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\(\sqrt{\frac{3\sqrt{5}+1}{2\sqrt{5}-3}}\left(\sqrt{10}-\sqrt{2}\right)\)
\(=\left(\sqrt{10}-\sqrt{2}\right)\sqrt{3+\sqrt{5}}\)
?? :v
\(\sqrt{\frac{3\sqrt{5}+1}{2\sqrt{5}-3}}\left(\sqrt{10}-\sqrt{2}\right)\)
\(=\sqrt{\frac{\left(3\sqrt{5}+1\right)\left(2\sqrt{5}+3\right)}{\left(2\sqrt{5}-3\right)\left(2\sqrt{5}+3\right)}}.\left(\sqrt{10}-\sqrt{2}\right)\)
\(=\sqrt{3+\sqrt{5}}.\sqrt{2}.\left(\sqrt{5}-1\right)\)
\(=\sqrt{6+2\sqrt{5}}.\left(\sqrt{5}-1\right)\)
\(=\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)=4\)
b) \(\sqrt{16x}-5\left(\sqrt{x}-2\right)-\sqrt{79x}-5\)
\(=\sqrt{4^2x}-5\sqrt{x}+10-\sqrt{79x}-5\)
\(=4\sqrt{x}-5\sqrt{x}-\sqrt{79x}+5\)
\(=-\sqrt{x}-\sqrt{79x}+5\)
\(=-\sqrt{x}-\sqrt{79}.\sqrt{x}+5\)
\(=\sqrt{x}\left(-1-\sqrt{79}\right)+5\)
\(=\sqrt{3-\sqrt{5}}.\sqrt{2}.\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)\)
\(=\sqrt{6-2\sqrt{5}}\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)\)
\(=\sqrt{\left(\sqrt{5}-1\right)^2}\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)\)
\(=\left(\sqrt{5}-1\right)\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)\)
\(=\left(\sqrt{5}-1\right)^2\left(3+\sqrt{5}\right)\)
\(=\left(6-2\sqrt{5}\right)\left(3+\sqrt{5}\right)\)
\(=2\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)\)
\(=2\left(9-\left(\sqrt{5}\right)^2\right)\)
\(=2.4=8\)
Chỉ vậy thôi nha bạn ^_^
\(C=\sqrt{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}.\sqrt{3+\sqrt{5}.}\sqrt{2}\left(\sqrt{5}-1\right)\)
\(C=\sqrt{4}.\sqrt{6+2\sqrt{5}}\left(\sqrt{5}-1\right)\)
\(C=2.\sqrt{\left(\sqrt{5}+1\right)^2}.\left(\sqrt{5}-1\right)\)
\(C=2.\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)=2.4=8\)
b) \(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\left(2+\sqrt{3}\right)}}}\)\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}\)\(=\sqrt{4+\sqrt{5\sqrt{3}+5\left(5-\sqrt{3}\right)}}=\sqrt{4+\sqrt{5\sqrt{3}+25-5\sqrt{3}}}=\sqrt{4+5}=3\)
\(13-4\sqrt{3}=\left(2\sqrt{3}\right)^2-2.2\sqrt{2}.1+1^2=\left(2\sqrt{3}-1\right)^2\)
a) \(\left(\sqrt{5}+\sqrt{3}\right)\sqrt{8-2\sqrt{15}}=\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)=5-3=2\)
câu này \(\sqrt{15}\)đúng hơn \(\sqrt{5}\)
b) \(\sqrt{3-\sqrt{5}}-\sqrt{3+\sqrt{5}}=\frac{\sqrt{6-2\sqrt{5}}-\sqrt{6+2\sqrt{5}}}{\sqrt{2}}=\frac{\sqrt{5}-1-\sqrt{5}-1}{\sqrt{2}}=\frac{-2}{\sqrt{2}}=-\sqrt{2}\)c) \(\sqrt{5-2\sqrt{6}}-\sqrt{5+2\sqrt{6}}=\sqrt{3}-\sqrt{2}-\sqrt{3}-\sqrt{2}=-2\sqrt{2}\)
\(5\sqrt{x}-\frac{\left(x+10\sqrt{x}+25\right)\left(\sqrt{x}-5\right)}{x-25}=5\sqrt{x}-\frac{\left(\sqrt{x}+5\right)^2\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\)
\(=5\sqrt{x}-\left(\sqrt{x}+5\right)=4\sqrt{x}-5\)
\(\frac{\sqrt{x^2-4x+4}}{x-2}=\frac{\sqrt{\left(x-2\right)^2}}{x-2}=\frac{\left|x-2\right|}{x-2}=\orbr{\begin{cases}\frac{x-2}{x-2}\left(x>2\right)\\\frac{2-x}{x-2}\left(x< 2\right)\end{cases}=\orbr{\begin{cases}1\left(x>2\right)\\-1\left(x< 2\right)\end{cases}}}\)
a) Ta có:
\(5\sqrt{a}-4b\sqrt{25a^3}+5a\sqrt{16ab^2}-2\sqrt{9a}\)
\(=5\sqrt{a}-4b.5a\sqrt{a}+5a.4b\sqrt{a}-2.3\sqrt{a}\)
\(=5\sqrt{a}-20ab\sqrt{a}+20ab\sqrt{a}-6\sqrt{a}\) \(=-\sqrt{a}\)
b) Ta có:
\(5a\sqrt{64ab^3}-\sqrt{3}.\sqrt{12a^3b^3}+2ab\sqrt{9ab}\) \(-5b\sqrt{81a^3b}\)
\(=5a.8b\sqrt{ab}-\sqrt{3.12a^3b^3}+2ab.3\sqrt{ab}\) \(-5b.9a\sqrt{ab}\)
\(=40ab\sqrt{ab}-6ab\sqrt{ab}+6ab\sqrt{ab}-45ab\)\(\sqrt{ab}\)
\(=-5ab\sqrt{ab}\)
a) Đặt \(A=\sqrt{5-2\sqrt{6}}+\sqrt{5+2\sqrt{6}}\)
\(A^2=5-2\sqrt{6}+2\sqrt{\left(5-2\sqrt{6}\right)\left(5+2\sqrt{6}\right)}+5+2\sqrt{6}\)
\(=10+2\sqrt{25-4.6}=10+2\sqrt{1}=10+2=12\)
\(\Rightarrow A=\sqrt{12}\)
b)\(\frac{\sqrt{10}-\sqrt{2}}{\sqrt{5}-1}+\frac{2-\sqrt{2}}{\sqrt{2}-1}=\frac{\sqrt{2}.\sqrt{5}-\sqrt{2}}{\sqrt{5}-1}+\frac{\sqrt{2}.\sqrt{2}-\sqrt{2}}{\sqrt{2}-1}\)
\(=\frac{\sqrt{2}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}+\frac{\sqrt{2}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}=\sqrt{2}+\sqrt{2}=2\sqrt{2}\)
= 0,2.10.√3 + 2|√3 - √5|
= 2√3 + 2(√5 - √3)
= 2√3 + 2√5 - 2√3 = 2√5