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a/ x - ( 25 - x) = x - 15
=> x - 25 + x = x - 15
=> 2x - x = -15 + 25
=> x = 10
b/ x - 8 = -13 - 8
=> x = -13 - 8 + 8 = -13
c/ x - (-15) = 5 - (-19)
=> x + 15 = 5 + 19 => x = 5 + 19 - 15 = 9
d/ (x + 35) = 2x + 15 + (13 - 4)
=> x + 35 = 2x + 15 + 13 - 4
=> x - 2x = 15 + 13 - 4 - 35
=> -x = -11 => x = 11
e/ 5 - 2x = 15-(-9) - 3x
=> 5 - 2x = 15 + 9 - 3x
=> -2x + 3x = 15 + 9 - 5
=> x = 19
g/ x + 135 = 3x - 15
=> x - 3x = -15 - 135
=> -2x = -150
=> x = -150/ - 2 = 75
\(\left(3x-15\right)\left(2x-6\right)=2x-6\)
\(\Leftrightarrow\left(3x-15\right)\left(2x-6\right)-\left(2x-6\right)=0\)
\(\Leftrightarrow\left(2x-6\right)\left(3x-16\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x-6=0\\3x-16=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=\frac{16}{3}\end{cases}}\)
(3x-15).(2x-6)=2x-6
3x-15 =(2x-6):(2x-6)
3x-15 =1
3x =1+15
3x =16
x =16:3
x = 8
Vậy x =8
a) 3x – 15 = 25 – 5x
=> 3x + 5x = 25 + 15
=> 8x = 40
=> x = 5
b) 3x - 17 = 2x – 7
=> 3x - 2x = -7 + 17
=> x = 10
c) 2x – 17 = – (3x – 18)
=> 2x - 17 = -3x + 18
=> 2x + 3x = 18 + 17
=> 5x = 35
=> x = 7
d) 3x – 14 = 2(x – 9) + 1
=> 3x - 14 = 2x - 18 + 1
=> 3x - 2x = -18 + 1 + 14
=> x = -3
f) (x – 5)2 = 9
\(\Rightarrow\left[{}\begin{matrix}x-5=3\\x-5=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=8\\x=2\end{matrix}\right.\)
a) Ta có: \(3x-15=25-5x\)
\(\Leftrightarrow3x-15-25+5x=0\)
\(\Leftrightarrow8x-40=0\)
\(\Leftrightarrow8x=40\)
hay x=5
Vậy: x=5
b) Ta có: \(3x-17=2x-7\)
\(\Leftrightarrow3x-17-2x+7=0\)
\(\Leftrightarrow x-10=0\)
hay x=10
Vậy: x=10
c) Ta có: \(2x-17=-\left(3x-18\right)\)
\(\Leftrightarrow2x-17=-3x+18\)
\(\Leftrightarrow2x-17+3x-18=0\)
\(\Leftrightarrow5x-35=0\)
\(\Leftrightarrow5x=35\)
hay x=7
Vậy: x=7
d) Ta có: \(3x-14=2\left(x-9\right)+1\)
\(\Leftrightarrow3x-14=2x-18+1\)
\(\Leftrightarrow3x-14-2x+18-1=0\)
\(\Leftrightarrow x+3=0\)
\(\Leftrightarrow x=-3\)
Vậy: x=-3
f) Ta có: \(\left(x-5\right)^2=9\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=3\\x-5=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=2\end{matrix}\right.\)
Vậy: \(x\in\left\{2;8\right\}\)
1) 4x-(2x-5)=21
4x-2x+5=21
2x+5=21
2x=21 -5
2x=16
x=16/2
x=8
\(2x-14=3x+15-8\)
\(\Leftrightarrow0=3x-2x+15+14-8\)
\(\Leftrightarrow0=x+21\)
\(\Leftrightarrow x=-21\)
Vậy x = -21
a) Ta có: \(\left(2x-15\right)^5=\left(2x-15\right)^3\)
\(\Leftrightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\)
\(\Leftrightarrow\left(2x-15\right)^3.\left[\left(2x-15\right)^2-1\right]=0\)
+ \(2x-15=0\)\(\Leftrightarrow\)\(x=\frac{15}{2}\)
+\(\left(2x-15\right)^2-1=0\)\(\Leftrightarrow\)\(2x-15=\pm1\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=8\\x=7\end{cases}}\)
Vậy \(x\in\left\{\frac{15}{2};8;7\right\}\)
a, \(\left(2x-15\right)^5=\left(2x-15\right)^3\)
\(\Leftrightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\)
\(\Leftrightarrow\left(2x-15\right)^3\left[\left(2x-15\right)^2-1\right]=0\)
\(\Leftrightarrow\left(2x-15\right)^3\left(2x-16\right)\left(2x-14\right)=0\)
\(\Leftrightarrow x=\frac{15}{2};8;7\)
`2x + 3x = 15`
`x(2+3) =15`
`5x=15`
`x=15/5 =3`
`2x+3x=15`
`(2+3)x=15`
`5x=15`
`x=15:5=3`