Tìm x
A,102÷(2012-(x-12)=6
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(1-2+3-4+5-6+7-8+...+101-102+103) * x-120=2012
[ -1 + ( -1 ) ... ( -1 ) + 103] * x -120 = 2012
( -1 x 51 + 103 ) * x - 120 = 2012
( -51 +103) * x = 2012 + 120
52 * x =2132
x= 2132 : 52
x= 41
a) |5/3 - x| - |-5/6| = |-5/9|
=> |5/3 - x| - 5/6 = 5/9
=> |5/3 - x| = 5/9 + 5/6
=> |5/3 - x| = 25/18
=> \(\orbr{\begin{cases}\frac{5}{3}-x=\frac{25}{18}\\\frac{5}{3}-x=-\frac{25}{18}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{5}{18}\\x=\frac{55}{18}\end{cases}}\)
a, \(\left|\frac{5}{3}-x\right|-\left|-\frac{5}{6}\right|=\left|-\frac{5}{9}\right|\)
\(\Leftrightarrow\left|\frac{5}{3}-x\right|-\frac{5}{6}=\frac{5}{9}\Rightarrow\left|\frac{5}{3}-x\right|=\frac{5}{9}+\frac{5}{6}=\frac{25}{18}\)
\(\Rightarrow\orbr{\begin{cases}\frac{5}{3}-x=\frac{25}{18}\\\frac{5}{3}-x=-\frac{25}{18}\end{cases}\Rightarrow}x.\)
a, 71.2 – 6.(2x+5) = 10 5 : 10 3
71.2 – 6.(2x+5) = 10 2
6.(2x+5) = 71.2 – 100
6.(2x+5) = 42
x = 1
b, 5 x + 3 4 . 6 8 = 6 9 . 3 4
5 x + 3 4 . 6 8 = 6 8 . 6 . 3 4
5 x + 3 4 = 6 8 . 6 . 3 4 : 6 8 = 6 . 3 4
5x = 6 . 3 4 - 3 4 = 5 . 3 4
x = 3 4
c, 12:{390:[5. 10 2 – ( 5 3 + x . 7 2 )]} = 4
390:[5. 10 2 – ( 5 3 + x . 7 2 )] = 12:4 = 3
5. 10 2 – ( 5 3 + x . 7 2 ) = 390:3 = 130
5 3 + x . 7 2 = 5. 10 2 – 130 = 370
x . 7 2 = 370 – 5 3 = 245
x = 245: 7 2 = 5
d, 5 3 .(3x+2):13 = 10 3 : 13 5 : 13 4
5 3 .(3x+2):13 = 10 3 : 13
3x+2 = 10 3 : 13 : 5 3 .13 = 8
x = 2
- 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
a,A=0-2+4-6+...+2010-2012
A=-2+-2+...+-2+-2 (503 cặp)
A=-2.503
A=-1006
1: \(\Leftrightarrow x=UCLN\left(24;36;150\right)=6\)
2: \(\Leftrightarrow x\in\left\{24;48;72;...\right\}\)
mà 16<=x<=50
nên \(x\in\left\{24;48\right\}\)
3: \(\Leftrightarrow x\inƯ\left(6\right)\)
mà x>-10
nên \(x\in\left\{1;-1;2;-2;3;-3;6;-6\right\}\)
4: \(\Leftrightarrow x\in BC\left(4;5;8\right)\)
\(\Leftrightarrow x\in\left\{...;-40;0;40;80;120;160;200;...\right\}\)
mà -20<x<180
nên \(x\in\left\{0;40;80;120;160\right\}\)
\(102:\left[2012-\left(x-12\right)\right]=6\)
\(102:\left[2012-x+12\right]=6\)
\(102:\left(2024-x\right)=6\)
\(2024-x=102:6\)
\(2024-x=17\)
\(x=2007\)
Ta có \(102\div\left(2012-\left(x-12\right)\right)=6\)
\(\Rightarrow2012-\left(x-12\right)=102\div6\)
\(\Rightarrow2012-\left(x-12\right)=17\)
\(\Rightarrow x-12=2012-17\)
\(\Rightarrow x-12=1995\)
\(\Rightarrow x=1995+12\)
\(\Rightarrow x=2007\)