Tìm x biết (2.x-1)2=36
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\(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(\Rightarrow\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(\Rightarrow\frac{2}{6\cdot7}+\frac{2}{7\cdot8}+\frac{2}{8\cdot9}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(\Rightarrow2\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2}{9}\)
\(\Rightarrow2\left(\frac{1}{6}-\frac{1}{x+1}\right)=\frac{2}{9}\)
\(\Rightarrow\frac{1}{6}-\frac{1}{x+1}=\frac{2}{9}\div2\)
\(\Rightarrow\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{6}-\frac{1}{9}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{18}\)
\(\Rightarrow x+1=18\Rightarrow x=18-1\Rightarrow x=17\)
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=>-(1+2+...+x)=-36
=>1+2+...+x=36
\(\Leftrightarrow x=8\)
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Ta có : \(\frac{1}{21}+\frac{1}{28}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(\Rightarrow\frac{2}{42}+\frac{2}{56}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(\Rightarrow2\left(\frac{1}{42}+\frac{1}{56}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2}{9}\)
\(\Rightarrow2\left(\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2}{9}\)
\(\Rightarrow2\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2}{9}\)
\(\Rightarrow2\left(\frac{1}{6}-\frac{1}{x+1}\right)=\frac{2}{9}\)
\(\Rightarrow\frac{1}{6}-\frac{1}{x+1}=\frac{2}{9}:2=\frac{1}{9}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{6}-\frac{1}{9}=\frac{1}{18}\)
\(\Rightarrow x+1=18\Rightarrow x=17\)
Vậy x = 17
\(\Rightarrow\dfrac{2}{42}+\dfrac{2}{56}+\dfrac{2}{72}+.....+\dfrac{2}{x\left(x+1\right)}\Rightarrow2\left(\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+.....+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{2}{9}\\ \Rightarrow2\left(\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+....+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{2}{9}\\ \Rightarrow2\left(\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+....+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{2}{9}\\ \Rightarrow2\left(\dfrac{1}{6}-\dfrac{1}{x+1}\right)=\dfrac{2}{9}\\ \Rightarrow\dfrac{1}{6}-\dfrac{1}{x+1}=\dfrac{2}{9}:2\\ \Rightarrow\dfrac{1}{6}-\dfrac{1}{x+1}=\dfrac{2}{9}.\dfrac{1}{2}\\ \Rightarrow\dfrac{1}{6}-\dfrac{1}{x+1}=\dfrac{1}{9}\\ \Rightarrow\dfrac{1}{x+1}=\dfrac{1}{6}-\dfrac{1}{9}\\ \Rightarrow\dfrac{1}{x+1}=\dfrac{3}{54}\\ \Rightarrow x+1=\dfrac{54}{3}\\ \Rightarrow x=\dfrac{54}{3}-1=\dfrac{51}{3}\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \)
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`@` `\text {Ans}`
`\downarrow`
`1,`
`x^2 - 9 = 0`
`<=> x^2 = 0 + 9`
`<=> x^2 = 9`
`<=> x^2 = (+-3)^2`
`<=> x = +-3`
Vậy, `S = {3; -3}`
`2,`
`25 - x^2 = 0`
`<=> x^2 = 25 - 0`
`<=> x^2 = 25`
`<=> x^2 = (+-5)^2`
`<=> x = +-5`
Vậy,` S= {5; -5}`
`3,`
`-x^2 + 36 = 0`
`<=> -x^2 = 0 - 36`
`<=> -x^2 = -36`
`<=> x^2 = 36`
`<=> x^2 = (+-6)^2`
`<=> x = +-6`
Vậy, `S= {6; -6}`
`4,`
`4x^2 - 4 = 0`
`<=> 4x^2 = 0+4`
`<=> 4x^2 = 4`
`<=> x^2 = 4 \div 4`
`<=> x^2 = 1`
`<=> x^2 = (+-1)^2`
`<=> x = +-1`
Vậy, `S= {1; -1}`
`@` `\text {Kaizuu lv uuu}`
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+....+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(\frac{2}{6.7}+\frac{2}{7.8}+\frac{2}{8.9}+.....+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(2\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+....+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2}{9}\)
\(\frac{1}{6}-\frac{1}{x+1}=\frac{2}{9}:2=\frac{1}{9}\)
\(\frac{1}{x+1}=\frac{1}{6}-\frac{1}{9}=\frac{1}{18}\)
=> x+1 = 18
=> x = 18 - 1
=> x = 17
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2(x+1)=36-12
2(x+1)=24
x+1=24:2
x+1=12
x=12-1
x=11
Vay x=11
Ban cho minh dung nha