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Bài 2:
a: x+5/3=x-6/7
=>0x=-6/7-5/3(vô lý)
b: \(\Leftrightarrow x-10\left(\dfrac{2}{11\cdot13}+\dfrac{2}{13\cdot15}+...+\dfrac{2}{53\cdot55}\right)=\dfrac{3}{11}\)
\(\Leftrightarrow x-10\left(\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}+...+\dfrac{1}{53}-\dfrac{1}{55}\right)=\dfrac{3}{11}\)
\(\Leftrightarrow x-10\cdot\dfrac{4}{55}=\dfrac{3}{11}\)
hay x=3/11+8/11=1
![](https://rs.olm.vn/images/avt/0.png?1311)
a) x+15 = 20-4x
=> x=1
b) 17-x=7-6x
=> x=-2
c) -12+x=5x-20
=> x=2
d) 4x-5=15-x
=> x=4
e) 9x-7=20-6x
=>x= \(\frac{9}{5}\)
g)2.(x-10)=3.(x-20)=x-4
=> x thuộc ∅
h) (x^2+2).(x-3) <0
=> x=3,...
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x-\frac{20}{11.13}-\frac{20}{13.15}-\frac{20}{15.17}-...-\frac{20}{53.55}=\frac{3}{11}\)
\(x-\left(\frac{20}{11.13}+\frac{20}{13.15}+\frac{20}{15.17}+...+\frac{20}{53.55}\right)=\frac{3}{11}\)
\(x-\left[10.\left(\frac{2}{11.13}+\frac{2}{13.15}+\frac{2}{15.17}+...+\frac{2}{53.55}\right)\right]=\frac{3}{11}\)
\(x-\left[10.\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}+...+\frac{1}{53}-\frac{1}{55}\right)\right]=\frac{3}{11}\)
\(x-\left[10.\left(\frac{1}{11}-\frac{1}{55}\right)\right]=\frac{3}{11}\)
\(x-\left[\frac{10}{11}-\frac{10}{55}\right]=\frac{3}{11}\)
\(x-\left[\frac{10}{11}-\frac{2}{11}\right]=\frac{3}{11}\)
\(x-\frac{8}{11}=\frac{3}{11}\)
\(x=\frac{3}{11}+\frac{8}{11}=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Vì \(\dfrac{x+5}{3}\)= \(\dfrac{x-6}{7}\) nên 7(x+5) = 3(x-6)
=> 7x+ 35 = 3x - 18
7x - 3x = -18 -35
4x = -53
x = -53:4
x = \(\dfrac{-53}{4}\)
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Bài 2:
d, 43-(24-x)=20
<=>43-24+x=20
<=>19+x=20
<=>x=20-19=1
Vậy x=1
e, 20-(x+14)=5
<=> 20-x-14=5
<=>6-x=5
<=>x=6-5=1
Vậy x=1
f,15:(x-3)=3
<=> x-3=15:3=5
<=>x=5+3
<=>x=8
Vậy x=8
\(d,43-\left(24-x\right)=20\)
\(24-x=23\)
\(x=1\)
\(e,20-\left(x+14\right)=5\)
\(x+14=15\)
\(x=1\)
\(f,15:\left(x-3\right)=3\)
\(x-3=5\)
\(x=8\)
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a) \(x-\left(\frac{20}{11.13}+\frac{20}{13.15}+\frac{20}{15.17}+...+\frac{20}{53.55}\right)=\frac{3}{11}\)
\(x-\frac{20}{2}.\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}+...+\frac{1}{53}-\frac{1}{55}\right)=\frac{3}{11}\)
\(x-10.\left(\frac{1}{11}-\frac{1}{55}\right)=\frac{3}{11}\)
\(x-10.\frac{4}{55}=\frac{3}{11}\)
\(x-\frac{8}{11}=\frac{3}{11}\)
x = 1
b) \(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x.\left(x+1\right)}=\frac{2}{9}\)
\(\Rightarrow\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+...+\frac{2}{x.\left(x+1\right)}=\frac{2}{9}\) ( nhân cho cả tử và mẫu của các số hạng trên ( ngoại trừ 2/x.(x+1) ) là 2)
\(\frac{2}{6.7}+\frac{2}{7.8}+\frac{2}{8.9}+...+\frac{2}{x.\left(x+1\right)}=\frac{2}{9}\)
\(2.\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2}{9}\)
\(2.\left(\frac{1}{6}-\frac{1}{x+1}\right)=\frac{2}{9}\)
\(\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)
\(\frac{1}{x+1}=\frac{1}{18}\)
=> x + 1 = 18
x = 17
\(a,x-\left(\frac{20}{11.13}+\frac{20}{13.15}+...+\frac{20}{53.55}\right)=\frac{3}{11}\)
\(\Rightarrow x-10\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{53}-\frac{1}{55}\right)=\frac{3}{11}\)
\(\Rightarrow x-10\left(\frac{1}{11}-\frac{1}{55}\right)=\frac{3}{11}\)
\(\Rightarrow x-\frac{8}{11}=\frac{3}{11}\)
\(\Rightarrow x=1\)
\(b,\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(\frac{2}{42}+\frac{2}{56}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(2\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2}{9}\)
\(\Rightarrow\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)
\(\Rightarrow\frac{1}{18}=\frac{1}{x+1}\)
\(\Rightarrow x+1=18\Leftrightarrow x=17\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x-\left(\frac{20}{11.13}+\frac{20}{13.15}+\frac{20}{15.17}+...+\frac{20}{53.55}\right)=\frac{3}{11}.\)
\(x-\frac{20}{2}.\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}+...+\frac{1}{53}-\frac{1}{55}\right)=\frac{3}{11}\)
\(x-10.\left(\frac{1}{11}-\frac{1}{55}\right)=\frac{3}{11}\)
\(x-10\cdot\frac{4}{55}=\frac{3}{11}\)
\(x-\frac{8}{11}=\frac{3}{11}\)
x = 1
\(x-3=20\\ x=20+3\\ x=23\)
Vậy \(x=23\)
|x-3|=20
=>\(\left[{}\begin{matrix}x-3=20\\x-3=-20\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=23\\x=-17\end{matrix}\right.\)