\(\sqrt{4+\sqrt{6}}=..........\)
\(\sqrt{4+\sqrt{6}}=..........\)
\(\sqrt{4+\sqrt{6}}=..........\)
\(\sqrt{4+\sqrt{6}}=..........\)
\(\sqrt{4+\sqrt{6}}=..........\)
\(\sqrt{4+\sqrt{6}}=..........\)
\(\sqrt{4+\sqrt{6}}=..........\)
\(\sqrt{4+\sqrt{6}}=..........\)
\(\sqrt{4+\sqrt{6}}=..........\)
\(\sqrt{4+\sqrt{6}}=..........\)
Chứng minh rằng :
a) \(\frac{\left(a+b\right)^2}{2}+\frac{a+b}{4}\ge a\sqrt{b}+b\sqrt{a}\)với \(a,b\ge0\)
b) \(\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}>2\)với \(a,b,c>0\)
A=(\(\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\)-\(\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\)):(\(\frac{2\left(\sqrt{x}-1\right)^2}{x-1}\)
=\(\left(\frac{x+\sqrt{x}+1-\left(x-\sqrt{x}+1\right)}{\sqrt{x}}\right).\frac{\sqrt{x}+1}{2\left(\sqrt{x}-1\right)}\)
=\(\frac{2\sqrt{x}}{\sqrt{x}}.\frac{\sqrt{x}+1}{2\left(\sqrt{x}-1\right)}\)=\(\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(y=\frac{1}{x^2+\sqrt{x}}\sqrt[]{}\sqrt{\sqrt[]{}\sqrt[]{}\frac{\frac{\frac{\frac{ }{ }}{ }}{ }}{ }}\)
\(\sqrt{36+\sqrt{25+\sqrt{49+\sqrt{81}}}}\) = ?
tinh gia tri bieu thuc a = \(\sqrt{4+\sqrt[3]{8}+\frac{2}{3}}+3^{10}\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{99\times100}\)
tinh gia tri bieu thuc a = \(\sqrt{4+\sqrt[3]{8+\frac{2}{3}+3^{10}\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{99\times100}}}\)
tinh gia tri bieu thuc a = \(\sqrt{4+\sqrt[3]{8+\frac{2}{3}+3^{10}\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{99\times100}}}\)
tìm ab biết :
ab + ba = 132
\(\sqrt{a}\) a-b=4
Câu 10:
Cho \(\frac{1}{\sqrt{2}+x}\cdot\frac{1}{\sqrt{2}+y}=a\)
Và \(\frac{1}{\sqrt{2}+x}+\frac{1}{\sqrt{2+y}}=b\)
Tìm x + y ( theo a và b )
