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\(\left(-3x+2\right)-\left(5-3x\right)=-3\)
\(\Rightarrow-3x+2-5+3x=-3\)
\(\Rightarrow-3x+3x=-3+5-2\)
\(\Rightarrow0x=0\Rightarrow x\in Z\)
\(3+x-\left(3x-1\right)=6-2x\)
\(\Rightarrow3+x-3x+1=6-2x\)
\(\Rightarrow x-3x+2x=6-1-3\)
\(\Rightarrow0x=2\left(loại\right)\)
\(\left(x-5\right)\left(3x+4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\3x+4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=5\\x=-\frac{4}{3}\end{cases}}}\)
\(7x\left(2x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}7x=0\\2x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{2}\end{cases}}}\)
\(\left(3x-1\right)2x=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x-1=0\\2x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=0\end{cases}}}\)
a: Ta có: \(7x+25=144\)
\(\Leftrightarrow7x=119\)
hay x=17
b: Ta có: \(33-12x=9\)
\(\Leftrightarrow12x=24\)
hay x=2
c: Ta có: \(128-3\left(x+4\right)=23\)
\(\Leftrightarrow3\left(x+4\right)=105\)
\(\Leftrightarrow x+4=35\)
hay x=31
d: Ta có: \(71+\left(726-3x\right)\cdot5=2246\)
\(\Leftrightarrow5\left(726-3x\right)=2175\)
\(\Leftrightarrow726-3x=435\)
\(\Leftrightarrow3x=291\)
hay x=97
e: Ta có: \(720:\left[41-\left(2x+5\right)\right]=40\)
\(\Leftrightarrow41-\left(2x+5\right)=18\)
\(\Leftrightarrow2x+5=23\)
\(\Leftrightarrow2x=18\)
hay x=9
1) (3x + 9)(3x - 6) = 0
=> \(\orbr{\begin{cases}3x+9=0\\3x-6=0\end{cases}}\)
=> \(\orbr{\begin{cases}3x=-9\\3x=6\end{cases}}\)
=> \(\orbr{\begin{cases}x=-3\\x=2\end{cases}}\)
Vậy ...
b) (2x + 15) - 25 = 47 - (10 - x)
=> 2x - 10 = 37 + x
=> 2x - x = 37 + 10
=> x = 47
3, tương tự
4) |4 - 3x| = 8
=> \(\orbr{\begin{cases}4-3x=8\\4-3x=-8\end{cases}}\)
=> \(\orbr{\begin{cases}3x=-4\\3x=12\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{4}{3}\\x=4\end{cases}}\)
Vì x là số nguyên nên ...
còn lại tương tự
`a)|2x-15|=13`
`**2x-15=13`
`<=>2x=28`
`<=>x=14.`
`**2x-15=-13`
`<=>2x=-2`
`<=>x=-1.`
`b)|7x+3|=66`
`**7x+3=66`
`<=>7x=63`
`<=>x9`
`**7x+3=-66`
`<=>7x=-69`
`<=>x=-69/7`
`c)|5x-2|=0`
`<=>5x-2=0`
`<=>5x=2`
`<=>x=2/5`
\(a,\Leftrightarrow\left[{}\begin{matrix}2x-5=13\\2x-5=-13\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-4\end{matrix}\right.\)
Vậy ...
\(b,\Leftrightarrow\left[{}\begin{matrix}7x+3=66\\7x+3=-66\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-\dfrac{69}{7}\end{matrix}\right.\)
Vậy ...
\(c,\Leftrightarrow5x-2=0\)
\(\Leftrightarrow x=\dfrac{5}{2}\)
Vậy ...
a)2x-9=1
=> 2x=10
=> x=5
b)-3x+5=12
=>-3x=17
=> x=-17/3
c)-7x+9=2x
=> -7x-2x=-9
=> 9x=9
=>x=1
d)(x-8)+x(x-8)=0
=> (1+x).(x-8)=0
=>__1+x=0=>x=-1
|__x-8=0=>x=8
e)\(\frac{x-9}{x-7}=\frac{x-7-2}{x-7}=1-\frac{2}{x-7}\)
để (x-9)chia hết cho(x-7) thì (x-7) phải thuộc Ư(2)
\(\Rightarrow\left(x-7\right)\in\left\{-2;-2;1;2\right\}\)
\(\Rightarrow x\in\left\{5;6;8;9\right\}\)
a) \(12\left(x-5\right)=7x-5\)
\(12x-60=7x-5\)
\(12x-7x=60-5\)
\(5x=55\)
\(x=11\)
a, 12(x-5)=7x-5
suy ra 12x-60-7x+5=0
suy ra 5x-55=0
suy ra x=55/5=11
vay x=11
b, ta có 5+2!3x-1/2!=6
suy ra 2!3x-1/2!=6-5=1
suy ra !3x-1/2!=1/2
xet th1: 3x-1/2=1/2
suy ra x=1/3
xet th2 3x-1/2=-1/2
suy ra x=0
vạy x=0 hoac x=1/3
c, (2x-3)^2010=(2x-3)^2012
xet th1 2x-3=1 suy ra x=2
xet th2 2x-*3=0 suy ra x=3/2
vạy x=2 hoac x=3/2
a,-3
b,-201
c,421
d,90
a) 2x + 3x = -150
x.(`2+3)=-150
x.5=-150
x=-150:5
x=-30
Vậy x=-30
b) -24x + 7x = 3417
x.(-24+7)=3417
x.(-17)=3417
x=3417:(-17)
x=-201
Vậy x=-201
c) -3x + x = -842
-3x+1x=-842
x.(-3+1)=-842
x.(-2)=-842
x=-842:(-2)
x=421
Vậy x=421
d) -7x + 2x = -450
x.(-7+-2)=-450
x.(-9)=-450
x=-450:(-9)
x=50
Vậy x=50