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a) \(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}< 1\)
\(\Rightarrow A< 1\)
b) \(B=\frac{1}{3}+\left(\frac{1}{3}\right)^2+...+\left(\frac{1}{3}\right)^{100}\)
\(\Rightarrow3B=1+\frac{1}{3}+...+\left(\frac{1}{3}\right)^{99}\)
\(\Rightarrow3B-B=1-\left(\frac{1}{3}\right)^{100}\)
\(\Rightarrow2B=1-\left(\frac{1}{3}\right)^{100}< 1\)
\(\Rightarrow2B< 1\)
\(\Rightarrow B< \frac{1}{2}\)
Ta có công thức :
\(\frac{1}{k\left(k+1\right)}=\frac{\left(k+1\right)-k}{k\left(k+1\right)}=\frac{k+1}{k\left(k+1\right)}-\frac{k}{k\left(k+1\right)}=\frac{1}{k}-\frac{1}{k+1}\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{n-1}-\frac{1}{n}\)
\(=1-\frac{1}{n}=\frac{n-1}{n}\)
Vì GTTĐ luôn lớn hơn hoặc bằng 0 với mọi x
\(\Rightarrow\left|x+\frac{1}{1\cdot2}\right|+\left|x+\frac{1}{2\cdot3}\right|+...+\left|x+\frac{1}{99\cdot100}\right|\ge0\)
\(\Rightarrow100x\ge0\)
\(\Rightarrow x\ge0\)
Từ điều kiện trên ta có :
\(x+\frac{1}{1\cdot2}+x+\frac{1}{2\cdot3}+...+x+\frac{1}{99\cdot100}=100x\)
\(50x+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)=100x\)
\(50x=1-\frac{1}{100}\)
\(50x=\frac{99}{100}\)
\(x=\frac{99}{5000}\)
Do \(\left|a\right|\ge0\forall a\) nên:
\(A=\left|x+\frac{1}{1.2}\right|+\left|x+\frac{1}{2.3}\right|+...+\left|x+\frac{1}{99.100}\right|\ge0\forall x\)
\(\Leftrightarrow100x\ge0\) hay \(x\ge0\)
Do vậy ta có: \(A=\left(x+x+...+x\right)+\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)=100x\) ( 50 chữ số x)
\(\Leftrightarrow A=50x+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)=100x\)
\(\Leftrightarrow50x+\left(1-\frac{1}{100}\right)=100x\Leftrightarrow50x+\frac{99}{100}=100x\)
\(\Leftrightarrow50x=\frac{99}{100}\Leftrightarrow x=\frac{99}{100.50}=\frac{99}{5000}\)
Ta có : 1/ 1.2 + 1/ 2.3 + 1/ 3.4 + ... + 1/ n.( n + 1 ) .
= 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ..... + 1/n - 1/ n+1 .
= 1 - 1/ n + 1 .
= n+1 / n+1 - 1/ n+1 .
= n/ n+1 .
Đáp sô : n/ n+1
phúc hơi phức tạp
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x.\left(x+1\right)}=\frac{2008}{2009}\)
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2008}{2009}\)
\(1-\frac{1}{x+1}=\frac{2008}{2009}\)
\(\frac{1}{x+1}=1-\frac{2008}{2009}\)
\(\frac{1}{x+1}=\frac{1}{2009}\)
\(\Rightarrow x+1=2009\)
\(x=2009-1\)
\(x=2008\)
Vậy \(x=2008\)
Tự làm bước biến đổi nhé tui lm lẹ luôn =v
\(\frac{1}{1}-\frac{1}{x+1}=\frac{2008}{2009}\)
\(\frac{x+1}{x+1}-\frac{1}{x+1}=\frac{2008}{2009}\)
\(\frac{x}{x+1}=\frac{2008}{2009}\)
\(=>x=2008\)
Vậy x = 2008
a, \(\left(\frac{1}{2}\right)^x+\left(\frac{1}{2}\right)^{x+4}=17\)
\(\Rightarrow\frac{1}{2^x}+\frac{1}{2^x}\cdot\frac{1}{16}=17\)
\(\Rightarrow\frac{1}{2^x}\left(1+\frac{1}{16}\right)=17\)
\(\Rightarrow\frac{1}{2^x}\cdot\frac{17}{16}=17\)
\(\Rightarrow\frac{1}{2^x}=17:\frac{17}{16}=\frac{1}{16}=\frac{1}{2^4}\)
=> x = 4
b, Ta có: \(\left|x+\frac{1}{1.2}\right|\ge0;\left|x+\frac{1}{2.3}\right|\ge0;....;\left|x+\frac{1}{99.100}\right|\ge0\)
\(\Rightarrow\left|x+\frac{1}{1.2}\right|+\left|x+\frac{1}{2.3}\right|+...+\left|x+\frac{1}{99.100}\right|\ge0\)
\(\Rightarrow100x\ge0\Rightarrow x\ge0\)
\(\Rightarrow x+\frac{1}{1.2}+x+\frac{1}{2.3}+...+x+\frac{1}{99.100}=100x\)
\(\Rightarrow\left(x+x+...+x\right)+\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)=100x\)
\(\Rightarrow99x+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}=100x\)
\(\Rightarrow100x-99x=1-\frac{1}{100}\)
\(\Rightarrow x=\frac{99}{100}\)
Thay n = a nha / lúc trước có giải r nên ko giải lại rắc rối
\(N=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\cdot\cdot\cdot+\frac{1}{a\left(a+1\right)}\)
\(\Rightarrow N=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\cdot\cdot\cdot+\frac{1}{a}-\frac{1}{a-1}\)
\(\Rightarrow N=1-\frac{1}{a-1}\)
\(\Rightarrow N=\frac{a-1-1}{a-1}\)
\(\Rightarrow N=\frac{a-2}{a-1}\)
Ta có : \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{a\left(a+1\right)}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{a}-\frac{1}{a+1}\)
\(=1-\frac{1}{a+1}\)
\(=\frac{a+1}{a+1}-\frac{1}{a+1}=\frac{a}{a+1}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{a\left(a+1\right)}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{a}-\frac{1}{a+1}\)
\(=1-\frac{1}{a+1}\)
\(=\frac{a+1}{a+1}-\frac{1}{a+1}\)
\(=\frac{a}{a+1}\)