2/32+2/42+....+2/20182
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Cho: \(A=\dfrac{2}{2^2}+\dfrac{2}{3^2}+\dfrac{2}{4^2}+....+\dfrac{2}{100^2}\)
\(A=2\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}\right)\)
Và cho \(B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}\)
Mà:
\(\dfrac{1}{2^2}< \dfrac{1}{1\cdot2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2\cdot3}\)
....
\(\dfrac{1}{100^2}< \dfrac{1}{99\cdot100}\)
Nên: \(B< \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+...+\dfrac{1}{99\cdot100}\)
\(\Rightarrow B< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(\Rightarrow B< 1-\dfrac{1}{100}\)
\(\Rightarrow B< \dfrac{99}{100}\)
Mà: \(\dfrac{99}{100}< 1\) (tử nhỏ hơn mẫu)
\(\Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}< 1\)
\(\Rightarrow A=2\cdot\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+..+\dfrac{1}{100^2}\right)< 2\) (đpcm)
\(\dfrac{2}{2^2}+\dfrac{2}{3^2}+\dfrac{2}{4^2}+...+\dfrac{2}{100^2}\)
\(=2\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}\right)\)
mà \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}< 1\)
\(\Rightarrow dpcm\)
\(42-2\left(32-2^{x+1}\right)=10\)
\(\Leftrightarrow2\left(32-2^{x+1}\right)=42-10=32\)
\(\Leftrightarrow32-2^{x+1}=16\)
\(\Leftrightarrow2^{x+1}=32-16=16\)
\(\Leftrightarrow2^{x+1}=2^4\)
\(\Leftrightarrow x+1=4\)
\(\Rightarrow x=3\)
\(\left[\left(46-32\right)^2-\left(54-42\right)^2\right]\times36-1872\)
\(=\left(14^2-12^2\right)\times36-1872\)
\(=\left(196-144\right)\times36-1872\)
\(=52\times36-1872\)
\(=1872-1872\)
\(=0\)
a) => (x+32) - 17 = 42:2 = 21
=> x+32 = 21+17 = 38
=> x=38-32=6
b) => 61+(53-x) = 1785:17=105
=> 53-x = 105-61=44
=> x = 53-44 =9
c) => x^2 +54 -32 = 244:2 = 122
=> x^2 +22 = 122
=> x^2 = 122-2=100
=> x= 10 hoặc -10
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