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Câu hỏi của Lê Tiến Cường - Toán lớp 6 - Học toán với OnlineMath
\(A=\frac{3}{2}+\frac{7}{6}+\frac{13}{12}+...+\frac{10101}{10100}=\frac{2+1}{2}+\frac{6+1}{6}+\frac{12+1}{12}+...+\frac{10100+1}{10100}\)
\(A=\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{6}\right)+\left(1+\frac{1}{12}\right)+....+\left(1+\frac{1}{10100}\right)\)
\(A=\left(1+\frac{1}{1\times2}\right)+\left(1+\frac{1}{2\times3}\right)+\left(1+\frac{1}{3\times4}\right)+...+\left(1+\frac{1}{100\times101}\right)\)
\(A=\left(1+1+1+....+1\right)+\left(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{100\times101}\right)\)
\(A=100+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{100}-\frac{1}{101}\right)\)
\(A=100+1-\frac{1}{101}=101-\frac{1}{101}< 101=B\)
\(\Rightarrow A< B\)
So easy
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C=\(\frac{101+100+...+3+2+1}{101-100+...+3-2+1}\)
=\(\frac{\left(101+1\right).101:2}{\left(101-100\right)+...+\left(3-2\right)+1}\) (nhóm 2 số hạng ở MS thì sẽ có 51 nhóm và dư 1 số hang )
=\(\frac{102.101:2}{1+...+1+1}\) ( Ms có 51 số 1)
=\(\frac{51.101}{51}\)=101
D=\(\frac{3737.43-4343.37}{2+4+6+...+100}\)
= \(\frac{37.101.43-43.101.37}{2+4+6+..+100}\)
= \(\frac{0}{2+4+6+...+100}\)
=0
Tick mik nha, thks bạn
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ĐKXĐ : 101x \(\ge\)0 nên x \(\ge\)0
khi đó : \(\left|x+\frac{1}{101}\right|+\left|x+\frac{2}{101}\right|+...+\left|x+\frac{100}{101}\right|=101x\)
\(\Leftrightarrow\left(x+\frac{1}{101}\right)+\left(x+\frac{2}{101}\right)+...+\left(x+\frac{100}{101}\right)=101x\)
\(\Leftrightarrow100x+\frac{5050}{101}=101x\Leftrightarrow x=50\)
*ĐK : 101x\(\ge\) 0 => x\(\ge\)0
=> \(x+\frac{1}{101}\ge\frac{1}{101}>0\Rightarrow\left|x+\frac{1}{101}\right|=x+\frac{1}{101}\)
\(x+\frac{2}{101}\ge\frac{2}{101}>0\Rightarrow\left|x+\frac{2}{101}\right|=x+\frac{2}{101}\)
...
\(x+\frac{100}{101}\ge\frac{100}{101}>0\Rightarrow\left|x+\frac{100}{101}\right|=x+\frac{100}{101}\)
Ta có :
\(x+\frac{1}{101}+x+\frac{2}{101}+...+x+\frac{100}{101}=101x\)
<=> \(100x+\left(\frac{1+2+...+100}{101}\right)=101x\)
<=> \(100x+\frac{5050}{101}=101x\)
<=> \(100x-101x=\frac{-5050}{101}\)
<=> -x = -50
=> x = 50 ( t/m x >/ 0)
![](https://rs.olm.vn/images/avt/0.png?1311)
D=\(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^2}+...+\frac{100}{3^{100}}+\frac{101}{3^{101}}\)
D=\(\frac{1}{3}+\frac{101}{3^{101}}\)
D=\(\frac{1}{3}\)
\(\frac{1}{3}và\frac{3}{4}\)
\(\frac{1}{3}=\frac{4}{12}\)
\(\frac{3}{4}=\frac{9}{12}\)
Vì\(\frac{4}{12}< \frac{9}{12}Vậy\frac{1}{3}< \frac{3}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(M=\frac{101^{102}+1}{101^{103}+1}\)
\(101M=\frac{101^{103}+1+100}{101^{103}+1}=1+\frac{100}{101^{103}+1}\)
Ta lại có:
\(N=\frac{101^{103}+1}{101^{104}+1}\)
\(101N=\frac{101^{104}+1+100}{101^{104}+1}=1+\frac{100}{101^{104}+1}\)
Vì \(\frac{100}{101^{104}+1}< \frac{100}{101^{103}+1}\Rightarrow101N< 101M\Rightarrow N< M\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: M =\(\frac{101^{102}+1}{101^{103}+1}=\frac{101^{103}+101}{101^{104}+101}=\frac{101^{103}+1+100}{101^{104}+1+100}\)
Mà : N = \(\frac{101^{103}+1}{101^{104}+1}\)< M = \(\frac{101^{103}+1+100}{101^{104}+1+100}\)
\(\Rightarrow N< M\)