3}}.\left[-\frac{7-3\sqrt{3}}{2\left(\sqrt{3}-2\right)}\right]\)
\(=-\frac{7-3\sqrt{3}}{2\left(\sqrt{3}-2\right)}.\frac{1}{\sqrt{3}+5}\)
\(=-\frac{7-3\sqrt{3}}{2\left(\sqrt{3}-2\right)\left(5+\sqrt{3}\right)}\) (làm tắt)
\(=-\frac{7-3\sqrt{3}}{6\sqrt{3}-14}\)
\(=-\left(-\frac{1}{2}\right)=\frac{1}{2}\)