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với n >0, ta có :
\(\left(\sqrt{n+1}+\sqrt{n}\right)\left(\sqrt{n+1}-\sqrt{n}\right)=n+1-n=1\Rightarrow\frac{1}{\sqrt{n+1}-\sqrt{n}}=\sqrt{n+1}+\sqrt{n}\)
Gọi biểu thức đã cho là A
\(A=\frac{1}{-\left(\sqrt{2}-\sqrt{1}\right)}-\frac{1}{-\left(\sqrt{3}-\sqrt{2}\right)}+...+\frac{1}{-\left(\sqrt{8}-\sqrt{7}\right)}-\frac{1}{-\left(\sqrt{9}-\sqrt{8}\right)}\)
\(A=-\frac{1}{\sqrt{2}-\sqrt{1}}+\frac{1}{\sqrt{3}-\sqrt{2}}-...-\frac{1}{\sqrt{8}-\sqrt{7}}+\frac{1}{\sqrt{9}-\sqrt{8}}\)
\(A=-\left(\sqrt{2}+\sqrt{1}\right)+\left(\sqrt{3}+\sqrt{2}\right)-...-\left(\sqrt{8}+\sqrt{7}\right)+\left(\sqrt{9}+\sqrt{8}\right)\)
\(A=-\sqrt{1}+\sqrt{9}=2\)
Đặt \(\left\{{}\begin{matrix}\sqrt{\sqrt{2}+1}=a\\\sqrt{\sqrt{2}-1}=b\end{matrix}\right.\) \(\Rightarrow a^2+b^2=2\sqrt{2}=\sqrt{8}\)
\(\Rightarrow\sqrt{a^2+b^2}=\sqrt[4]{8}\)
Do đó:
\(A=\dfrac{\sqrt{\sqrt{a^2+b^2}+b}-\sqrt{\sqrt{a^2+b^2}-b}}{\sqrt{\sqrt{a^2+b^2}-a}}>0\)
\(\Rightarrow A^2=\dfrac{2\sqrt{a^2+b^2}-2\sqrt{a^2+b^2-b^2}}{\sqrt{a^2+b^2}-a}=\dfrac{2\left(\sqrt{a^2+b^2}-a\right)}{\sqrt{a^2+b^2}-a}=2\)
\(\Rightarrow A=\sqrt{2}\)
Đặt \(A=\frac{T}{M}\), ta có T>0 => \(T=\sqrt{T^2}\). Xét
\(T^2=\left(\sqrt[4]{8}+\sqrt{\sqrt{2}-1}\right)-2\sqrt{\left(\sqrt[4]{8}+\sqrt{\sqrt{2}-1}\right)}+\left(\sqrt[4]{8}-\sqrt{\sqrt{2}-1}\right)\)
\(=2\sqrt[4]{8}-2\sqrt{\sqrt{8}-\left(\sqrt{2}-1\right)}\)
\(=2\sqrt[4]{8}-2\sqrt{\sqrt{2}+1}\)
\(=2\left(\sqrt[4]{8}-\sqrt{\sqrt{2}+1}\right)\)
\(\Rightarrow T=\sqrt{2}\cdot\sqrt{\sqrt[4]{8}-2\sqrt{2}+1}\)
\(\Rightarrow A=\sqrt{2}\)
Phân tích mỗi hạng tử theo kiểu như dưới đây
\(\frac{\sqrt{1}+\sqrt{2}}{\left(\sqrt{1}\right)^2-\left(\sqrt{2}\right)^2}\)
\(\frac{\sqrt{2}+\sqrt{3}}{\left(\sqrt{2}\right)^2-\left(\sqrt{3}\right)^2}\)
Khi đó mọi mẫu đều bằng -1
Bạn tiếp tục làm và kết quả nhận được là \(1-\sqrt{9}\)
A>0
Bình phương A được :
\(A^2=\frac{\sqrt[4]{8}+\sqrt{\sqrt{2}-1}-2\sqrt{\left(\sqrt[4]{8}+\sqrt{\sqrt{2}-1}\right)\left(\sqrt[4]{8}-\sqrt{\sqrt{2}-1}\right)}+\sqrt[4]{8}-\sqrt{\sqrt{2}-1}}{\sqrt[4]{8}-\sqrt{\sqrt{2}+1}}\)
