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a) \(70-\left(x-3\right)=45\)
\(x-3=70-45\)
\(x-3=25\)
\(x=25+3\)
\(x=28\)
b) \(12+\left(5+x\right)=20\)
\(5+x=20-12\)
\(5+x=8\)
\(x=8-5\)
\(x=3\)
c) \(130-\left(100+x\right)=25\)
\(100+x=130-25\)
\(100+x=105\)
\(x=105-100\)
\(x=5\)
d) \(175+\left(30-x\right)=200\)
\(30-x=200-175\)
\(30-x=25\)
\(x=30-25\)
\(x=5\)
e) \(\left(x+12\right)+22=92\)
\(x+12=92-22\)
\(x+12=70\)
\(x=70-12\)
\(x=58\)
f) \(95-\left(x+2\right)=45\)
\(x+2=95-45\)
\(x+2=50\)
\(x=50-2\)
\(x=48\)
a)
70 - (x - 3) = 45
x - 3 = 70 - 45 = 25
x = 25 + 3 = 28
Vậy x = 28
b)
12 + (5 + x) = 20
5 + x = 20 - 12 = 8
x = 8 - 5 = 3
Vậy x = 3
c)
130 - (100 + x) = 25
100 + x = 130 - 25 = 115
x = 115 - 100 = 15
Vậy x = 15
d)
175 + (30 - x) = 200
30 - x = 200 - 175 = 25
x = 30 - 25 = 5
Vậy x = 5
e)
(x + 12) + 22 = 92
x + 12 = 92 - 22 = 70
x = 70 - 12 = 58
Vậy x = 58
f)
95 - (x + 2) = 45
x + 2 = 95 - 45 = 50
x = 50 - 2 = 48
Vậy x = 48
a) \(70-5\left(x-3\right)=45\)
\(5\left(x-3\right)=25\)
\(x-3=5\)
\(\Rightarrow x=8\)
b) \(12+\left(5+x\right)=20\)
\(5+x=8\)
\(\Rightarrow x=3\)
c) \(130-\left(100+x\right)=25\)
\(100+x=105\)
\(\Rightarrow x=5\)
d) \(175+\left(30-x\right)=200\)
\(30-x=25\)
\(\Rightarrow x=5\)
e) \(5\left(x+12\right)+22=92\)
\(5\left(x+12\right)=70\)
\(x+12=14\)
\(\Rightarrow x=2\)
a. 70 - 5.(x - 3) = 45
5.(x - 3) = 70 - 45
x - 3 = 25 : 5
x = 5 + 3
x = 8
b. 12 + (5 + x) = 20
5 + x = 20 - 12
x = 8 - 5
x = 3
c. 130 - (100 + x) = 25
100 + x = 130 - 25
x = 105 - 100
x = 5
d. 175 + (30 - x) = 200
30 - x = 200 - 175
x = 30 - 25
x = 5
e. 5.(x + 12) + 22 = 92
5.(x + 12) = 92 - 22
x + 12 = 70 : 5
x = 14 - 12
x = 2
a) 12 + ( 5 + x ) = 20
⇔ 5 + x = 8
⇔ x = 3
Vậy x = 3
b) 120 + ( 50 + x ) = 180
⇔ 50 + x = 60
⇔ x = 10
Vậy x = 10
c) 175 + ( 30 - x ) = 200
⇔ 30 - x = 25
⇔ x = 5
Vậy x = 5
d) 130 - ( 100 + x ) = 25
⇔ 100 + x = 105
⇔ x = 5
Vậy x = 5
e) 145 - ( 125 + x ) = 12
⇔ 125 + x = 133
⇔ x = 8
Vậy x = 8
j) 5 ( x + 12 ) + 22 = 92
⇔ 5x + 60 + 22 = 92
⇔ 5x + 82 = 92
⇔ 5x = 10
⇔ x = 2
Vậy x = 2
g) 3 ( x + 23 ) + 6 = 96
⇔ 3x + 69 + 6 = 96
⇔ 3x + 75 = 96
⇔ 3x = 21
⇔ x = 7
Vậy x = 7
Tìm x :
a) 12 + ( 5 + x ) = 20
5 + x =20 - 12
5 + x = 8
x = 3
b) 120 + ( 50 + x ) = 180
50 + x = 180-120
50 + x = 60
x = 60 - 50
x = 10
c) 175 + ( 30 - x ) = 200
30 -x = 200 - 175
30 - x = 25
x = 30-25
x=5
d) 130- ( 100 +x ) =25
100 + x =130 - 25
100+x= 105
x=105-100
x=5
e) 145 - (125+x) = 12
125+x = 145-12
125+x=133
x= 133-125
x=8
j) 5 (x+12) +22 = 92
5 ( x+12) = 92 -22
5 (x+12)=70
x+12=70 : 5
x + 12 = 14
x = 14-12
x=2
g) 2 (x+23) + 6 =96
2 (x+23) = 96 - 6
2 (x+23)= 90
x+23=90:2
x+23=45
x= 45-23
x=22
CHÚC BẠN HỌC TỐT
Bài 3:
a: Ta có: 60-3(x-2)=51
\(\Leftrightarrow x-2=3\)
hay x=5
b: Ta có: \(4x-20=25:2^2\)
\(\Leftrightarrow4x=\dfrac{25}{4}+20=\dfrac{105}{4}\)
hay \(x=\dfrac{105}{16}\)
c: Ta có: \(8\cdot6+288:\left(x-3\right)^2=50\)
