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29 tháng 7 2018
A) 7/38 x 9/11 +7/38 x 4/11 -7/38 x 2/11
=7/38.(9/11+4/11-2/11)
=7/38
B) 5/31 x 21/25 + 5/31 x -7/10 - 5/31 x 9/20
=5/31.(21/25-7/10-9/20)
=5/31.(-31/100)
=-1/20
26 tháng 3 2022
\(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x\left(x+2\right)}=\frac{11}{75}\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{11}{75}\)
\(\Leftrightarrow\frac{1}{3}-\frac{1}{x+2}=\frac{11}{75}:\frac{1}{2}=\frac{22}{75}\Leftrightarrow\frac{1}{x+2}=\frac{1}{25}\Leftrightarrow x=23\)
TT
2
4 tháng 8 2021
a) Ta có: \(3\left(x+5\right)-3=2\left(x+1\right)+7\)
\(\Leftrightarrow3x+15-3=2x+2+7\)
\(\Leftrightarrow3x+12=2x+9\)
hay x=-3
b) Ta có: \(15-\left(x+2\right)^2=-1\)
\(\Leftrightarrow\left(x+2\right)^2=16\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=4\\x+2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-6\end{matrix}\right.\)
c) Ta có: \(75-\left(2x+1\right)^3=11\)
\(\Leftrightarrow\left(2x+1\right)^3=64\)
\(\Leftrightarrow2x+1=4\)
hay \(x=\dfrac{3}{2}\)
LN
3
TA
4 tháng 8 2021
`3(x+5)-3=2(x+1)+7`
`3x+15-3=2x+2+7`
`x=-3`
.
`15-(x+2)^2=-1`
`(x+2)^2=16`
`(x+2)^2=4^2=(-4)^2`
`[(x+2=4),(x+2=-4):}`
`[(x=2),(x=-6):}`
.
`75-(2x+1)^3=11`
`(2x+1)^3=64`
`(2x+1)^2=4^3`
`2x+1=4`
`x=3/2`
TN
5
14 tháng 9 2017
[x+1]:25=3
x+1=25*3
x+1=75
x=75-1=74
12:[x-3]=6x-3=12:6
x-3=2
x=3+2
x=5
4[x-1]:5=8
4[x-1]=5.8
4[x-1]=40
x-1=40:4
x-1=10
x=10+1=11
9[x+3]-25=11
9[x+3]=25+11=36
x+3=36:9=4
x=4-3=1
75:[x-2]=1
x-2=75:1=75
x=75+2=77
[2x-5]-7=6
2x-5=7+6=13
2x=13+5=18
x=18:2=9
k mk nha mk cầu xin bn đó công mk làm mà
10 tháng 5 2023
2: \(=\dfrac{-2}{75}+\dfrac{5}{39}=\dfrac{33}{325}\)
3: \(=\dfrac{6}{11}\left(\dfrac{4}{9}+\dfrac{5}{9}\right)=\dfrac{6}{11}\)
4: \(=\dfrac{7}{19}\left(\dfrac{5}{13}+\dfrac{8}{13}-1\right)=-2\cdot\dfrac{7}{19}=-\dfrac{14}{19}\)
5: \(=\dfrac{2}{7}\left(\dfrac{4}{23}-\dfrac{27}{23}+1\right)=0\)
6: \(=\dfrac{3}{8}\left(\dfrac{3}{7}+\dfrac{4}{7}\right)+\dfrac{11}{8}=\dfrac{3}{8}+\dfrac{11}{8}=\dfrac{14}{8}=\dfrac{7}{4}\)
\(\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+\dfrac{1}{7\cdot9}+...+\dfrac{1}{x\left(x+2\right)}=\dfrac{11}{75}\left(x\ne0;x\ne-2\right)\)
\(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+...+\dfrac{2}{x\left(x+2\right)}=\dfrac{22}{75}\)
\(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{x}-\dfrac{1}{x+2}=\dfrac{22}{75}\)
\(\dfrac{1}{3}-\dfrac{1}{x+2}=\dfrac{22}{75}\)
\(\dfrac{1}{x+2}=\dfrac{1}{3}-\dfrac{22}{75}\)
\(\dfrac{1}{x+2}=\dfrac{1}{25}\)
suy ra
\(x+2=25\\ x=23\left(tm\right)\)
=>2/3*5+2/5*7+...+2/x(x+2)=22/75
=>1/3-1/5+1/5-1/7+...+1/x-1/x+2=22/75
=>1/3-1/x+2=22/75
=>1/x+2=25/75-22/75=3/75=1/25
=>x=23