Tìm x, biết:
a) 2 . x + 12 = 36
b) (x + 21) : 8 + 12 = 21
c) (3 . x - 18) . (x - 9) = 0
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Bài 2 : a, x = -36/9 = -4
b, đề sai
c, <=> -2 =< x =< -3 => x = -1
Bài 1:
a: 2/8=9/36; 2/9=8/36; 8/2=36/9; 9/2=36/8
b: -2/4=9/-18; -2/9=4/-18; 4/-2=-18/9; 9/-2=-18/4
Bài 2:
a: =>x/3=-4/3
hay x=-4
Câu b đề sai rồi bạn
a)\(x-15=3\Leftrightarrow x=3+15\Leftrightarrow x=18\)
b) \(2x+12=36\Leftrightarrow2x=24\Leftrightarrow x=12\)
c)\(\left(x+21\right):8+12=21\)
\(\Leftrightarrow\)\(\frac{x+21}{8}=9\)
\(\Leftrightarrow x+21=72\)
\(\Leftrightarrow x=51\)
d)\(\left(3x+18\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}3x-18=0\\x-9=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=6\\x=9\end{array}\right.\)
a: \(3\left(x-3\right)-6x=0\)
=>\(3x-9-6x=0\)
=>-3x-9=0
=>3x+9=0
=>3x=-9
=>\(x=-\dfrac{9}{3}=-3\)
b: Đề thiếu vế phải rồi bạn
c: \(2\left(x-3\right)+3x=9\)
=>2x-6+3x=9
=>5x-6=9
=>5x=6+9=15
=>x=15/5=3
d: \(x\left(x-11\right)+2\left(x-11\right)=0\)
=>\(\left(x-11\right)\left(x+2\right)=0\)
=>\(\left[{}\begin{matrix}x-11=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=11\\x=-2\end{matrix}\right.\)
e: \(x\left(x+2\right)+8=x^2\)
=>\(x^2+2x+8=x^2\)
=>2x+8=0
=>2x=-8
=>x=-8/2=-4
f: \(8\left(x+1\right)+2x=-2\)
=>\(8x+8+2x=-2\)
=>10x=-2-8=-10
=>\(x=-\dfrac{10}{10}=-1\)
g: 12-3(x+2)=0
=>3(x+2)=12
=>x+2=12/3=4
=>x=4-2=2
a) \(3\sqrt{x-3}=12\left(đk:x\ge3\right)\)
\(\Leftrightarrow\sqrt{x-3}=4\)
\(\Leftrightarrow x-3=16\Leftrightarrow x=19\left(tm\right)\)
b) \(\sqrt{16\left(1-2x\right)}-8=0\left(đk:x\le\dfrac{1}{2}\right)\)
\(\Leftrightarrow4\sqrt{1-2x}=8\Leftrightarrow\sqrt{1-2x}=2\)
\(\Leftrightarrow1-2x=4\Leftrightarrow x=-\dfrac{3}{2}\left(tm\right)\)
c) \(\sqrt{4\left(9-6x+x^2\right)}-12=0\)
\(\Leftrightarrow2\sqrt{\left(x-3\right)^2}=12\)
\(\Leftrightarrow\left|x-3\right|=6\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=6\\x-3=-6\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-3\end{matrix}\right.\)
a: ta có: \(3\sqrt{x-3}=12\)
\(\Leftrightarrow x-3=16\)
hay x=19
b: Ta có: \(\sqrt{16\left(1-2x\right)}-8=0\)
\(\Leftrightarrow1-2x=4\)
\(\Leftrightarrow2x=-3\)
hay \(x=-\dfrac{3}{2}\)
`a)`
`x-15 = -21`
`<=> x = -21 + 15`
`<=> x = -6`
`b)`
`42 - x = -7`
`<=> x = 42 - (-7)`
`<=> x = 42 + 7`
`<=> x = 49`
`c)`
`12 - (30-x) = -23`
`30-x = 12 - (-23)`
`30-x = 35`
`x = 30-35`
`x = -5`
`d)`
`31 - (17+x)=18`
`<=> 17+x = 31-18`
`<=> 17+x =13`
`<=> x = 13-17`
`<=> x = -4`
a) \(x-15=-21\)
\(\Rightarrow x=-21+15=-6\)
b) \(42-x=-7\)
\(\Rightarrow x=42+7=49\)
c) \(12-\left(30-x\right)=-23\)
\(\Rightarrow30-x=12+23=35\)
\(\Rightarrow x=30+35=65\)
d) \(31-\left(17+x\right)=18\)
\(\Rightarrow17+x=31-18=13\)
\(\Rightarrow x=13-17=-4\)
1) 5.( x - 6 ) - 2.( x + 9 ) = 21
5x - 30 - 2x - 18 = 21
3x - 48 = 21
3x = 21 + 48
3x = 69
x = 23
2) 2.( x + 3 ) + 3.( x + 1 ) = 15 - ( - 9 )
2x + 6 + 3x + 3 = 24
5x + 9 = 24
5x = 24 - 9
5x = 15
x = 3
3) ( - x + 5 ).(3 - x ) = 0
=> - x + 5 = 0 hoặc 3 - x = 0
=> x = 5 hoặc x = 3
4) ( x - 12 ) - 15 = ( 20 - 7 ) - ( 18 + x )
x - 12 - 15 = 13 - 18 - x
x - 27 = - 5 - x
x + x = - 5 + 27
2x = 22
x = 11
5) x - ( 17 - 8 ) = 5 + ( 10 - 3x )
x - 9 = 5 + 10 - 3x
x + 3x = 15 + 9
4x = 24
x = 6
a) 2 . x + 12 = 36
2 . x = 36 - 12
2 . x = 24
x = 24 : 2
x = 12
Vậy x = 12
b) (x + 21) : 8 + 12 = 21
( x + 21 ) : 8 = 21 - 12
( x + 21 ) : 8 = 9
x + 21 = 9 . 8
x + 21 = 72
x = 72 - 21
x = 51
Vậy x = 51
c) (3 . x - 18) . (x - 9) = 0
\(\Rightarrow\begin{cases}3.x-18=0\\x-9=0\end{cases}\)
\(\Rightarrow\begin{cases}3.x=18\\x=0+9\end{cases}\)
\(\Rightarrow\begin{cases}x=18:3=6\\x=9\end{cases}\)
Vậy \(x\in\left\{6;9\right\}\)
a)
\(2x+12=36\)
\(\Rightarrow2\left(x+6\right)=36\)
\(\Rightarrow x+6=18\)
=> x = 12
Vậy x = 12
b)
\(\left(x+21\right):8+12=21\)
\(\Rightarrow\left(x+21\right):8=9\)
\(\Rightarrow x+21=72\)
=> x = 51
Vậy x = 51
c)
\(\left(3x-18\right)\left(x-9\right)=0\)
\(\Rightarrow3\left(x-6\right)\left(x-9\right)=0\)
\(\Rightarrow\left(x-6\right)\left(x-9\right)=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}x-6=0\\x-9=0\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=6\\x=9\end{array}\right.\)
Vậy x = 6 ; x = 9