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1) 3x-(4+2x)=11
3x - 4 - 2x = 11
3x - 2x = 11+4
x = 15
Vậy x = 15.
2) 4x+5=11+2x
4x - 2x = 11 - 5
2x = 6
x = 6 : 2
x = 3
Vậy x = 3.
3) 7x-17=31+x
7x - x = 31 + 17
6x = 48
x = 48 : 6
x = 8
Vậy x = 8
# HOK TỐT #
\(3x-\left(4+2x\right)=11\)
\(3x-4-2x=11\)
\(x-15=0\)
\(x=15\)
\(4x+5=11+2x\)
\(2x-6=0\)
\(2x=6\)
\(x=3\)
Bn lm phần c đi , cố lên !
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
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x(7-2)=6^2+4
x5=40
x=8
vậy x=8
7x-x=5^21:5^19+3.2^2-7
x(7-1)=5^2+12-7
x6=30
x=5
vậy x=5
nhớ h cho mình
\(7x-2x=6^{17}:6^{15}+44:11\)
\(5x=6^2+4\)
\(5x=36+4\)
\(5x=40\)
\(x=40:5\)
\(x=8\)
\(7x-x=5^{21}:5^{19}+3.2^2-7\)
\(6x=5^2+3.4-7\)
\(6x=25+12-7\)
\(6x=37-7\)
\(6x=30\)
\(x=30:6\)
\(x=5\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)
\(\left(2x-15\right)^5=\left(2x-15\right)^3\\ \Leftrightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\\ \Leftrightarrow\left(2x-15\right)^3.\left[\left(2x-15\right)^2-1\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-15=0\\\left(2x-15\right)^2-1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x-15=0\\\left(2x-15-1\right).\left(2d-15+1\right)=0\end{matrix}\right.\\\Leftrightarrow\left[{}\begin{matrix}2x-15=0\\2x-16=0\\2x-14=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=8\\x=7\end{matrix}\right. \)
b) \(\left(7x-11\right)^3=\left(-3\right)^2.15+208\\ \Leftrightarrow\left(7x-11\right)^3=343=7^3\\ \Leftrightarrow7x-11=7\\ \Leftrightarrow x=\dfrac{18}{7}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
7x - 2x = 617 : 615 + 44 : 11
x(7 - 2) = 62 + 4
5x = 36 + 4
5x = 40
x = 40 : 5 = 8
![](https://rs.olm.vn/images/avt/0.png?1311)
`(11-x):(-5)=15`
`=> 11-x=-75`
`=> x=86`
Vậy `x = 86`
`((2x+3)^2022) . (-7x+84)=0`
`=> (2x+3)^2022 = 0` hoặc `-7x + 84 = 0`
`=> 2x+3=0` hoặc `-7x = -84`
`=> x = -3/2` hoặc `x = 12`
Vậy `x = -3/2` hoặc `x = 12`
`(x-3)(x+1) < 0`
Ta có: `x - 3 < x + 1`
nên: `x - 3 < 0` và `x + 1 > 0`
`=> x < 3 và x > -1`
`=> -1 < x < 3`
Vậy `-1 <x < 3`
#\(N\)
`a, (11-x)`\(\div\)`(-5)=15`
`11-x=15.-5`
`11-x=-75`
`x=11-(-75)`
`x=11+75=86`
`b, (2x-3)^2022.(-7x+84)=0`
`=>`\(\left\{{}\begin{matrix}\left(2x-3\right)^{2022}=0\\\left(-7x+84\right)=0\end{matrix}\right.\)
`=>` \(\left\{{}\begin{matrix}2x-3=0\\-7x+84=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2x=3+0\\-7x=0-84\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2x=3\\-7x=-84\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=3\div2\\x=-84\div-7\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{3}{2}\\x=12\end{matrix}\right.\)
`c, (x-3) (x+1) <0`
`-> (x-3) < x+1`
`-> (x-3)<0 , (x+1)>0`
`-> x < 3 , x> (-1)`
`-> 3 > x > -1`