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\(a)3-\left(17-x\right)=-24+12\)
\(\Leftrightarrow3-17+x=-12\)
\(\Leftrightarrow-14+x=-12\)
\(\Leftrightarrow x=-12+14=2\)
a) 3 - (17 - x) = -24 + 12
=> 3 - 17 + x = -12
=> -14 + x = -12
=> x = -12 + 14
=> x = 2
b) -5 + (16 - x) = 22.3 - 52
=> -5 + 16 - x = 12 - 25
=> 11 - x = -13
=> x = 11 + 13
=> x = 24
c) -26 - (x - 7) = 34 - 92
=> -26 - x + 7 = 81 - 81
=> -19 - x = 0
=> x = -19 - 0
=> x = -19
d) -36 - (4 - x) = 37 + (-5)
=> -36 - 4 + x = 32
=> -40 + x = 32
=> x = 32 + 40
=> x = 72
a)
2x - 138 = 23 . 32
2x - 138 = 8 . 9
2x - 138 = 72
2x = 72 + 138
2x = 110
x = 110 : 2
x = 55
còn lại tương tự
a. 5(x-7)-4(x+5)=3.5-12
5x - 5.7 -4x - 4.5 = 15 -12
x(5-4) -35-20 = 3
x -55 = 3
x = 3+55
x = 58
vậy x = 58
b. (2x -15)5= (2x -15)3
vì (2x -15)5= (2x -15)3
=> (2x -15)5 = 1 hoặc (2x -15)5 = 0
=> (2x -15)5 = 15 hoặc (2x -15)5 = 05
=> 2x-15 =1 hoặc 2x-15 = 0
=> 2x = 1+15 hoặc 2x = 0+15
=> 2x =16 hoặc 2x = 15
=> x = 16:2 hoặc x = 15:2
=> x= 8 hoặc x = 7,5
Vì x \(\in\)N => x = 8
vậy x = 8
a) \(\left(\left|x\right|+3\right):5-3=12\Leftrightarrow\left|x\right|+3=45\Leftrightarrow\orbr{\begin{cases}x=42\\x=-42\end{cases}}\)
b) \(86:\left[2\left(2x-1\right)^2-7\right]+4^2=2\cdot3^2\Leftrightarrow2\left(2x-1\right)^2-7=43\Leftrightarrow\left(2x-1\right)^2=25\Leftrightarrow\orbr{\begin{cases}2x-1=-5\\2x-1=5\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-2\\x=3\end{cases}}\)
a) \(\left(\left|x\right|+3\right)\div5-3=12\)
\(\left(\left|x\right|+3\right)\div5=12+3\)
\(\left(\left|x\right|+3\right)\div5=15\)
\(\left|x\right|+3=15.5\)
\(\left|x\right|+3=75\)
\(\left|x\right|=75-3\)
\(\left|x\right|=72\)
\(\Rightarrow\orbr{\begin{cases}x=72\\x=-72\end{cases}}\)
Vậy \(x\in\left\{72;-72\right\}\)
b) \(86\div\left[2,\left(2x-1\right)^2-7\right]+4^2=2.3^2\)
\(86\div\left[2.\left(2x-1\right)^2-7\right]+16=18\)
\(86\div\left[2.\left(2x-1\right)^2-7\right]=18-16\)
\(86\div\left[2.\left(2x-1\right)^2-7\right]=2\)
\(2.\left(2x-1\right)^2-7=86\div2\)
\(2.\left(2x-1\right)^2-7=43\)
\(2.\left(2x-1\right)^2=43+7\)
\(2.\left(2x-1\right)^2=50\)
\(\left(2x-1\right)^2=50\div2\)
\(\left(2x-1\right)^2=25\)
\(\left(2x-1\right)^2=5^2\)
\(\Rightarrow2x-1=5\)
\(2x=5+1\)
\(2x=6\)
\(x=6\div2\)
\(x=3\)
b: =>x-4=-4
hay x=0