tìm y : ( 1/2x3=1/3x4+ ........1/9x10):y=4/5
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\(\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{9\cdot10}\right)\cdot100-\left[\frac{5}{2}:\left(x+\frac{206}{100}\right)\right]:\frac{1}{2}=89\)
\(\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\right)\cdot100-\left[\frac{5}{2}:\left(x+\frac{103}{50}\right)\right]\cdot2=89\)
\(\left(1-\frac{1}{10}\right)\cdot100-\frac{5}{2}:\left(x+\frac{103}{50}\right)\cdot2=89\)
\(\frac{9}{10}\cdot100-\frac{5}{2}\cdot2:\left(x+\frac{103}{50}\right)=89\)
\(90-5\cdot\left(x+\frac{103}{50}\right)=89\)
\(5\cdot\left(x+\frac{103}{50}\right)=1\)
\(x+\frac{103}{50}=\frac{1}{5}\)
\(x=-\frac{93}{50}\)
1/1 x 2 + 1/2 x 3 + 1/3 x 4 + .... + 1/9 x 10
= 1 - 1/2 + 1/2 - 1/3 +1/3 - 1/4 + ... + 1/9 - 1/10
= 1 - 1/10
= 9/10
\(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+...+\dfrac{1}{9\times10}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\\ =1-\dfrac{1}{10}\\ =\dfrac{9}{10}\)
\(\Leftrightarrow2\left(x-\dfrac{1}{3}\right)\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)=\dfrac{3}{4}\)
\(\Leftrightarrow2\left(x-\dfrac{1}{3}\right)\left(1-\dfrac{1}{10}\right)=\dfrac{3}{4}\Leftrightarrow\dfrac{9}{10}\left(x-\dfrac{1}{3}\right)=\dfrac{3}{8}\)
\(\Leftrightarrow x-\dfrac{1}{3}=\dfrac{5}{12}\Leftrightarrow x=\dfrac{5}{12}+\dfrac{1}{3}=\dfrac{9}{12}=\dfrac{3}{4}\)
Ta đặt biểu thức là:
A = 1/1 x 2 + 1/2 x 3 + 1/3 x 4 + .... + 1/9 x 10
A = 1 - 1/2 + 1/2 - 1/3 +1/3 - 1/4 + ... + 1/9 - 1/10
A = 1 - 1/10
A = 9/10
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}\)
\(=\frac{9}{10}\)
1/2x3 + 1/3x4 + 1/4x5 + 1/5x6 + 1/ 6x7 + 1/7x8 + 1/8x9 + 1/9x10
= 2/5
bạn theo công thức nay nè
\(\frac{n}{a\left(a+n\right)}=\frac{1}{a}-\frac{1}{a+n}\)là giải đc thôi
\(\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+...+\frac{1}{9x10}\) \(=\) \(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\)
\(=\frac{1}{2}-\frac{1}{10}\)
\(=\frac{2}{5}\)
\(A=\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}....\frac{1}{9x10}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}...+\frac{1}{9}-\frac{1}{10}=\frac{9}{10}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}=\frac{1}{1}-\frac{1}{10}=\frac{9}{10}\)
y=1/2. Li ke nha!:))