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\(\text{Ta có: }\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+.....+\frac{3}{\left(x+2\right)\left(x+5\right)}=\frac{3}{20}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+.....+\frac{1}{\left(x+2\right)}-\frac{1}{\left(x+5\right)}=\frac{3}{20}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{\left(x+5\right)}=\frac{3}{20}\)
\(\Rightarrow\frac{1}{\left(x+5\right)}=\frac{1}{2}-\frac{3}{20}\)
ta có: \(2x-1=2\left(x-3\right)+5\)
để \(2x-1⋮x-3\Rightarrow2\left(x-3\right)+5⋮x-3\\ m\text{à }x.nguy\text{ê}n\Rightarrow x-3nguy\text{ê}n\\ \Rightarrow x-3\in\text{Ư}\left(5\right)=\left\{-5;5;1;-1\right\}\)
ta có bảng sau :
| x-3 | -5 | 5 | -1 | 1 |
| x | -2 | 2 | 4 | 8 |
\(\Leftrightarrow2.\left(x-3\right)+5⋮x-3\)
\(do2.\left(x-3\right)⋮x-3\)
\(\Leftrightarrow5⋮x-3\)
\(\Leftrightarrow x-3\inƯ\left(5\right)=\left\{-5;-1;1;5\right\}\)
\(\Leftrightarrow x\in\left\{-2;2;4;8\right\}\)
a,(x+1)-(x+2)-(x+3)=24
=>x+1-x-2-x-3 =24
=>(x-x-x)+(1-2-3) =24
=> -x-4 =24
=> -x =24+4
=> -x =28
=> x =-28
Vậy x=-28
b,4x+2-3(x-1)=3x-5
=>4x+2-3x+3=3x-5
=>3x-4x+3x =2+3+5
=>2x =10
=>x =5
Vậy x=5
c,x-1-2(x-2)=x-11
=>x-1-2x+4=x-11
=>x-2x-x =-11+1-4
=>-2x =-14
=>x =7
Vậy x = 7
a) \(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+100\right)=5750\)
\(\Rightarrow\left(x+x+x+...+x\right)+\left(1+2+3+..+100\right)=5750\Rightarrow x.100+\left(100+1\right)\cdot100:2=5750\)\
\(\Rightarrow x.100+5050=5750\Rightarrow x.100=700\Rightarrow x=7\)
b) \(\frac{x+1}{2}=\frac{8}{x+1}\Rightarrow\left(x+1\right)\left(x+1\right)=2.8\)
\(\Rightarrow\left(x+1\right)^2=16\Rightarrow\left(x+1\right)^2=4^2\)
\(\Leftrightarrow x+1=4\Rightarrow x=3\)
1.\(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+100\right)=5750\)
\(\Leftrightarrow\left(x+x+x+...+x\right)+\left(1+2+3+...+100\right)=5750\)
\(\Leftrightarrow100x+5050=5750\)
\(\Leftrightarrow100x=5750-5050=700\)
\(\Leftrightarrow x=700:100=7\)
2. \(\frac{x+1}{2}=\frac{8}{x+1}\)
\(\Leftrightarrow\left(x+1\right).\left(x+1\right)=8.2\)
\(\Leftrightarrow\left(x+1\right).\left(x+1\right)=16\)
\(\Leftrightarrow\left(x+1\right)^2=16\)
\(\Leftrightarrow\left(x+1\right)=16:2\)
\(\Leftrightarrow\left(x+1\right)=8\)
\(\Leftrightarrow x=8-1=7\)
\(a,x+1-\dfrac{1}{2}=\dfrac{-1}{3}\)
\(x+1=\dfrac{-1}{3}+\dfrac{1}{2}\)
\(x+1=\dfrac{-2}{6}+\dfrac{3}{6}\)
\(x+1=\dfrac{1}{6}\)
\(x=\dfrac{1}{6}-1\)
\(x=\dfrac{1}{6}-\dfrac{6}{6}\)
\(x=\dfrac{-5}{6}\)
Vậy \(x=\dfrac{-5}{6}\)
x + 1/2 = -1/3 x = -1/3 - 1/2 x = -5/6