A.(x-13)-(2x+31)=-8-15
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a) 2x - 31 + 42 = - 3x - 45
2x + 11 = - 3x - 45
2x + 3x = - 45 - 11
5x = - 56
x = - 56/5
b) - 25 + 2x = 15 + 17 - x
2x + x = 32 + 25
3x = 57
x = 19
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
a) \(2x-13=7\)
\(\Leftrightarrow2x=7+13=20\)
\(\Leftrightarrow x=\frac{20}{2}=10\)
b) \(23-x=31-\left(-9\right)\)
\(\Leftrightarrow23-x=40\)
\(\Leftrightarrow x=23-40=-17\)
c) \(\left|x+13\right|-4=16\)
\(\Leftrightarrow\left|x+13\right|=16+4=20\)
\(\Leftrightarrow\left[\begin{matrix}x+13=20\Rightarrow x=7\\x+13=-20\Rightarrow x=-33\end{matrix}\right.\)
d) Vì 45 \(⋮x\) mà x > 15
\(\Rightarrow\) x = 45
a, 2x = 7 + 13 =20 => x = 20 : 2 =10
b, 23 - x = 31 + 9 = 40 => x = 23 - 40 = -17
c, x = 7 ; x = -33
d, x = 22,5
a, x = -8 - 3 = -11
b, 35 + x = 27 + 12 = 39
=> x = 39 - 35 = 4
c, 2x = 31 - 15 = 16
=> x = 4
a.\(3+x=-8\)
\(x=-8-3\)
\(x=-11\)
b.\(\left(35+x\right)-12=27\)
\(35+x=27+12\)
\(35+x=39\)
\(x=39-35\)
\(x=4\)
c.\(2^x+15=31\)
\(2^x=31-15\)
\(2^x=16\)
\(2^x=2^4\)
\(x=4\)
2/5 + 11/15 = 17/15
7/8 - 7/9 = 7/72
11/13 x 26/31 = 22/31
16/24 : 4 = 1/6
30 : 6/5 = 25
9/16 : 4 = 9/64
1/2 : 1/3 x 8/15 = 3/2 x 8/15 = 4/5
a, ( x + 17 ) : 21 + 3 = 7
( x + 17 ) : 21 = 7 - 3
( x + 17 ) : 21 = 4
x + 17 = 4 x 21
x + 17 = 84
x = 84 - 17
x = 67
b, 3x+3 - 13 = 230
=> 3x + 3 = 230 + 13
=> 3x+3 = 243
=> 3x+3 = 35
=> x + 3 = 5
=> x = 5 -3
=> x = 2
c, 2x - 15 = 23
2x = 23 + 15 = 38
x = 38 : 2
x = 19
d, ( x + 16 ) : 31 - 3 = 7
( x + 16 ) : 31 = 10
x + 16 = 10 x 31 = 310
x = 310 - 16
x = 294
1.
b) \(3^x+3^{x+2}=2430\)
\(\Rightarrow3^x.1+3^x.3^2=2430\)
\(\Rightarrow3^x.\left(1+3^2\right)=2430\)
\(\Rightarrow3^x.10=2430\)
\(\Rightarrow3^x=2430:10\)
\(\Rightarrow3^x=243\)
\(\Rightarrow3^x=3^5\)
\(\Rightarrow x=5\)
Vậy \(x=5.\)
c) \(\left(2x-15\right)^5=\left(2x-15\right)^3\)
\(\Rightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\)
\(\Rightarrow\left(2x-15\right)^3.\left[\left(2x-15\right)^2-1\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x-15=0\\\left(2x-15\right)^2=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=15\\2x-15=\pm1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=15:2\\2x-15=1\\2x-15=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{15}{2}\\2x=16\\2x=14\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{15}{2}\\x=8\\x=7\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{15}{2};8;7\right\}.\)
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