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\(\frac{A}{\sqrt{2}}=\frac{1+\sqrt{7}}{2+\sqrt{8+2\sqrt{7}}}+\frac{1-\sqrt{7}}{2-\sqrt{8-2\sqrt{7}}}\)
\(=\frac{1+\sqrt{7}}{2+1+\sqrt{7}}+\frac{1-\sqrt{7}}{2-\sqrt{7}+1}\)
\(=\frac{1+\sqrt{7}}{3+\sqrt{7}}+\frac{1-\sqrt{7}}{3-\sqrt{7}}\)
=\(\frac{\left(1+\sqrt{7}\right)\left(3-\sqrt{7}\right)+\left(1-\sqrt{7}\right)\left(3+\sqrt{7}\right)}{\left(3+\sqrt{7}\right)\left(3-\sqrt{7}\right)}\)
\(=\frac{-8}{2}=-4\)
\(\Rightarrow A=-4\sqrt{2}\)
Bài 32:
a) P= \(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
= \(\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\left(\sqrt{4}+\sqrt{6}+\sqrt{8}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
= \(\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
= \(\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)\left(1+\sqrt{2}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
= \(1+\sqrt{2}\)
b) Có: \(x^2-2y^2=xy\)
\(\Leftrightarrow x^2-y^2-y^2-xy=0\)
\(\Leftrightarrow\left(x-y\right)\left(x+y\right)-y\left(y+x\right)\)
\(\Leftrightarrow\left(x+y\right)\left(x-y-y\right)=0\)
\(\Leftrightarrow\left(x+y\right)\left(x-2y\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+y=0\\x-2y=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-y\\x=2y\end{cases}}}\)
Thay x=-y ta có: Q=\(\frac{-y-y}{-y+y}\)=\(\frac{-2y}{0}\)(loại )
Thay x=2y ta có : Q=\(\frac{2y-y}{2y+y}=\frac{y}{3y}=\frac{1}{3}\)
Có \(x=\frac{2}{2\sqrt[3]{2}+2+\sqrt[3]{4}}=\frac{2}{\sqrt[3]{16}+\sqrt[3]{8}+\sqrt[3]{4}}=\frac{\sqrt[3]{8}}{\sqrt[3]{4}\left(\sqrt[3]{4}+\sqrt[3]{2}+1\right)}=\frac{\sqrt[3]{2}}{\sqrt[3]{4}+\sqrt[3]{2}+1}\)
\(y=\frac{2}{2\sqrt[3]{2}-2+\sqrt[3]{4}}=\frac{\sqrt[3]{8}}{\sqrt[3]{16}-\sqrt[3]{8}+\sqrt[3]{4}}=\frac{\sqrt[3]{8}}{\sqrt[3]{4}\left(\sqrt[3]{4}-\sqrt[3]{2}+1\right)}=\frac{\sqrt[3]{2}}{\sqrt[3]{4}-\sqrt[3]{2}+1}\)
Đặt \(\sqrt[3]{2}=a\)
=> \(x=\frac{a}{a^2+a+1}\) ,\(y=\frac{a}{a^2-a+1}\)
Có: \(x+y=\frac{a}{a^2+a+1}+\frac{a}{a^2-a+1}=\frac{a^3-a^2+a+a^3+a^2+a}{\left(a^2+a+1\right)\left(a^2-a+1\right)}=\frac{2a^3+2a}{a^4+a^2+1}\)
\(x-y=\frac{a}{a^2+a+1}-\frac{a}{a^2-a+1}=\frac{a^3-a^2+a-a^3-a^2-a}{\left(a^2+a+1\right)\left(a^2-a+1\right)}=\frac{-2a^2}{a^4+a^2+1}\)
Có x2-y2= (x-y)(x+y)=\(\frac{2a^3+2a}{a^4+a^2+1}.\frac{-2a^2}{a^4+a^2+1}=\frac{-2a^2.2a\left(a^2+1\right)}{\left(a^4+a^2+1\right)^2}=\frac{-4a^3\left(a^2+1\right)}{\left(a^4+a^2+1\right)^2}=\frac{-4.2\left(\sqrt[3]{4}+1\right)}{\left(\sqrt[3]{16}+\sqrt[3]{4}+1\right)^2}\)
=\(\frac{-8\left(\sqrt[3]{4}+1\right)}{\left(\sqrt[3]{16}+\sqrt[3]{4}+1\right)^2}\)