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\(A=\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2.10^2}\)
\(A=\dfrac{2^2-1^2}{1^2.2^2}+\dfrac{3^2-2^2}{2^2.3^2}+\dfrac{4^2-3^2}{3^2.4^2}+...+\dfrac{10^2-9^2}{9^2.10^2}\)
\(A=\dfrac{2^2}{1^2.2^2}-\dfrac{1^2}{1^2.2^2}+\dfrac{3^2}{2^2.3^2}-\dfrac{2^2}{2^2.3^2}+...+\dfrac{10^2}{9^2.10^2}-\dfrac{9^2}{9^2.10^2}\)\(A=1-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+...+\dfrac{1}{9^2}-\dfrac{1}{10^2}\)
\(A=1-\dfrac{1}{10^2}< 1\left(đpcm\right)\)
\(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2.10^2}\)
\(=\left(\dfrac{1}{1^2}-\dfrac{1}{2^2}\right)+\left(\dfrac{1}{2^2}-\dfrac{1}{3^2}\right)+\left(\dfrac{1}{3^2}-\dfrac{1}{4^2}\right)+...+\left(\dfrac{1}{9^2}-\dfrac{1}{100^2}\right)\)
\(=\dfrac{1}{1^2}-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+\dfrac{1}{3^2}-\dfrac{1}{4^2}+...+\dfrac{1}{9^2}-\dfrac{1}{10^2}\)
\(=\dfrac{1}{1}-\dfrac{1}{100}\)
\(=\dfrac{99}{100}\)
\(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+...+\dfrac{19}{9^2.10^2}\)
\(=\dfrac{3}{1.4}+\dfrac{5}{4.9}+...+\dfrac{19}{81.100}\)
\(=\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{9}+...+\dfrac{1}{81}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}=\dfrac{99}{100}\)
Ta có:
\(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2.10^2}\)
= \(\dfrac{2^2-1^2}{1^2.2^2}+\dfrac{3^2-2^2}{2^2.3^2}+\dfrac{4^2-3^2}{3^2.4^2}+...+\dfrac{10^2-9^2}{9^2.10^2}\)
= \(\dfrac{1}{1^2}-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+\dfrac{1}{3^2}-\dfrac{1}{4^2}+...+\dfrac{1}{9^2}-\dfrac{1}{10^2}\)
= \(1-\dfrac{1}{10^2}\)
Mà \(1-\dfrac{1}{10^2}< 1\) nên:
\(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2.10^2}\) < 1 (đpcm).
Chứng minh:
\(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2.10^2}< 1\)
\(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2+10^2}\)
\(=\left(\dfrac{1}{1^2}-\dfrac{1}{2^2}\right)+\left(\dfrac{1}{2^2}-\dfrac{1}{3^2}\right)+...+\left(\dfrac{1}{9^2}-\dfrac{1}{10^2}\right)\)
=\(\dfrac{1}{1}-\dfrac{1}{10^2}\)
\(=1-\dfrac{1}{100}\)
Mà \(1-\dfrac{1}{100}< 1\)
\(\Rightarrow\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+...+\dfrac{19}{9^2.10^2}< 1\) (đpcm)
1)\(\dfrac{A}{B}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}}{\dfrac{2008}{1}+\dfrac{2007}{2}+\dfrac{2006}{3}+...+\dfrac{2}{2007}+\dfrac{1}{2008}}\)
\(\dfrac{A}{B}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}}{2008+\dfrac{2007}{2}+\dfrac{2006}{3}+...+\dfrac{2}{2007}+\dfrac{1}{2008}}\)
\(\dfrac{A}{B}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}}{1+\left(\dfrac{2007}{2}+1\right)+\left(\dfrac{2006}{3}+1\right)+...+\left(\dfrac{2}{2007}+1\right)+\left(\dfrac{1}{2008}+1\right)}\)
\(\dfrac{A}{B}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}}{\dfrac{2009}{2009}+\dfrac{2009}{2}+\dfrac{2009}{3}+...+\dfrac{2009}{2007}+\dfrac{2009}{2008}}\)
\(\dfrac{A}{B}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}}{2009\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}\right)}\)
\(\dfrac{A}{B}=\dfrac{1}{2009}\)
2) \(A=\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2.10^2}\)
\(A=\dfrac{2^2-1^2}{1^2.2^2}+\dfrac{3^2-2^2}{2^2.3^2}+\dfrac{4^2-3^2}{3^2.4^2}+...+\dfrac{10^2-9^2}{9^2.10^2}\)
\(A=1-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+\dfrac{1}{3^2}-\dfrac{1}{4^2}+...+\dfrac{1}{9^2}-\dfrac{1}{10^2}\)
