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\(\text{a) x. (x + 2)= 0}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x+2=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=-2\end{cases}}\)
vậy_____
\(d.\left(x-5\right)\left(x^2+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-5=0\\x^2+1=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=5\\x\in\varnothing\end{cases}}\)
Mình làm mẫu câu a còn các câu khác tương tự nha
a, x.(x+2) = 0
=> x=0 hoặc x+2=0
=> x=0 hoặc x=-2
Vậy x thuộc {-2;0}
Tk mk nha
![](https://rs.olm.vn/images/avt/0.png?1311)
- 12 . ( x - 5 ) + 7 . ( 3 - x ) = 5
=> - 12x - 12 . 5 + 7 . 3 - 7x = 5
=> - 12x - 60 + 21 - 7x = 5
=> ( - 12 - 7 )x + ( 60 + 21 ) = 5
=> - 19x + 81 = 5
=> - 19x = - 76
=> x = 4
30 . ( x + 2 ) - 6 . ( x + 5 ) - 24x = 100
=> 30x + 30 . 2 - 6x + 6 . 5 - 24x = 100
=> 30x + 60 - 6x + 30 - 24x = 100
=> 0 . x = 100
=> Không có giá trị x
( x + 1 ) + ( x + 2 ) + . . . + ( x + 99 ) = 0
=> x . 99 + ( 1 + 2 + . . . + 99 ) = 0
=> x . 99 + 4950 = 0
=> x . 99 = - 4950
=> x = - 50
![](https://rs.olm.vn/images/avt/0.png?1311)
a) x-14=3x + 18
x - 3x = 18 + 14
-2x = 32
=> x = -16
b) (x+7)(x-9)=0
=> TH1: x+7=0 => x = -7
=> TH2: x-9=0 => x = 9
c) x(x+3) =0
=> TH1: x=0
=> TH2: x+3 =0 => x = -3
d) (x-2)(5-x)=0
=> TH1: x-2=0 => x=2
=> Th2: 5-x=0 => x=5
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,-12\left(x-5\right)+7\left(3-x\right)=5\)
\(-12x+60+21-7x=5\)
\(-12x-7x=5-60-21\)
\(-19x=-76\Leftrightarrow x=4\)
\(b,30\left(x+2\right)-6\left(x-5\right)-24x=100\)
\(30x+60-6x+30-24x=100\)
\(30x-6x-24x=100-60-30\)
\(0x=10\left(vl\right)\)
Vậy pt vô nghiệm
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(5\left(x-7\right)=0\)
\(\Rightarrow x-7=0\)
\(\Rightarrow x=7\)
b) \(25\left(x-4\right)=0\)
\(\Rightarrow x-4=0\)
\(\Rightarrow x=4\)
c) \(\left(34-2x\right)\left(2x-6\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}34-2x=0\\2x-6=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=34\\2x=6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=17\\x=3\end{matrix}\right.\)
d) \(\left(2019-x\right)\left(3x-12\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2019-x=0\\3x-12=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2019\\3x=12\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2019\\x=\dfrac{12}{3}=4\end{matrix}\right.\)
e) \(57\left(9x-27\right)=0\)
\(\Rightarrow9x-27=0\)
\(\Rightarrow9\left(x-3\right)=0\)
\(\Rightarrow x-3=0\)
\(\Rightarrow x=3\)
a) 5.(x-7)=0⇔x-7=0⇔x=7
b) 25(x-4)=0⇔x-4=0⇔x=4
c) (34-2x).(2x-6)=0
⇔ 34-2x=0 hoặc 2x-6=0
⇔2x=34 hoặc 2x=6
⇔ x=17 hoặc x=3
d) (2019-x).(3x-12)=0
⇔ 2019-x=0 hoặc 3x-12=0
⇔ x=2019 hoặc x=4
e) 57.(9x-27)=0
⇔ 9x-27=0
⇔ x=3
f) 25+(15-x)=30
⇔ 15-x=5
⇔ x=10
g) 43-(24-x)=20
⇔ 24-x=23
⇔ x=1
h) 2.(x-5)-17=25
⇔ 2(x-5)=42
⇔x-5=21
⇔ x=26
i) 3(x+7)-15=27
⇔ 3(x+7)=42
⇔ x+7=14
⇔ x=7
j) 15+4(x-2)=95
⇔ 4(x-2)=80
⇔ x-2=20
⇔ x=22
k) 20-(x+14)=5
⇔ x+14=15
⇔ x=1
l) 14+3(5-x)=27
⇔ 3(5-x)=13
⇔ 5-x=13/3
⇔ x=5-13/3
⇔ x=2/3
![](https://rs.olm.vn/images/avt/0.png?1311)
a)-12.(x-5)+7.(3-x)=15
-12x+60+21-7x=15
-19x+81=15
-19x=15-81
-19x=-66
=>x=66/19
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,x\left(x+3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x+3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=-3\end{cases}}}\)
\(b,\left(x-2\right)\left(5-x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-2=0\\5-x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=5\end{cases}}}\)
\(c,\left(x-1\right)\left(x^2+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\x^2+1=0\end{cases}\Rightarrow x=1}\)
\(d,-12\left(x-5\right)+7\left(3-x\right)=15\)
\(-12x+60+21-7x=15\)
\(-19x+81=15\)
\(-19x=15-81\)
\(-19x=-66\)
\(x=\frac{66}{19}\)
\(e,30\left(x+2\right)-6\left(x-5\right)-24x=100\)
\(30x+60-6x+30-24x=100\)
\(0x+90=100\)
\(0x=10\) ( vô lí )
=> không có giá trị x nào thõa mãn
a) x(x + 3) = 0
=> \(\orbr{\begin{cases}x=0\\x+3=0\end{cases}}\)
Mà x < x + 3
=> x = 0
b)( x - 2 )( 5 - x ) = 0
=> \(\orbr{\begin{cases}x-2=0\Rightarrow x=2\\5-x=0\Rightarrow x=5\end{cases}}\)
=> \(x\in\left\{2,5\right\}\)
c) ( x - 1 )( x2 + 1 ) = 0
=> \(\orbr{\begin{cases}x-1=0\Rightarrow x=1\\x^2+1=0\Rightarrow x^2=-1\end{cases}}\)
Vì x2 không thể bằng -1 => x = 1
d)-12 ( x - 5 ) + 7 ( 3 - x ) = 15
=> -12x - (-60) + 21 - 7x = 15
=> -12x + 60 + 21 + (-7x) = 15
=>[-12x + (-7x)] + 81 = 15
=> -19x = -66
=> \(x\in\varphi\)
a: x(x+2)=0
=>x=0 hoặc x+2=0
=>x=0 hoặc x=-2
c: \(\Leftrightarrow x\left(x-2\right)=0\)
=>x=0 hoặc x=2
e: \(\Leftrightarrow-12x+60+21-7x=5\)
=>-19x=-76
hay x=4
g: =>30x+60+6x-30-24x=102
=>12x+30=102
=>12x=72
hay x=6