=\(\frac{2\sqrt[4]{8}-2\sqrt{\sqrt{8}-\sqrt{2}+1}}{\sqrt[4]{8}-\sqrt{\sqrt{2}+1}}\)=\(\frac{2\sqrt[4]{8}-2\sqrt{2\sqrt{2}-\sqrt{2}+1}}{\sqrt[4]{8}-\sqrt{\sqrt{2}+1}}=\frac{2\sqrt[4]{8}-2\sqrt{\sqrt{2}+1}}{\sqrt[4]{8}-\sqrt{\sqrt{2}+1}}=2\)
=> A= \(\sqrt{2}\)
@Vũ Minh Tuấn @Lê Thị Thục Hiền @No choice teen
\(b,\frac{2+\sqrt{3}}{1-\sqrt{4-2\sqrt{3}}}+\frac{2-\sqrt{3}}{1+\sqrt{4+2\sqrt{3}}}\)
\(=\frac{2+\sqrt{3}}{1-\sqrt{3-2\sqrt{3}+1}}+\frac{2-\sqrt{3}}{1+\sqrt{3+2\sqrt{3}+1}}\)
\(=\frac{2+\sqrt{3}}{1-\sqrt{\left(\sqrt{3}-1\right)^2}}+\frac{2-\sqrt{3}}{1+\sqrt{\left(\sqrt{3}+1\right)^2}}\)
\(=\frac{2+\sqrt{3}}{1-\left(\sqrt{3}-1\right)}+\frac{2-\sqrt{3}}{1+\sqrt{3}+1}\)
\(=\frac{2+\sqrt{3}}{2-\sqrt{3}}+\frac{2-\sqrt{3}}{2+\sqrt{3}}\)
\(=\frac{\left(2+\sqrt{3}\right)^2}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}+\frac{\left(2-\sqrt{3}\right)^2}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}\)
\(=\frac{4+4\sqrt{3}+3+4-4\sqrt{3}+3}{4-3}\)
\(=14\)
\(a,\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\sqrt{2}+\sqrt{3}+4+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+2}\)
\(=\frac{\sqrt{2}+\sqrt{3}+2}{\sqrt{2}+\sqrt{3}+2}+\frac{\sqrt{2}.\sqrt{2}+\sqrt{2}.\sqrt{3}+\sqrt{2}.2}{\sqrt{2}+\sqrt{3}+2}\)
\(=1+\frac{\sqrt{2}\left(\sqrt{2}+\sqrt{3}+2\right)}{\sqrt{2}+\sqrt{3}+2}\)
\(=1+\sqrt{2}\)
`c)root{3}{4}.root{3}{1-sqrt3}.root{6}{(sqrt3+1)^2}`
`=root{3}{4(1-sqrt3)}.root{3}{1+sqrt3}`
`=root{3}{4(1-sqrt3)(1+sqrt3)}`
`=root{3}{4(1-3)}=-2`
`d)2/(root{3}{3}-1)-4/(root{9}-root{3}{3}+1)`
`=(2(root{3}{9}+root{3}{3}+1))/(3-1)-(4(root{3}{3}+1))/(3+1)`
`=root{3}{9}+root{3}{3}+1-root{3}{3}-1`
`=root{3}{9}`
`a)root{3}{8sqrt5-16}.root{3}{8sqrt5+16}`
`=root{3}{(8sqrt5-16)(8sqrt5+16)}`
`=root{3}{320-256}`
`=root{3}{64}=4`
`b)root{3}{7-5sqrt2}-root{6}{8}`
`=root{3}{1-3.sqrt{2}+3.2.1-2sqrt2}-root{6}{(2)^3}`
`=root{3}{(1-sqrt2)^3}-sqrt2`
`=1-sqrt2-sqrt2=1-2sqrt2`
Nhân cả tử và mẫu với biểu thức liên hợp của mẫu (câu a mẫu cuối kì kì)
\(A=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\frac{1}{4}=\sqrt{3}-\frac{3}{4}\)
\(B=-\left(\sqrt{1}+\sqrt{2}-\sqrt{2}-\sqrt{3}+\sqrt{3}+\sqrt{4}-...+\sqrt{7}+\sqrt{8}-\sqrt{8}-\sqrt{9}\right)\)
\(B=-\left(\sqrt{1}-\sqrt{9}\right)=2\)