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\(LINH_1=\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+....+\dfrac{1}{49.50}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
\(=\left(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{49}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+...+\dfrac{1}{50}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+....+\dfrac{1}{49}+\dfrac{1}{50}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+...+\dfrac{1}{50}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+...+\dfrac{1}{49}+\dfrac{1}{50}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{25}\right)\)\(=\dfrac{1}{26}+\dfrac{1}{27}+\dfrac{1}{28}+...+\dfrac{1}{50}=LINH_2\left(đpcm\right)\)
Lời giải:
\(\text{VT}=\frac{1}{1.2}+\frac{1}{3.4}+....+\frac{1}{49.50}\)
\(=\frac{2-1}{1.2}+\frac{4-3}{3.4}+....+\frac{50-49}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}....+\frac{1}{50}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{49}+\frac{1}{50}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}....+\frac{1}{50}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{49}+\frac{1}{50}-\left(1+\frac{1}{2}+\frac{1}{3}....+\frac{1}{25}\right)\)
\(=\frac{1}{26}+\frac{1}{27}+....+\frac{1}{49}+\frac{1}{50}\)
Do đó ta có đpcm.
Bài 1:
a: \(A=\left(-\dfrac{1}{5}\right)^{33}:\left(-\dfrac{1}{5}\right)^{32}=\dfrac{-1}{5}\)
c: \(C=\dfrac{2^{12}\cdot3^{10}+3^9\cdot2^9\cdot2^3\cdot3\cdot5}{2^{12}\cdot3^{12}+2^{11}\cdot3^{11}}\)
\(=\dfrac{2^{12}\cdot3^{10}\left(1+5\right)}{2^{11}\cdot3^{11}\cdot7}=\dfrac{2}{3}\cdot\dfrac{6}{7}=\dfrac{12}{21}=\dfrac{4}{7}\)
Bài 2:
a: =>x^2=60
=>\(x=\pm2\sqrt{15}\)
b: =>2^2x+3=2^3x
=>3x=2x+3
=>x=3
c: \(\Leftrightarrow\sqrt{\dfrac{1}{2}x-2}\cdot\dfrac{1}{2}=1\)
\(\Leftrightarrow\sqrt{\dfrac{1}{2}x-2}=2\)
=>1/2x-2=4
=>1/2x=6
=>x=12
Vì \(\left|x+\dfrac{1}{1\cdot2}\right|+\left|x+\dfrac{1}{2\cdot3}\right|+...+\left|x+\dfrac{1}{99\cdot100}\right|\ge0\forall x\)
\(\Rightarrow100x\ge0\Rightarrow x\ge0\)
\(\Rightarrow\left|x+\dfrac{1}{1\cdot2}\right|+...+\left|x+\dfrac{1}{99\cdot100}\right|=x+\dfrac{1}{1\cdot2}+...+x+\dfrac{1}{99\cdot100}\)
\(\Rightarrow\left(x+x+...+x\right)+\left(\dfrac{1}{1\cdot2}+...+\dfrac{1}{99\cdot100}\right)=100x\)
\(\Rightarrow99x+\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\right)=100x\)
\(\Rightarrow\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}=x\)
\(\Rightarrow1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}=x\)
\(\Rightarrow x=1-\dfrac{1}{100}=\dfrac{99}{100}\)
\(a)\dfrac{1}{4}-\dfrac{3}{4}:\left(\dfrac{-5}{8}\right)\)
\(=\dfrac{1}{4}-\dfrac{3}{4}.\dfrac{-8}{5}\)
\(=\dfrac{1}{4}-\dfrac{-6}{5}\)
\(=\dfrac{5}{20}+\dfrac{24}{20}\)
\(=\dfrac{29}{20}\)
\(b)3-\left(\dfrac{-6}{7}\right)^0+\sqrt{\dfrac{1}{16}}:2\)
\(=3-1+\sqrt{\left(\dfrac{1}{4}\right)^2}:2\)
\(=2+\dfrac{1}{4}.\dfrac{1}{2}\)
\(=\dfrac{16}{8}+\dfrac{1}{8}\)
\(=\dfrac{17}{8}\)
\(c)\dfrac{9^5.2^6}{4^3.3^8}=\dfrac{\left(3^2\right)^5.2^6}{\left(2^2\right)^3.3^8}=\dfrac{3^{10}.2^6}{2^6.3^8}=3^2=9\)
= \(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+.....+\dfrac{1}{5}-\dfrac{1}{6}\)
= 1 - \(\dfrac{1}{6}\)
= \(\dfrac{5}{6}\)
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