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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)
\(S=\frac{1}{\sqrt{2}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{6}}+...+\frac{1}{\sqrt{100}+\sqrt{102}}\)
\(S=\frac{\sqrt{4}-\sqrt{2}}{(\sqrt{4}+\sqrt{2})(\sqrt{4}-\sqrt{2})}+\frac{\sqrt{6}-\sqrt{4}}{(\sqrt{6}+\sqrt{4})(\sqrt{6}-\sqrt{4})}+...+\frac{\sqrt{102}-\sqrt{100}}{(\sqrt{102}+\sqrt{100})(\sqrt{102}-\sqrt{100})}\)
\(S=\frac{\sqrt{4}-\sqrt{2}}{4-2}+\frac{\sqrt{6}-\sqrt{4}}{6-4}+....+\frac{\sqrt{102}-\sqrt{100}}{102-100}\)
\(S=\frac{\sqrt{4}-\sqrt{2}+\sqrt{6}-\sqrt{4}+\sqrt{8}-\sqrt{6}+...+\sqrt{102}-\sqrt{100}}{2}\)
\(S=\frac{\sqrt{102}-\sqrt{2}}{2}\)
......................?
mik ko biết
mong bn thông cảm
nha ................
ĐKXĐ: \(x\ge0,x\ne1\)
\(A=\left(1+\dfrac{\sqrt{x}}{x+1}\right):\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2\sqrt{x}}{x\sqrt{x}+\sqrt{x}-x-1}\right)-1\)
= \(\dfrac{x+\sqrt{x}+1}{x+1}:\left(\dfrac{x+1-2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}-1\right)}\right)-1\)
= \(\dfrac{\left(x+\sqrt{x}+1\right)\left(x+1\right)\left(\sqrt{x}-1\right)}{\left(x+1\right)\left(\sqrt{x}-1\right)^2}-1\)
= \(\dfrac{x+\sqrt{x}+1}{\sqrt{x}-1}-1\)
= \(\dfrac{x+\sqrt{x}+1-\sqrt{x}+1}{\sqrt{x}-1}\)
= \(\dfrac{x+2}{\sqrt{x}-1}\)
\(a,\dfrac{3}{\sqrt{12x-1}}\) xác định \(\Leftrightarrow12x-1>0\Leftrightarrow12x>1\Leftrightarrow x>\dfrac{1}{12}\)
\(b,\sqrt{\left(3x+2\right)\left(x-1\right)}\) xác định \(\Leftrightarrow\left(3x+2\right)\left(x-1\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}3x+2\ge0\\x-1\ge0\end{matrix}\right.\\\left[{}\begin{matrix}3x+2\le0\\x-1\le0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x\ge-\dfrac{2}{3}\\x\ge1\end{matrix}\right.\\\left[{}\begin{matrix}x\le-\dfrac{2}{3}\\x\le1\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x\le-\dfrac{2}{3}\\x\ge1\end{matrix}\right.\)
\(c,\sqrt{3x-2}.\sqrt{x-1}\) xác định \(\Leftrightarrow\left[{}\begin{matrix}3x-2\ge0\\x-1\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x\ge\dfrac{2}{3}\\x\ge1\end{matrix}\right.\) \(\Leftrightarrow x\ge1\)
\(d,\sqrt{\dfrac{-2\sqrt{6}+\sqrt{23}}{-x+5}}\) xác định \(\Leftrightarrow-x+5>0\Leftrightarrow x< 5\)
2/
a) Ta có:
\(3\sqrt{2}=\sqrt{3^2\cdot2}=\sqrt{9\cdot2}=\sqrt{18}\)
\(2\sqrt{3}=\sqrt{2^2\cdot3}=\sqrt{4\cdot3}=\sqrt{12}\)
Mà: \(12< 18\Rightarrow\sqrt{12}< \sqrt{18}\Rightarrow2\sqrt{3}< 3\sqrt{2}\)
b) Ta có:
\(4\sqrt[3]{5}=\sqrt[3]{4^3\cdot5}=\sqrt[3]{320}\)
\(5\sqrt[3]{4}=\sqrt[3]{5^3\cdot4}=\sqrt[3]{500}\)
Mà: \(320< 500\Rightarrow\sqrt[3]{320}< \sqrt[3]{500}\Rightarrow4\sqrt[3]{5}< 5\sqrt[3]{4}\)
3/
a)ĐKXĐ: \(x\ne1;x\ge0\)
b) \(A=\left(1-\dfrac{x-\sqrt{x}}{\sqrt{x}-1}\right)\left(1+\dfrac{x+\sqrt{x}}{\sqrt{x}+1}\right)\)
\(A=\left[1-\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right]\left[1+\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right]\)
\(A=\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)\)
\(A=1^2-\left(\sqrt{x}\right)^2\)
\(A=1-x\)
\(a)A=\dfrac{5+3\sqrt{5}}{\sqrt{5}}+\dfrac{3+\sqrt{3}}{\sqrt{3}+1}-\left(\sqrt{5}+3\right)\\ =\dfrac{\sqrt{5}\left(\sqrt{5}+3\right)}{\sqrt{5}}+\dfrac{\sqrt{3}\left(\sqrt{3}+1\right)}{\sqrt{3}+1}-\left(\sqrt{5}+3\right)\\ =\sqrt{5}+3+\sqrt{3}-\left(\sqrt{5}+3\right)\\ =\sqrt{3}\)
\(b)B=\left(5+\sqrt{21}\right)\left(\sqrt{14}-\sqrt{6}\right)\sqrt{5-\sqrt{21}}\\ =\left(5+\sqrt{21}\right)\left(\sqrt{7}-\sqrt{3}\right)\sqrt{10-2\sqrt{21}}\\ =\left(5+\sqrt{21}\right)\left(\sqrt{7}-\sqrt{3}\right)\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}\\ =\left(5+\sqrt{21}\right)\left(\sqrt{7}-\sqrt{3}\right)^2\\ =\left(5+\sqrt{21}\right)\left(10-2\sqrt{21}\right)\\ =2\left(5+\sqrt{21}\right)\left(5-\sqrt{21}\right)\\ =2\left(25-21\right)=8\)