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Đặt \(x=\sqrt[3]{\sqrt{5}+2}-\sqrt[3]{\sqrt{5}-2}.\)
\(\Rightarrow x^3=\sqrt{5}+2-3\sqrt[3]{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}\left(\sqrt[3]{\sqrt{5}+2}-\sqrt[3]{\sqrt{5}-2}\right)-\sqrt{5}+2\)
\(=4-3\sqrt[3]{5-4}.x\)( Vì \(x=\sqrt[3]{\sqrt{5}+2}-\sqrt[3]{\sqrt{5}-2}\))
\(=4-3x\)
\(\Rightarrow x^3+3x-4=0\Leftrightarrow\left(x^3-1\right)+\left(3x-3\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+4\right)=0\Leftrightarrow x-1=0\)( Vì \(x^2+x+4=\left(x+\frac{1}{2}\right)^2+\frac{15}{4}>0\))
\(\Leftrightarrow x=1\)hay \(\sqrt[3]{\sqrt{5}+2}-\sqrt[3]{\sqrt{5}-2}=1\)
Bài 1 : Với : \(x>0;x\ne1\)
\(P=\left(1+\frac{1}{\sqrt{x}-1}\right)\frac{1}{x-\sqrt{x}}=\left(\frac{\sqrt{x}}{\sqrt{x}-1}\right).\sqrt{x}\left(\sqrt{x}-1\right)=x\)
Thay vào ta được : \(P=x=25\)
Bài 2 :
a, Với \(x\ge0;x\ne1\)
\(A=\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{2}{\sqrt{x}+1}-\frac{2}{x-1}=\frac{x+\sqrt{x}-2\sqrt{x}+2-2}{x-1}\)
\(=\frac{x-\sqrt{x}}{x-1}=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}}{\sqrt{x}+1}\)
Thay x = 9 vào A ta được : \(\frac{3}{3+1}=\frac{3}{4}\)
a, \(\sqrt{11-2\sqrt{10}}=\sqrt{\left(\sqrt{10}\right)^2-2\sqrt{10}+1}=\sqrt{\left(\sqrt{10}+1\right)^2}\)
\(=\left|\sqrt{10}+1\right|=\sqrt{10}+1\)
b, \(\sqrt{27-10\sqrt{2}}=\sqrt{5^2-10\sqrt{2}+\left(\sqrt{2}\right)^2}=\sqrt{\left(5-\sqrt{2}\right)^2}\)
\(=\left|5-\sqrt{2}\right|=5-\sqrt{2}\)
c, \(\sqrt{4+2\sqrt{3}}=\sqrt{\left(\sqrt{3}\right)^2+2\sqrt{3}+1}=\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=\left|\sqrt{3}+1\right|=\sqrt{3}+1\)
làm nốt 2 câu cuối nhé, cách làm y trên
d/\(\sqrt{9+4\sqrt{5}}\)
= \(\sqrt{2^2+4\sqrt{5}+\left(\sqrt{5}\right)^2}\)
=\(\sqrt{\left(2+\sqrt{5}\right)^2}\)
= \(\left|2+\sqrt{5}\right|\)
= \(2+\sqrt{5}\)
e/ \(\sqrt{21+4\sqrt{5}}\)
= \(\sqrt{20+4\sqrt{5}+1}\)
=\(\sqrt{\left(2\sqrt{5}\right)^2+2.2\sqrt{5}+1^2}\)
=\(\sqrt{\left(2\sqrt{5}+1\right)^2}\)
= \(\left|2\sqrt{5}+1\right|\)
= \(2\sqrt{5}+1\)
\(ĐKXĐ:x\ne16\)
\(Q=\frac{1+3\sqrt{x}-12}{\sqrt{x}-4}.\frac{\left(\sqrt{x}-4\right)\left(\sqrt{x}+4\right)}{3-\sqrt{x}-11}\)
\(=\frac{\left(3\sqrt{x}-11\right)\left(\sqrt{x}+4\right)}{-\sqrt{x}-8}\)
\(\left(\frac{1}{\sqrt{x}-4}+3\right).\frac{x-16}{3-\sqrt{x}-11}=\left(\frac{1}{\sqrt{x}-4}+\frac{3\left(\sqrt{x}-4\right)}{\sqrt{x}-4}\right).\frac{\left(\sqrt{x}-4\right)\left(\sqrt{x}+4\right)}{-\sqrt{x}-8}\)
\(=\frac{1+3\left(\sqrt{x}-4\right)}{\sqrt{x}-4}.\frac{\left(\sqrt{x}-4\right)\left(\sqrt{x}+4\right)}{-\sqrt{x}-8}=\frac{1+3\sqrt{x}-12}{\sqrt{x}-4}.\frac{\left(\sqrt{x}-4\right)\left(\sqrt{x}+4\right)}{-\sqrt{x}-8}\)
\(=\frac{3\sqrt{x}-11}{\sqrt{x}-4}.\frac{\left(\sqrt{x}-4\right)\left(\sqrt{x}+4\right)}{-\left(\sqrt{x}+8\right)}=\frac{\left(3\sqrt{x}-11\right)\left(\sqrt{x}+4\right)}{-\left(\sqrt{x}+8\right)}\)
=\(\sqrt{3-\sqrt{5}}\)\(\sqrt{2}\)(\(\sqrt{5}-1\)) (\(3+\sqrt{5}\))
=\(\sqrt{6-2\sqrt{5}}\)(\(\sqrt{5}-1\)) (\(3+\sqrt{5}\))
=\(\sqrt{\left(\sqrt{5}+1\right)^2}\)(\(\sqrt{5}-1\))(\(3+\sqrt{5}\))
=(\(\sqrt{5}+1\))(\(\sqrt{5}-1\))(\(3+\sqrt{5}\))
=4(\(3+\sqrt{5}\))
=12+4\(\sqrt{5}\)
a: \(=6\sqrt{2}-12\sqrt{3}-10\sqrt{2}+12\sqrt{3}=-4\sqrt{2}\)
b: \(=\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}=\sqrt{4-3}=1\)