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Tìm x thuoc z:
1) \(26-\left|x+9\right|=-13\)
\(\Leftrightarrow\left|x+9\right|=26-\left(-13\right)\)
\(\Leftrightarrow\left|x+9\right|=39\)
\(\Leftrightarrow\left[{}\begin{matrix}x+9=39\\x+9=-39\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=39-9=30\\x=-39-9=-48\end{matrix}\right.\)
Vậy: \(x\in\left\{30;-48\right\}\)
2) \(\left|x+7\right|-13=25\)
\(\Leftrightarrow\left|x+7\right|=25+13=38\)
\(\Leftrightarrow x+7\in\left\{38;-38\right\}\)
\(\Leftrightarrow x\in\left\{31;-45\right\}\)
Vậy:.................
tim x biet
\(1)123-3.\left(x+4\right)=23\)
\(\Leftrightarrow3\left(x+4\right)=123-23\)
\(\Leftrightarrow3\left(x+4\right)=100\)
\(\Leftrightarrow x+4=\frac{100}{3}\)
\(\Leftrightarrow x=\frac{100}{3}-4=\frac{100-12}{3}=\frac{88}{3}\)
Vậy:................
2) Tương tự
720 : [ 41 - ( 2x - 5 ) ] = 23 . 5
720 : [ 41 - ( 2x - 5 ) ] = 8 . 5
720 : [ 41 - ( 2x - 5 ) ] = 40
41 - ( 2x - 5 ) = 720 : 40
41 - ( 2x - 5 ) = 18
2x - 5 = 41 - 18
2x - 5 = 23
2x = 23 + 5
2x = 28
=> x = 28 : 2
=> x = 14
720 : [ 41 - ( 2 x - 5 ) ] = 40
41 - ( 2 x - 5 ) = 720 : 40
41 - ( 2 x - 5 ) = 18
2 x - 5 = 41 - 18
2 x - 5 = 23
2 x = 23 + 5
2 x = 28
x = 28 : 2
x = 14
720:[41-(2x-5)]=2^3.5
720:[41-2x+5]=40
[41+5-2x]=720:40
46-2x=18
2x=46-18
2x=28
x=14
a, 2x.4=128
2x=128:4
2x=32
2x=25
x=5
Vay x=5
b, x15=x
x=1
Vay x=1
720 : [ 41 - (2x - 5)] = 23 . 5
=> 720 : [ 41 - ( 2x - 5)] = 40
=> 41 - (2x - 5) = 18
=> 2x - 5 = 23
=> 2x = 28
=> x = 14
Vậy x = 14
720 :[ 41 -(2x -5)]=23 . 5.
=>720 :[41 -(2x-5)]=40.
=>[41- (2x -5)]=720:40.
=>[41-(2x-5)]=18.
=>(2x-5)=41-18.
=>2x-5=23.
=>2x=23+5.
=>2x=28.
=>x=28:2.
=>x=14.
720 : [ 41 - ( 2x-5)] = 2^3 .5
=> 720 :[ 41- (2x-5)] = 40
=> 41 - (2x-5) = 18
=> 2x -5 = 23
=> 2x = 28
=> x=14