Không sử dụng máy tính cầm tay, tính A=\(\dfrac{1}{1+\sqrt{2}} + \dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{99}+\sqrt{100}}\).
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So sánh(không dùng bảng số hay máy tính cầm tay)
a)\(\dfrac{1}{7}\sqrt{51}\) với \(\dfrac{1}{9}\sqrt{150}\)
b)\(\sqrt{2017}-\sqrt{2016}\) với \(\sqrt{2016}-\sqrt{2015}\)
b: \(\sqrt{2017}-\sqrt{2016}=\dfrac{1}{\sqrt{2016}+\sqrt{2017}}\)
\(\sqrt{2016}-\sqrt{2015}=\dfrac{1}{\sqrt{2016}+\sqrt{2015}}\)
mà \(\sqrt{2016}+\sqrt{2017}< \sqrt{2016}+\sqrt{2015}\)
nên \(\sqrt{2017}-\sqrt{2016}>\sqrt{2016}-\sqrt{2015}\)
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Tính tổng sau: \(S=\dfrac{1}{2+\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+\dfrac{1}{4\sqrt{3}+3\sqrt{4}}+...+\dfrac{1}{100\sqrt{99}+99\sqrt{100}}\)
Ta có: \(S=\dfrac{1}{2+\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+\dfrac{1}{4\sqrt{3}+3\sqrt{4}}+...+\dfrac{1}{100\sqrt{99}+99\sqrt{100}}\)
\(=1-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{99}}-\dfrac{1}{10}\)
\(=1-\dfrac{1}{10}=\dfrac{9}{10}\)
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Giải pt, ko sử dụng mt cầm tay
\(\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)=\(\dfrac{1}{2}\)
\(\Leftrightarrow2\sqrt{x}+2=\sqrt{x}+3\)
=>\(\sqrt{x}=1\)
hay x=1
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\(\dfrac{\sqrt{x+1}}{\sqrt{x+3}}=\dfrac{1}{2}\)
\(\Leftrightarrow2\left(\sqrt{x+1}\right)=\sqrt{x}+3\)
\(\Leftrightarrow2\sqrt{x}+2=\sqrt{x}+3\)
\(\Leftrightarrow2\sqrt{x}+2-\sqrt{x}-3=0\)
\(\Leftrightarrow\sqrt{x}-1=0\Leftrightarrow\sqrt{x}=1\Leftrightarrow x=1\)
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\(\dfrac{\sqrt{x}+1}{\sqrt{x}+3}=\dfrac{1}{2}\)
=> \(2\left(\sqrt{x}+1\right)=\sqrt{x}+3\)
\(2\sqrt{x}+2=\sqrt{x}+3\)
\(2\sqrt{x}-\sqrt{x}=3-2\)
\(\sqrt{x}=1\)
=> x=1
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tính A= \(\sqrt{3+2\sqrt{2}}\) - \(\dfrac{1}{1+\sqrt{2}}\)( tính ko sử dụng máy tính)
\(A=\sqrt{2}+1-\sqrt{2}+1=2\)
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B1. ko sử dụng máy tính, rút gọnDdfrac{1}{2}sqrt{48}-dfrac{sqrt{33}}{sqrt{11}}+5sqrt{1dfrac{1}{3}}Esqrt{dfrac{3-sqrt{5}}{3+sqrt{5}}}-sqrt{dfrac{3+sqrt{5}}{3-sqrt{5}}}Fsqrt{3+sqrt{5}}+sqrt{7-3sqrt{5}}-sqrt{2}B2.Gdfrac{sqrt{x}}{x-sqrt{x}+1}so sánh G với 1B3. giải ptleft{{}begin{matrix}left(x-3right)left(2y+5right)left(2x+7right)left(y-1right)left(4x+1right)left(3y-6right)left(6x-1right)left(2y+3right)end{matrix}right.left{{}begin{matrix}left(x+yright)left(x-1right)left(x-yright)left(x+1right)+2xyl...
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B1. ko sử dụng máy tính, rút gọn
\(D=\dfrac{1}{2}\sqrt{48}-\dfrac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\dfrac{1}{3}}\)
\(E=\sqrt{\dfrac{3-\sqrt{5}}{3+\sqrt{5}}}-\sqrt{\dfrac{3+\sqrt{5}}{3-\sqrt{5}}}\)
\(F=\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}-\sqrt{2}\)
B2.
