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a: \(=\dfrac{1}{\sqrt{6}-1+1}-\dfrac{1}{\sqrt{6}+1-1}\)
\(=\dfrac{1}{\sqrt{6}}-\dfrac{1}{\sqrt{6}}\)
=0
b: \(=\dfrac{3+\sqrt{7}-3+\sqrt{7}}{2}=\dfrac{2\sqrt{7}}{2}=\sqrt{7}\)
c: \(=\sqrt{\left(3\sqrt{2}+\sqrt{3}\right)^2}+\sqrt{\left(3\sqrt{2}-\sqrt{3}\right)^2}\)
\(=3\sqrt{2}+\sqrt{3}+3\sqrt{2}-\sqrt{3}=6\sqrt{2}\)
Câu 1:
\(\Leftrightarrow1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n}-\dfrac{1}{n+1}=\dfrac{2999}{3000}\)
\(\Leftrightarrow1-\dfrac{1}{n+1}=\dfrac{2999}{3000}\)
=>n+1=3000
hay n=2999
a)
\(S=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+....+\frac{1}{\sqrt{100}+\sqrt{101}}\)
\(S=\frac{\sqrt{2}-\sqrt{1}}{(\sqrt{2}+\sqrt{1})(\sqrt{2}-\sqrt{1})}+\frac{\sqrt{3}-\sqrt{2}}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}+....+\frac{\sqrt{101}-\sqrt{100}}{(\sqrt{101}+\sqrt{100})(\sqrt{101}-\sqrt{100})}\)
\(S=\frac{\sqrt{2}-\sqrt{1}}{2-1}+\frac{\sqrt{3}-\sqrt{2}}{3-2}+...+\frac{\sqrt{101}-\sqrt{100}}{101-100}\)
\(S=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{101}-\sqrt{100}\)
\(S=\sqrt{101}-1\)
b)
\(S=\frac{1}{\sqrt{2}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{6}}+...+\frac{1}{\sqrt{100}+\sqrt{102}}\)
\(S=\frac{\sqrt{4}-\sqrt{2}}{(\sqrt{4}+\sqrt{2})(\sqrt{4}-\sqrt{2})}+\frac{\sqrt{6}-\sqrt{4}}{(\sqrt{6}+\sqrt{4})(\sqrt{6}-\sqrt{4})}+...+\frac{\sqrt{102}-\sqrt{100}}{(\sqrt{102}+\sqrt{100})(\sqrt{102}-\sqrt{100})}\)
\(S=\frac{\sqrt{4}-\sqrt{2}}{4-2}+\frac{\sqrt{6}-\sqrt{4}}{6-4}+....+\frac{\sqrt{102}-\sqrt{100}}{102-100}\)
\(S=\frac{\sqrt{4}-\sqrt{2}+\sqrt{6}-\sqrt{4}+\sqrt{8}-\sqrt{6}+...+\sqrt{102}-\sqrt{100}}{2}\)
\(S=\frac{\sqrt{102}-\sqrt{2}}{2}\)
ĐKXĐ: \(x\ge0,x\ne1\)
\(A=\left(1+\dfrac{\sqrt{x}}{x+1}\right):\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2\sqrt{x}}{x\sqrt{x}+\sqrt{x}-x-1}\right)-1\)
= \(\dfrac{x+\sqrt{x}+1}{x+1}:\left(\dfrac{x+1-2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}-1\right)}\right)-1\)
= \(\dfrac{\left(x+\sqrt{x}+1\right)\left(x+1\right)\left(\sqrt{x}-1\right)}{\left(x+1\right)\left(\sqrt{x}-1\right)^2}-1\)
= \(\dfrac{x+\sqrt{x}+1}{\sqrt{x}-1}-1\)
= \(\dfrac{x+\sqrt{x}+1-\sqrt{x}+1}{\sqrt{x}-1}\)
= \(\dfrac{x+2}{\sqrt{x}-1}\)
a: \(A=\left(\dfrac{\sqrt{3}\left(x-\sqrt{3}\right)+3}{\left(x-\sqrt{3}\right)\left(x^2+x\sqrt{3}+3\right)}\right)\cdot\dfrac{x^2+3+x\sqrt{3}}{x\sqrt{3}}\)
\(=\dfrac{x\sqrt{3}}{\left(x-\sqrt{3}\right)\left(x^2+x\sqrt{3}+3\right)}\cdot\dfrac{x^2+x\sqrt{3}+3}{x\sqrt{3}}\)
\(=\dfrac{1}{x-\sqrt{3}}\)
b: \(B=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}+x+1\)
\(=x-\sqrt{x}-x-\sqrt{x}+x+1\)
\(=x-2\sqrt{x}+1\)
c: \(C=\left(\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}-\dfrac{\sqrt{x}-2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\cdot\dfrac{x\left(\sqrt{x}+1\right)-\left(\sqrt{x}+1\right)}{\sqrt{x}}\)
\(=\dfrac{x+\sqrt{x}-2-\left(x-\sqrt{x}-2\right)}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}}=2\)
1) Đk: \(x\ge4\)
\(\dfrac{\sqrt{x^2-16}}{\sqrt{x-3}}+\sqrt{x-3}=\dfrac{7}{\sqrt{x-3}}\)
\(\Leftrightarrow\dfrac{\sqrt{x^2-16}}{\sqrt{x-3}}+\dfrac{x-3}{\sqrt{x-3}}=\dfrac{7}{\sqrt{x-3}}\)
\(\Leftrightarrow\dfrac{\sqrt{x^2-16}+x-10}{\sqrt{x-3}}=0\)
\(\Leftrightarrow\sqrt{x^2-16}+x-10=0\)
\(\Leftrightarrow\sqrt{x^2-16}=10-x\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-16=100-20x+x^2\\x\le10\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}20x=116\\x\le10\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{29}{5}\left(N\right)\\x\le10\end{matrix}\right.\)
Kl: x= 29/5
2) Đk: \(x\ge-1\)
\(x^2-5x+14=4\sqrt{x+1}\)
\(\Leftrightarrow x^4+25x^2+196-10x^3-140x+28x^2=16x+16\)
\(\Leftrightarrow x^4-10x^3+53x^2-156x+180=0\)
\(\Leftrightarrow\left(x-3\right)\left(x^3-7x^2+32x-60\right)=0\)
\(\Leftrightarrow\left(x-3\right)^2\left(x^2-4x+20\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x^2-4x+20=0\left(vn\right)\end{matrix}\right.\)
\(\Leftrightarrow x=3\left(N\right)\)
Kl: x=3
B=\(2\sqrt{18}-4\sqrt{32}+\sqrt{72}+3\sqrt{8}\)
\(=6\sqrt{2}-16\sqrt{2}+6\sqrt{2}+6\sqrt{2}\)
\(=2\sqrt{2}\)
\(B=2\sqrt{18}-4\sqrt{32}+\sqrt{72}+3\sqrt{8}\)
\(=6\sqrt{2}-16\sqrt{2}+6\sqrt{2}+6\sqrt{2}\)
\(=\sqrt{2}\left(6-16+6+6\right)\)
\(=2\sqrt{2}\)