\(A=1-\dfrac{1}{10^2}< 1\left(đpcm\right)\)
1)\(B=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2017}{2018}\)
\(B=\dfrac{1}{2018}\)
2)a)\(x^2-2x-15=0\)
\(\Leftrightarrow x^2-2x+1-16=0\)
\(\Leftrightarrow\left(x-1\right)^2-16=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)
3)\(\dfrac{a}{b}=\dfrac{d}{c}\)
\(\Rightarrow\dfrac{a^2}{b^2}=\dfrac{d^2}{c^2}=\dfrac{a}{b}\cdot\dfrac{d}{c}=\dfrac{ad}{bc}\)
Lại có:\(\dfrac{a^2}{b^2}=\dfrac{d^2}{c^2}=\dfrac{a^2+d^2}{b^2+c^2}\)
\(\Rightarrow\dfrac{a^2+d^2}{b^2+c^2}=\dfrac{ad}{bc}\)
4)Ta có:\(g\left(x\right)=-x^{101}+x^{100}-x^{99}+...+x^2-x+1\)
\(g\left(x\right)=-x^{101}+\left(x^{100}-x^{99}+...+x^2-x+1\right)\)
\(g\left(x\right)=-x^{101}+f\left(x\right)\)
\(\Rightarrow f\left(x\right)-g\left(x\right)=f\left(x\right)+x^{101}-f\left(x\right)=x^{101}\)
Tại x=0 thì f(x)-g(x)=0
Tại x=1 thì f(x)-g(x)=1
câu E
\(\left\{{}\begin{matrix}x\ne\dfrac{5}{2}\\\left(2x-5\right)\left(5-2x\right)=-\left(\dfrac{3}{2}\right)^4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{5}{2}\\\left|2x-5\right|=\left(\dfrac{3}{2}\right)^2\end{matrix}\right.\)
\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x< \dfrac{5}{2}\\2x-5=-\left(\dfrac{3}{2}\right)^2\Rightarrow x=\dfrac{11}{8}< \dfrac{5}{2}\left(n\right)\end{matrix}\right.\\\left\{{}\begin{matrix}x>\dfrac{5}{2}\\2x-5=\left(\dfrac{3}{2}\right)^2\Rightarrow x=\dfrac{29}{8}>\dfrac{5}{2}\left(n\right)\end{matrix}\right.\end{matrix}\right.\)
câu F (bạn cho vào lớp 7.2=lớp 14 nhé. )
a/ \(\dfrac{x+1}{100}+\dfrac{x+2}{99}=\dfrac{x+3}{98}+\dfrac{x+4}{97}\)
\(\Leftrightarrow\left(\dfrac{x+1}{100}+1\right)+\left(\dfrac{x+2}{99}+1\right)=\left(\dfrac{x+3}{98}+1\right)+\left(\dfrac{x+4}{97}+1\right)\)
\(\Leftrightarrow\dfrac{x+101}{100}+\dfrac{x+101}{99}=\dfrac{x+101}{98}+\dfrac{x+101}{97}\)
\(\Leftrightarrow\dfrac{x+101}{100}+\dfrac{x+101}{99}-\dfrac{x+101}{98}-\dfrac{x+101}{97}=0\)
\(\Leftrightarrow\left(x+101\right)\left(\dfrac{1}{100}+\dfrac{1}{99}-\dfrac{1}{98}-\dfrac{1}{97}\right)=0\)
Mà \(\dfrac{1}{100}+\dfrac{1}{99}-\dfrac{1}{98}-\dfrac{1}{97}\ne0\)
\(\Leftrightarrow x+101=0\)
\(\Leftrightarrow x=-101\)
Vậy...
b/ Đặt :
\(A=\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+.........+\dfrac{19}{9^2.10^2}\)
\(=\dfrac{2^2-1^2}{1^2.2^2}+\dfrac{3^2-2^2}{2^2.3^2}+....+\dfrac{10^2-9^2}{9^2.10^2}\)
\(=\dfrac{2^2}{1^2.2^2}-\dfrac{1^2}{1^2.2^2}+\dfrac{3^2}{2^2.3^2}-\dfrac{2^2}{2^2.3^2}+....+\dfrac{10^2}{9^2.10^2}-\dfrac{9^2}{9^2.10^2}\)
\(=1-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+...+\dfrac{1}{9^2}-\dfrac{1}{10^2}\)
\(=1-\dfrac{1}{10^2}< 1\)
\(\Leftrightarrow A< 1\left(đpcm\right)\)
Vậy...
c/ Với mọi x ta có :
\(\left|x-5\right|=\left|5-x\right|\)
\(\Leftrightarrow\left|x-10\right|+\left|x-5\right|=\left|x-10\right|+\left|5-x\right|\)
\(\Leftrightarrow A=\left|x-10\right|+\left|5-x\right|\)
\(\Leftrightarrow A\ge\left|x-10+5-x\right|\)
\(\Leftrightarrow A\ge5\)
Dấu "=" xảy ra
\(\Leftrightarrow\left(x-10\right)\left(5-x\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-10\ge0\\5-x\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x-10\le0\\5-x\le0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge10\\5\ge x\end{matrix}\right.\\\left\{{}\begin{matrix}x\le10\\5\le x\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x\in\varnothing\\5\le x\le10\end{matrix}\right.\)
Vậy..
đcmcm