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Bài này giống tìm nghiệm quá :
1) \(5\left(x-7\right)=0\)
\(\left(x-7\right)=0\div5\)
\(\left(x-7\right)=0\)
\(x=0+7\)
\(x=7\)
2) \(25\left(x-4\right)=0\)
\(\left(x-4\right)=0\div25\)
\(\left(x-4\right)=0\)
\(x=0+4\)
\(x=4\)
3) \(34\left(2x-6\right)=0\)
\(\left(2x-6\right)=0\div34\)
\(\left(2x-6\right)=0\)
\(2x=0+6\)
\(2x=6\)
\(x=6\div2\)
\(x=3\)
4) \(2007\left(3x-12\right)=0\)
\(\left(3x-12\right)=0\div2007\)
\(\left(3x-12\right)=0\)
\(3x=0+12\)
\(3x=12\)
\(x=12\div3\)
\(x=4\)
5) \(47\left(5x-15\right)=0\)
\(\left(5x-15\right)=0\div47\)
\(\left(5x-15\right)=0\)
\(5x=0+15\)
\(5x=15\)
\(x=15\div5\)
\(x=3\)
6) \(13\left(4x-24\right)=0\)
\(\left(4x-24\right)=0\div13\)
\(\left(4x-24\right)=0\)
\(4x=0+24\)
\(4x=24\)
\(x=24\div4\)
\(x=6\)
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 , 3 - ( 17 - x ) = 289 - ( 36 + 289 )
3 - 17 + x = 0 - 36
-14 + x = -36
x = -36 - ( - 14 ) = -22
B, 25 - ( x + 5 ) = -415 - ( 15 - 415 )
25 - x - 5 = 0 - 15
20 - x = -15
x = 20 - ( - 15 ) = 35
C , 34 + ( 21 - x ) = ( 3747 - 30 ) - 3746
34 + 21 - x = 1 - 30
55 - x = -29
x = 55 - (-29 ) = 74
D , -2x - ( x -17 ) = 34 - ( -x + 25 )
- 2x - x + 17 = 34 - 25 + x
- 3x + 17 = 9 + x
- 3x - x = 9 - 17
-4x = -8
x = -8 : ( - 4 )
x = 2
E , 17x + ( -16x - 37 ) = x + 43
17x - 16x -37 = x + 43
x - 37 = x + 43
-37 - 43 = x - x
- 80 = 0 ( vô lý )
G , ( x + 12 ) . (x - 3 ) = 0
\(\hept{\begin{cases}x+12=0\\x-3=0\end{cases}}\)
\(\hept{\begin{cases}x=-12\\x=3\end{cases}}\)
Bài 1 Tìm x biết:
a)65-(29-x)=32
65 -29+x=31
x=31-65+29
x=-5
b)(x+5)-(x+23)=x-34
x+5 -x +23 = x-34
(x-x)+ (23+5)=x-34
0+28=x-34
28=x-34
28+34=x
62=x
=>x=62
c)(16-x)+(x-38)=x+44
16-x+x-38=x+44
-x+x-x=44-16+38
-x=36
=>x=-36
d)-12+3(-x+7)=-18
3(-x+7)=-18+12
3(-x+7)=-6
-x+7=-6:3
-x+7=-2
-x=-2-7
-x=-9
=>x=9
Baif 2
d)|7-x|=10
=> \(\left[{}\begin{matrix}7-x=10\\7-x=-10\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=7-10\\x=-10-7\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=-3\\x=-17\end{matrix}\right.\)
e)(x-6).(7-2x)=0
\(\Rightarrow\)\(\left[{}\begin{matrix}x-6=0\\7-2x=0\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=0+6\\2x=7\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=6\\x=7:2\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=6\\x=3,5\end{matrix}\right.\)
f)(9-x).(2x+8)=0
\(\Rightarrow\)\(\left[{}\begin{matrix}9-x=0\\2x+8=0\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=0+9\\2x=-8\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=9\\x=-4\end{matrix}\right.\)
g)x(-x+8).(-3x-18)=0
