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\(a,\left(2x-5\right)+17=6\\ \Rightarrow2x-5=-11\\ \Rightarrow2x=-6\\ \Rightarrow x=-3\\ b,10-2\left(4-3x\right)=-4\\ \Rightarrow2\left(4-3x\right)=14\\ \Rightarrow4-3x=7\\ \Rightarrow3x=-3\\ \Rightarrow x=-1\\ c,24:\left(3x-2\right)=-3\\ \Rightarrow3x-2=-8\\ \Rightarrow3x=-6\\ \Rightarrow x=-2\\ d,5-2x=-17+12\\ \Rightarrow5-2x=-5\\ \Rightarrow2x=10\\ \Rightarrow x=5\)
a: =>2x-5=-11
=>2x=-6
hay x=-3
b: =>2(4-3x)=14
=>4-3x=7
=>3x=-3
hay x=-1
c: =>3x-2=-8
=>3x=-6
hay x=-2
(27-X)+(15+x)=x-24
27-x+15+x=x-24
-x+x-x=-24-27-15
-x=-66
x=-(-66)
x=66
a) x – 5 = - 1 ;
=> x = -1 + 5 = 4
Vậy x = 4
b) x + 30 = - 4;
=> x = - 4 - 30 = - 34
Vậy x = - 34
c) x – ( - 24) = 3 ;
=> x + 24 = 3
=> x = 3 - 24
=> x = - 21
Vậy x = - 21
d) 22 – ( - x ) = 12;
=> 24 + x = 12
=> x = 12
Vậy x = 12
e) ( x + 5 ) + ( x – 9 ) = x + 2 ;
=> x + 5 +x - 9 = x + 2
=> 2x - x = 1 + 9 - 5
=> x = 5
Vậy x = 5
f) ( 27 – x ) + ( 15 + x ) = x – 24 .
=> 27 - x + 15 + x = x - 24
=> x + 42 = x - 24
=> x - x = 42 - 24
=> 0 = 8 ( vô lí)
Vậy x thuộc rỗng
Rảnh nhỉ
@@ Học tốt
## Chiyuki Fujito
a)
\(x-5=-1\)
\(x=-1+5\)
\(x=4\)
b)
\(x+30=4\)
\(x=4-30\)
\(x=-26\)
c)
\(x-(-24)=3\)
\(x+24=3\)
\(x=3-24\)
\(x=-21\)
d)
\(22-(-x)=12\)
\(22+x=12\)
\(x=12-22\)
\(x=-10\)
e)
\(( x + 5 ) + ( x - 9 ) = x + 2\)
\(x+5+x-9=x+2\)
\(x+x-x=2+9-5\)
\(x=6\)
f)
\(( 27 - x ) + ( 15 + x ) = x - 24\)
\(27-x+15+x=x-24\)
\(-x+x-x=-24-15-27\)
\(-x=-66\)
\(x=66\)
x - 5 = -1 x - (-24) = 3
x = -1 + 5 x + 24 = 3
x = 4 x = 3 - 24
x + 30 = 4 x = - 21
x = 4 - 30 22 - ( -x) = 12
x = - 26 22 + x = 12
x + 5 + ( x - 9) = x + 2 x = 12 - 22
x + 5 + x - 9 = x + 2 x = -10
2x - x = 2 - 5 + 9 ( 27 - x) + ( 15 + x) = x - 24
x = - 3 + 9 27 - x + 15 + x = x - 24
x = 6 27 + 15 = x - 24
x - 24 = 42
x = 42 + 24
x = 66
a, Ta có : \(x-5=-1\)
\(\Leftrightarrow x=\left(-1\right)+5\)
\(\Rightarrow x=4\)
Vậy \(x=4\)
b, Ta có :\(x+30=-4\)
\(\Leftrightarrow\left(-4\right)-30\)
\(\Leftrightarrow x=\left(-34\right)\)
Vậy \(x=\left(-34\right)\)
c, Ta có : \(x-\left(-24\right)=3\)
\(x=3+\left(-24\right)\)
\(\Leftrightarrow x=\left(-21\right)\)
Vậy \(x=\left(-21\right)\)
d,Ta có : \(22-\left(-x\right)=12\)
\(\Leftrightarrow\left(-x\right)=22-12\)
\(\Leftrightarrow\left(-x\right)=10\)
Vậy ....
Sai thì sửa,chửa thì đẻ
a, Ta có :
Vậy
b, Ta có :
Vậy
c, Ta có :
Vậy
d,Ta có :
Vậy ....
e,Ta có :
Vậy x=6
a: =>x/-3=3
hay x=-9
b: =>x/9=-1/9
hay x=-1
c: =>x+1/5=-1/3
hay x=-8/15
d: =>-7/x=-7/9
hay x=9
a, \(\dfrac{x}{-3}=3\Leftrightarrow x=-9\)
b, \(\dfrac{x}{9}=-\dfrac{1}{9}\Rightarrow x=-1\)
c, \(\dfrac{x+3}{15}=-\dfrac{6}{15}\Rightarrow x=-9\)
d, \(\dfrac{42}{-54}=-\dfrac{42}{6x}\Rightarrow6x=54\Leftrightarrow x=9\)
e) (x+5) + (x-9) = x+2
x+5 + x -9 = x+2
2x + 5 - 9 = x+2
2x - 4 = x +2
3x = 2 +4
3x =6
x = 2
f) (27-x) + ( 15+x) = x-24
27-x + 15 +x = x-24
27 + 15 - (x - x) = x - 24
42 = x-24
x = 24 + 42
x = 66
a)\(x-5=-1\)
⇔\(x=4\)
b)\(x+30=-4\)
⇔\(x=-34\)
c)\(x-\left(-24\right)=3\)
⇔\(x+24=3\)
⇔\(x=-21\)
e)\(\left(x+5\right)+\left(x-9\right)=x+2\)
⇔\(x+5+x-9-x-2=0\)
⇔\(x-6=0\)
⇔\(x=6\)
f)\(\left(27-x\right)+\left(15+x\right)=x-24\)
⇔\(27-x+15+x-x+24=0\)
⇔\(66-x=0\)
⇔\(x=66\)
\(a.x-5=-1\) \(b.x+30=-4\)
\(x=\left(-1\right)+5\) \(x=\left(-4\right)-30\)
\(x=4\) \(x=-34\)
\(c.x-\left(-24\right)=3\) \(e.\left(x+5\right)+\left(x-9\right)=x+2\)
\(x=3+\left(-24\right)\) \(x+5+x-9=x+2\)
\(x=-21\) \(2x-4=x+2\)
\(2x-x=2+4\)
\(x=6\)
\(f.\left(27-x\right)+\left(15+x\right)=x-24\)
\(27-x+15+x=x-24\)
\(27+15=x-24\)
\(42=x-24\)
\(x=24+42\)
\(x=66\)