Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\frac{A}{\sqrt{2}}=\frac{1+\sqrt{7}}{2+\sqrt{8+2\sqrt{7}}}+\frac{1-\sqrt{7}}{2-\sqrt{8-2\sqrt{7}}}\)
\(=\frac{1+\sqrt{7}}{2+1+\sqrt{7}}+\frac{1-\sqrt{7}}{2-\sqrt{7}+1}\)
\(=\frac{1+\sqrt{7}}{3+\sqrt{7}}+\frac{1-\sqrt{7}}{3-\sqrt{7}}\)
=\(\frac{\left(1+\sqrt{7}\right)\left(3-\sqrt{7}\right)+\left(1-\sqrt{7}\right)\left(3+\sqrt{7}\right)}{\left(3+\sqrt{7}\right)\left(3-\sqrt{7}\right)}\)
\(=\frac{-8}{2}=-4\)
\(\Rightarrow A=-4\sqrt{2}\)
\(x=\dfrac{1}{2}\cdot\sqrt{\left(\sqrt{2}-1\right)^2}=\dfrac{\sqrt{2}-1}{2}\)
\(A=\left[4\cdot\left(\dfrac{\sqrt{2}-1}{2}\right)^4+4\cdot\left(\dfrac{\sqrt{2}-1}{2}\right)^3-5\cdot\left(\dfrac{\sqrt{2}-1}{2}\right)^2+5\cdot\dfrac{\sqrt{2}-1}{2}-2\right]^{2015}+2016\)
=-1,13+2016=2014,87
Bạn xem lại đề bài 1 và 2.b nhé !
2/ \(A=\sqrt{\left(3-5\sqrt{2}\right)^2}-\sqrt{51+10\sqrt{2}}\)
\(A=5\sqrt{2}-3-\sqrt{\left(5\sqrt{2}+1\right)^2}\)
\(A=5\sqrt{2}-3-5\sqrt{2}-1\)
\(A=-4\)
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)=\(\frac{2a^3+2a}{a^4+a^2+1}.\frac{-2a^2}{a^4+a^2+1}=\frac{-2a^2.2a\left(a^2+1\right)}{\left(a^4+a^2+1\right)^2}=\frac{-4a^3\left(a^2+1\right)}{\left(a^4+a^2+1\right)^2}=\frac{-4.2\left(\sqrt[3]{4}+1\right)}{\left(\sqrt[3]{16}+\sqrt[3]{4}+1\right)^2}\)
=\(\frac{-8\left(\sqrt[3]{4}+1\right)}{\left(\sqrt[3]{16}+\sqrt[3]{4}+1\right)^2}\)