\(\Leftrightarrow288:\left(x-3\right)^2=50-48=2\)
\(\Leftrightarrow\left(x-3\right)^2=144\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=12\\x-3=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=15\\x=-9\end{matrix}\right.\)
\(a,5x-16=40-1\)
<=>\(5x=40-1+16\)
<=>\(5x=55\)
<=>\(x=11\)
\(b,x-10=-25\)
<=>\(x=\left(-25\right)-10\)
<=>\(x=-35\)
\(c,-12+x=-30\)
<=>\(x=\left(-30\right)+12\)
<=>\(x=-18\)
a) 5x-16=40-1
5x-16=39
5x=39+16
5x=55
x=55:5
x=11
b)x-10=(-25)
x=(-25)+10
x=(-15)
c) -12+x= -30
x= -30-(-12)
x= -30+12
x= -18
d) 2x +12 = -40+ 6
2x+12=(-34)
2x=(-34)-12
2x=(-46)
x=(-46):2
x=(-23)
e)125:(3x-13)=25
3x-13=125:25
3x-13=5
3x=5+13
3x=18
x=18:3
x=6
f)541+(218-x)=735
218-x=735-541
218-x=194
x=218-194
x= 24
g) 3(2x+1)-19=14
3(2x+1)=14+19
3(2x+1)=33
2x+1=33:3
2x+1=11
2x=11-1
2x=10
x=10:2
x=5
h)175-5(x+3)=85
5(x+3)=175-85
5(x+3)=90
x+3=90:5
x+3=18
x=18-3
x=15
i)8x+(-3)=39
8x=39-(-3)
8x=42
x=42:8
x=42/8
L)2x+4x=36+(-6)
6x=30
x=30:6
x=5
`@` `\text {Ans}`
`\downarrow`
`c)`
`( 34 - 2x ) . ( 2x - 6 ) = 0`
`=>`\(\left[{}\begin{matrix}34-2x=0\\2x-6=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}2x=34\\2x=6\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=34\div2\\x=6\div2\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=17\\x=3\end{matrix}\right.\)
Vậy, `x \in {17; 3}`
`d)`
`( 2019 - x ) . ( 3x - 12 ) =0` `?`
`=>`\(\left[{}\begin{matrix}2019-x=0\\3x-12=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=2019-0\\3x=12\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=2019\\x=12\div3\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=2019\\x=4\end{matrix}\right.\)
Vậy, `x \in {2019; 4}`
`e) `
`57 . ( 9x - 27 ) = 0`
`=>`\(9x-27=0\div57\)
`=> 9x - 27 = 0`
`=> 9x = 27`
`=> x = 27 \div 9`
`=> x = 3`
Vậy, `x = 3`
`f)`
`25 + ( 15 - x ) = 30`
`=> 15 - x = 30 - 25`
`=> 15 - x = 5`
`=> x = 15 -5 `
`=> x = 10`
Vậy, `x = 10`
`g) `
`43 - ( 24 - x ) = 20`
`=> 24 - x = 43 - 20`
`=> 24 - x = 23`
`=> x = 24 - 23`
`=> x = 1`
Vậy, `x = 1`
`h) `
`2 . ( x - 5 ) - 17 = 25`
`=> 2 ( x - 5) = 25+17`
`=> 2 ( x - 5) = 42`
`=> x - 5 = 42 \div 2`
`=> x - 5 = 21`
`=> x = 21 + 5`
`=> x = 26`
Vậy, `x = 26`
`i)`
`3 . ( x + 7 ) - 15 = 27`
`=> 3(x + 7) = 27 + 15`
`=> 3(x + 7) = 42`
`=> x +7 = 42 \div 3`
`=> x + 7 = 14`
`=> x = 14 - 7`
`=> x = 7`
Vậy, `x = 7`
`j)`
`15 + 4 . ( x - 2 ) = 95`
`=> 4(x - 2) = 95 - 15`
`=> 4(x - 2) = 80`
`=> x - 2 = 80 \div 4`
`=> x - 2 = 20`
`=> x = 20 + 2`
`=> x = 22`
Vậy, `x = 22`
`k)`
`20 - ( x + 14 ) = 5`
`=> x + 14 = 20 - 5`
`=> x + 14 = 15`
`=> x = 15 - 14`
`=> x = 1`
Vậy, `x = 1`
`l) `
`14 + 3 . ( 5 - x ) = 27`
`=> 3(5 - x) = 27 - 14`
`=> 3(5 - x) = 13`
`=> 5 - x = 13 \div 3`
`=> 5 - x = 13/3`
`=> x = 5- 13/3`
`=> x = 2/3`
Vậy, `x = 2/3.`
`@` `\text {Kaizuu lv uuu}`
\(\frac{2x-5}{\left(175-25.6\right).20}=100\)
<=> \(\frac{2x-5}{500}\)\(=100\)
=> \(2x-5=50000\)
<=>\(x=\frac{50005}{2}=25002,5\)