\(\Rightarrow\) \(\left[{}\begin{matrix}x=0\\-x+8=0\\-3x-18=0\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=0\\-x=0+8\\-3x=0+18\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=0\\-x=8\\-3x=18\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=0\\x=-8\\x=18:\left(-3\right)\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=0\\x=-8\\x=-6\end{matrix}\right.\)
h)(-x+8).(x-54).(-24-x)=0
\(\Rightarrow\)\(\left[{}\begin{matrix}-x+8=0\\x-54=0\\-24-x=0\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}-x=8\\x=0+54\\-x=0+24\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=8\\x=54\\-x=24\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=8\\x=54\\x=-24\end{matrix}\right.\)
mik chỉ ghi kết quả thôi nha
vì làm ra dài dòng lw
=> ta có:
x = 7
x = 4
x = 3
x = 4
x = 3
nha bn
5(x-7)=0
=>x-7=0
=>x=0+7
=>x=7
25(x-4)=0
=>x-4=0
=>x=0+4
=>x=4
34(2x-6)=0
=>2x-6=0
=>2x=0+6
=>2x=6
=>x=6:2
=>x=3
2016(3x-12)=0
=>3x-12=0
=>3x=0+12
=>3x=12
=>x=12:3
=>x=4
47(5x-15)=0
=>5x-15=0
=>5x=0+15
=>5x=15
=>x=15:5
=>x=3
a) 5.(x-7)=0
=> x-7 = 0
x = 7
b) 34.(2x-6)=0
=> 2x -6 = 0
2x = 6
x = 3
5 (x - 7) = 0
=> x - 7 = 0
=> x = 7
34 (2x - 6) = 0
=> 2x - 6 = 0
=> 2x = 6
=> x = 3
5 (x - 7) = 0
<=> x-7=0
<=> x=7
34 (2x - 6) = 0
<=> 2x-6=0
<=> 2x=6
<=> x=3
a, 5(x-7)=0
=> x-7 = 0
x=7
b,34(2x-6)=0
=>2x-6=0
2x=6
x=3
a)\(5\left(x-7\right)=0\)
\(\Rightarrow x-7=0\)
\(\Rightarrow x=7\)
b)\(34\left(2x-6\right)=0\)
\(\Rightarrow2x-6=0\)
\(\Rightarrow2x=6\)
\(\Rightarrow x=3\)
5 (x - 7) = 0
x - 7 = 0 : 5
x - 7 = 0
x = 0 + 7
x = 7
34 (2x - 6) = 0
2x - 6 = 0 : 34
2x - 6 =0
2x =0+6
2x =6
x =6:2
x =3
x - 7 = 5 x 0= 0
x = 0 + 7
x = 7
2.x - 6 = 34 . 0 = 0
2.x = 0 + 6 =6
x = 6 : 2
x = 3
k mình nha!!! mình đang cần điểm hỏi đáp!!!
a) \(5\left(x-7\right)=0\)
\(x-7=0\)
\(\Rightarrow x=7\)
b) \(34.\left(2x-6\right)=0\)
\(2x-6=0\)
\(2x=6\)
\(\Rightarrow x=3\)
5(x-7)= 0
=> X-7=0
X=0+7
X=7
5.(x-7)=0
=> x - 7 = 0
x = 7
phần cn lại tự lm ik
34(2 x-6)=0
2x-6=0:34
2 x-6=0
2 x=0+6
2 x=6
X= 6:2
X= 3
5(x-7)=0
x-7=0
x=7
34(2x-6)=0
2x-6=0
2x=6
x=3
\(5\cdot\left(x-7\right)=0\)
\(\Rightarrow x-7=0\)
\(\Rightarrow x=0+7\)
\(\Rightarrow x=7\)
Vậy \(x=7\)
\(34\cdot\left(2x-6\right)=0\)
\(\Rightarrow2x-6=0\)
\(\Rightarrow2x=0+6\)
\(\Rightarrow2x=6\)
\(x=6:2\)
\(x=3\)
Vậy \(x=3\)
\(5.\left(x-7\right)=0\)
\(\Rightarrow x-7=0\)
\(\Rightarrow x=7\)
Vậy \(x=7\)
\(32.\left(2x-6\right)=0\)
\(2x-6=0\)
\(2x=6\)
\(x=3\)
Vậy \(x=3\)
5·(x−7)=0
⇒x−7=0
⇒x=0+7
⇒x=7
Vậy x=7
34·(2x−6)=0
⇒2x−6=0
⇒2x=0+6
⇒2x=6
x=6:2
x=3
Vậy x=3
\(5\left(x-7\right)=0\)
\(x-7=0\)
\(x=0+7\)
\(\Rightarrow x=7\)
\(34\left(2x-6\right)=0\)
\(2x-6=0\)
\(2x=0+6\)
\(2x=6\)\(x=6:2\Rightarrow x=3\)
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