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\(a,\left(15x+165\right):15=11+x\)
\(=>x+11=11+x\)
\(=>x-x=11-11\)
\(=>0=0\)
=>x nhận mọi giá trị
![](https://rs.olm.vn/images/avt/0.png?1311)
Tìm x :
(x - 21*13) : 11 = 30
(627 - 56,5) : (x: 2) = 163
2/3 + 1/3 : x = 1
165 - (35 : x + 3) * 15 = 15
![](https://rs.olm.vn/images/avt/0.png?1311)
a, ( x - 21 . 13 ) : 11 = 30
x - 273 = 30 . 11
x - 273 = 330
x = 330 + 273
x = 603
b, ( 627 - 56,5 ) : ( x : 2 ) = 163
570,5 : ( x : 2 ) = 163
x : 2 = 570,5 : 163
x : 2 = 3,5
x = 3,5 . 2
x = 7
c, 2/3 + 1/3 : x = 1
1/3 : x = 1 - 2/3
1/3 : x = 1/3
x = 1/3 : 1/3
x = 1
d, 165 - ( 35 : x + 3 ) . 15 = 15
165 - ( 35 : x + 3 ) = 15 : 15
165 - ( 35 : x + 3 ) = 1
35 : x + 3 = 165 - 1
35 : x + 3 = 164
35 : x = 164 - 3
35 : x = 161
x = 35 : 161
x = 5/23
Mk tính nhẩm nên chắc có chỗ sai nên bn cứ kiểm tra lại nha
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165 - ( 35 : x + 3 ) x 15 = 15
=> ( 35 : x + 3) x 15 = 165 - 15 = 150
=> 35 : x + 3 = 150 : 15 = 10
=> 35 : x = 10 - 3 = 7
=> x = 35 : 7 = 5
Mik nghĩ bạn chép sai đề, có lẽ đề như thế này:
165-(35:x+3)x15=15
(35:x+3)x15=165-15
(35:x+3)x15=150
35:x+3 =150:15
35:x+3 =10
35:x =10-3
35:x =7
x =35:7
x =5
![](https://rs.olm.vn/images/avt/0.png?1311)
a)2012(x-9)=1-1
2012(x-9)=0
x-9=0:2012
x-9=0
x=0+9
x=9
b)15(3x+5)=165-45
15(3x+5)=120
3x+5=120:15
3x+5=8
3x=8-5
3x=3
x=3:3
x=1
1) 2012.[ x - 9 ] + 1 = 1
<=> 2012.[ x - 9 ] = <=> x - 9 = 0 : 2012 <=> x - 9 = 0 => x = 9
2) 165 - 15[ 3x + 5 ] = 5.9
<=>165 - 15[ 3x + 5 ] = 45 <=> 15[ 3x + 5 ] = 120 <=> 3x + 5 = 8 <=> 3x = 3 => x = 1
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x-\frac{11}{15}=x+\frac{3}{5}\)
\(\Rightarrow x-x=\frac{3}{5}+\frac{11}{15}\)
\(\Rightarrow0x=\frac{4}{3}\)
\(\Rightarrow x\in\varnothing\)
x - 11/15 = x + 3/5
=> x-x = 3/5 + 11/15
=> 0x = 4/3 ( vô lý )
Vậy không có x thỏa mãn
![](https://rs.olm.vn/images/avt/0.png?1311)
`[ ( 2 xx x - 11 ) : 3 + 1 ] xx 5 = 20`
`( 2 xx x - 11 ) : 3 + 1=20:5`
`( 2 xx x - 11 ) : 3 + 1=4`
`( 2 xx x - 11 ) : 3 =4-1`
`( 2 xx x - 11 ) : 3 =3`
`2xx x -11=3xx3`
`2xx x -11=9`
`2xx x =9+11`
`2 xx x=20`
`x=20:2`
`x=10`
Vậy `x=10`
`b, x - 96 = ( 443 - x ) - 15`
`x-96=443-x-15`
` x+x=443-15+96`
`2x=524`
`x=524:2`
`x= 262`
Vậy `x=262`
\(#Nqoc\)
`a)`
\([ ( 2 \times x - 11 ) \div 3 + 1 ] \times 5 = 20\)
`(2 \times x - 11) \div 3 + 1 = 20 \div 5`
`(2 \times x - 11) \div 3 + 1 = 4`
`(2 \times x - 11) \div 3 = 4 - 1`
`(2 \times x - 11) \div 3 = 3`
`2 \times x - 11 = 3 \times 3`
`2 \times x - 11 = 9`
`2 \times x = 9 + 11`
`2 \times x = 20`
`x = 20 \div 2`
`x = 10`
Vậy, `x = 10`
`b)`
\(x - 96 = ( 443 - x ) - 15\)
`x - 96 = 443 - x - 15`
`x - 96 = 428 - x`
`x = 428 - x + 96`
`x = 524 - x`
`x - 524 + x = 0`
`(x + x) - 524 = 0`
`2x - 524 = 0`
`2x = 524`
`x = 524 \div 2`
`x = 262`
Vậy, `x = 262.`
![](https://rs.olm.vn/images/avt/0.png?1311)
( 285 - x - 0,15 ) : 0,25 = 60
284,85 - x = 15
x = 269,85
2010 \(x\)x =4062220
x = 2021,004975