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a, => |15-x| = 2+3
=> |15-x| = 5
=> 15-x = -5 hoặc 15-x = 5
=> x = 10 hoặc x = 10
Vậy ......
Tk mk nha
a/|15-x|=|-2|+|-3|
=>|15-x|=2+3=5
=>\(15-x=\hept{\begin{cases}5\\-5\end{cases}}\)
=>\(x=\hept{\begin{cases}10\\15\end{cases}}\)
.........
k nha , thanks
a,3.x-12=125-5.5.5
3.x-12=125-125
3.x-12=0
3.x=0+12
3.x=12
x=12:3
x=4
b,x-2!+17=35
x-2=35-17
x-2=18
x=18+2
x=20
c,x.x-3.x=0
x.3-3=0
x.3=0+3
x.3=3
x=3:3
x=1
d, (2.x-4).(3.x-9)=0
khi2.x-4=0 thì 3.x-9=0
2.x-4=0 3.x-9=0
2.x=0+4 3.x=0+9
2.x=4 3.x=9
x=4:2 x=9:3
x=2 x=3
vậy x=2 hoạc x=3
h,(15-3.x).(2.x-7)=0
khi 15-3.x=0 thì 2.x-7=0
15-3.x=0 2.x-7=0
3.x=0+15 2.x=0+7
3.x=15 2.x=7
x=15:3 x=7:2
x=5 7ko chia hết cho 2 nên ko có x
vậy x =5
nhớ k cho mình nha
a) x = 4
b) x = 2 ; x = -4
c) x = 2 ; x = -15
d) x = 7 ; x = -19
e) x = -4 ; -3 ; -2 ; -1 ; 0
g) x = -1 ; - 2 ; 1 ; 2 ; 3 ; 4 ; ...
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a) -12x + 60 + 21 - 7x = 5
-12x - 7x + 60+21 = 5
-19x + 81 = 5
81-5 = 19x
19x = 76
x= 4
a) \(-12\left(x-5\right)+7\left(3-x\right)=5\\ < =>-12x+60+21-7x=5\\ < =>-12x-7x=-60-21+5\\ < =>-19x=-76\\ =>x=4\)
Vậy: x=4
b) \(\left(x-2\right)\left(x+4\right)=0\\ < =>\left[{}\begin{matrix}x-2=0\\x+4=0\end{matrix}\right.\\ =>\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)
Vậy: x=2 hoặc x=4
c) \(\left(x-2\right)\left(x+15\right)=0\\ < =>\left[{}\begin{matrix}x-2=0\\x+15=0\end{matrix}\right.=>\left[{}\begin{matrix}x=2\\x=-15\end{matrix}\right.\)
Vậy: x=2 hoặc x= -15
d) \(\left(7-x\right)\left(x+19\right)=0\\ < =>\left[{}\begin{matrix}7-x=0\\x+19=0\end{matrix}\right.=>\left[{}\begin{matrix}x=7\\x=-19\end{matrix}\right.\)
Vậy: x=7 hoặc x=-19
b) \(\left(x-2\right)\left(x+4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-2=0\\x+4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)
c) \(\left(x-2\right)\left(x+15\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-2=0\\x+15=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=-15\end{matrix}\right.\)
d) \(\left(7-x\right)\left(x+19\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}7-x=0\\x+19=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=7\\x=-19\end{matrix}\right.\)
a) −12(x−5)+7(3−x)=5<=>−12x+60+21−7x=5<=>−12x−7x=−60−21+5<=>−19x=−76=>x=4
Vậy: x=4
b) (x−2)(x+4)=0<=>[x−2=0x+4=0=>[x=2x=−4
Vậy: x=2 hoặc x=4
c) (x−2)(x+15)=0<=>[x−2=0x+15=0=>[x=2x=−15
Vậy: x=2 hoặc x= -15
d) (7−x)(x+19)=0<=>[7−x=0x+19=0=>[x=7x=−19
Vậy: x=7 hoặc x=-19
\(\left(x-2\right)\left(x+4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-2=0\\x+4=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=2\\x=-4\end{cases}}\)
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
a) \(\left(x-2\right)\left(x+4\right)=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}x-2=0\\x+4=0\end{array}\right.\) \(\Rightarrow\left[\begin{array}{nghiempt}x=0+2\\x=0-4\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=2\\x=-4\end{array}\right.\)
Vậy: \(x\in\left\{2;-4\right\}\)
b) \(\left(x-2\right)\left(x+15\right)=0\)
\(\Rightarrow\begin{cases}x-2=0\\x+15=0\end{cases}\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=0+2\\x=0-15\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=2\\x=-15\end{array}\right.\)
Vậy: \(x\in\left\{2;-15\right\}\)
c) \(\left(x-3\right)\left(x-5\right)< 0\)
\(\Rightarrow\left[\begin{array}{nghiempt}x-3< 0\\x-5< 0\end{array}\right.\) \(\Rightarrow\left[\begin{array}{nghiempt}x< 0+3\\x< 0+5\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x< 3\\x< 5\end{array}\right.\) \(\Rightarrow3< x< 5\)
Vậy: \(x\in\left\{4\right\}\)
a) \(\left(x-2\right).\left(x+4\right)=0\)
\(\Rightarrow\begin{cases}x-2=0\\x+4=0\end{cases}\)
\(\Rightarrow\begin{cases}x=0+2\\x=0-4\end{cases}\)
\(\Rightarrow\begin{cases}x=2\\x=-4\end{cases}\)
Vậy x thuộc { 2 ; - 4 }
b) \(\left(x-2\right).\left(x+15\right)=0\)
\(\Rightarrow\begin{cases}x-2=0\\x+15=0\end{cases}\)
\(\Rightarrow\begin{cases}x=0+2\\x=0-15\end{cases}\)
\(\Rightarrow\begin{cases}x=2\\x=-15\end{cases}\)
Vậy x thuộc { 2 ; - 15 }
c) \(\left(x-3\right).\left(x-5\right)< 0\)
=> \(x-3\) và \(x-5\) trái dấu
Mà \(x-3>x-5\Rightarrow x-3>0\) và \(x-5< 0\)
\(\Rightarrow\begin{cases}x-3>0\Rightarrow x>3\\x-5< 0\Rightarrow x< 5\end{cases}\)
=> \(3< x< 5\)
=> \(x=4\)
Vậy x = 4