\(G=\dfrac{\sqrt{x}}{x-\sqrt{x}+1}\)
so sánh G với 1
B3. giải pt
\(\left\{{}\begin{matrix}\left(x-3\right)\left(2y+5\right)=\left(2x+7\right)\left(y-1\right)\\\left(4x+1\right)\left(3y-6\right)=\left(6x-1\right)\left(2y+3\right)\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\left(x+y\right)\left(x-1\right)=\left(x-y\right)\left(x+1\right)+2xy\\\left(y-x\right)\left(y+1\right)=\left(y+x\right)\left(y-2\right)-2xy\end{matrix}\right.\)
Bài 1:
\(D=\dfrac{1}{2}\sqrt{48}-\dfrac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\dfrac{1}{3}}=\dfrac{1}{2}.4\sqrt{3}-\sqrt{3}+5.\dfrac{2\sqrt{3}}{3}=2\sqrt{3}-\sqrt{3}+\dfrac{10\sqrt{3}}{3}=\dfrac{3\sqrt{3}+10\sqrt{3}}{3}=\dfrac{13\sqrt{3}}{3}\)
\(E=\sqrt{\dfrac{3-\sqrt{5}}{3+\sqrt{5}}}-\sqrt{\dfrac{3+\sqrt{5}}{3-\sqrt{5}}}=\sqrt{\dfrac{\left(3-\sqrt{5}\right)^2}{9-5}}-\sqrt{\dfrac{\left(3+\sqrt{5}\right)^2}{9-5}}=\dfrac{3-\sqrt{5}}{2}-\dfrac{3+\sqrt{5}}{2}=-\sqrt{5}\)
\(F=\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}-\sqrt{2}=\sqrt{\left(\sqrt{\dfrac{5}{2}}+\sqrt{\dfrac{1}{2}}\right)^2}+\sqrt{\left(\dfrac{3}{\sqrt{2}}-\sqrt{\dfrac{5}{2}}\right)^2}-\sqrt{2}=\sqrt{\dfrac{5}{2}}+\sqrt{\dfrac{1}{2}}+\dfrac{3}{\sqrt{2}}-\sqrt{\dfrac{5}{2}}-\sqrt{2}=2\sqrt{2}-\sqrt{2}=\sqrt{2}\)
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Bài 2:
Ta có: G-1
\(=\dfrac{\sqrt{x}-x+\sqrt{x}-1}{x-\sqrt{x}+1}\)
\(=\dfrac{-\left(x-2\sqrt{x}+1\right)}{x-\sqrt{x}+1}\)
\(=\dfrac{-\left(\sqrt{x}-1\right)^2}{x-\sqrt{x}+1}\le0\forall x\) thỏa mãn ĐKXĐ
hay \(G\le1\)
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Tính gía trị của biểu thức \(T=\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{99}+\sqrt{100}}\)
Ta có: \(T=\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{99}+\sqrt{100}}\)
\(=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+...+\sqrt{100}-\sqrt{99}\)
=10-1
=9
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Bài 40: Chứng minh rằng:
a) \(A=\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{4}}+...+\dfrac{1}{\sqrt{99}+\sqrt{100}}=9\)
b) \(B=\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+\dfrac{1}{4\sqrt{3}+3\sqrt{4}}+...+\dfrac{1}{100\sqrt{99}+99\sqrt{100}}=\dfrac{9}{10}\)
a rút gọn biểu thức: T=\(\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+\dfrac{1}{4\sqrt{3}+3\sqrt{4}}+...+\dfrac{1}{100\sqrt{99}+99\sqrt{100}}\)
b tìm số tự nhiên n thỏa mãn
\(\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+\dfrac{1}{4\sqrt{3}+3\sqrt{4}}+...+\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\dfrac{4}{5}\)
Với n\(\in N\)* có: \(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\dfrac{1}{\sqrt{n\left(n+1\right)}\left(\sqrt{n+1}+\sqrt{n}\right)}\)\(=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}\left(n+1-n\right)}=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}}\)\(=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)
\(\Rightarrow\)\(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\) (*)
a) Áp dụng (*) vào T
\(\Rightarrow T=1-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{99}}-\dfrac{1}{\sqrt{100}}\)\(=1-\dfrac{1}{10}=\dfrac{9}{10}\)
b) Có \(VT=1-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)\(=1-\dfrac{1}{\sqrt{n+1}}=\dfrac{4}{5}\)
\(\Leftrightarrow\sqrt{n+1}=5\Leftrightarrow n=24\) (tm)
Vậy n=24.
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Cho \(x=\dfrac{\sqrt{2}-\sqrt{1}}{1+\sqrt{2}}+\dfrac{\sqrt{3}-\sqrt{2}}{2+\sqrt{3}}+\dfrac{\sqrt{4}-\sqrt{3}}{3+4}+...+\dfrac{\sqrt{100}-\sqrt{99}}{99+100}\). Chứng minh \(x< \dfrac{1}{2}\)