Tìm x: x+1 /4=12 ???
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2. 34 x 18 + 18 x 66
= 18 x ( 34 + 66)
= 18 x 100 = 1800
3. X × ( 8 + 12) = 160 + 20 × 12
= X x 20 = 160 + 240
= X x 20 = 400
X = 400 : 20 = 20
4. X x 12 - X x 2 = 2020
(12 - 2) x X = 2020
10 x X = 2020
X = 2020 : 10 = 202
Đáp án là B vì 12: -3 = -4; 12: -4 = -3; 12: -6 = -2;12: -12 = -1 và đáp ứng điều kiện a< -2
13 - ( 2 x X - 1) = 4 4 x X - 12 = 40
( 2 x X - 1) = 13 - 4 4 x X = 40 + 12
( 2x X -1) = 9 4 x X = 52
2 x X = 9 + 1 X = 52 : 4
2 x X = 10 X = 13
X = 10 : 2
X = 5
\(\frac{x}{12}=\frac{4}{3x-3+\frac{1}{12}}\)
đề như này à
- \(\dfrac{1}{4}\) < \(\dfrac{x}{12}\) < \(\dfrac{1}{6}\) tìm \(x\)
\(\dfrac{-3}{12}\) < \(\dfrac{x}{12}\) < \(\dfrac{2}{12}\)
-3 < \(x\) < 2
vì \(x\) \(\in\) Z nên \(x\) \(\in\) { -2; -1; 0; 1}
Kết luận: \(x\) \(\in\) {-2; -1; 0; 1}
`3/5 : x =1/3 +1/2`
` 3/5 : x= 2/6 +3/6`
` 3/5 : x= 5/6`
` x= 3/5 : 5/6`
` x= 3/5 xx 6/5`
` x= 18/25`
__
`x: 7/15 =2`
` x= 2xx 7/15`
` x= 14/15`
__
`x-3/2=11/4-5/4`
`x-3/2= 6/4`
`x= 3/2 +3/2`
`x= 6/2`
`x=3`
__
`x+5/4 = 3/2+7/12`
`x+5/4 = 18/12+7/12`
`x+5/4 = 25/12`
`x= 25/12-5/4`
`x= 25/12- 15/12`
`x= 10/12`
`x= 5/6`
a: x*3/4=1/5
=>x=1/5:3/4=1/5*4/3=4/15
b: x*3/7=2/5
=>x=2/5:3/7=2/5*7/3=14/15
c: 1/3+2/9=2/12x
=>1/6x=3/9+2/9=5/9
=>x=5/9*6=30/9=10/3
d: 4/15*x-2/3=1/5
=>4/15*x=2/3+1/5=10/15+3/15=13/15
=>4x=13
=>x=13/4
e: x:1/7=2/3
=>x=2/3*1/7=2/21
f: 1/9:x=7/3
=>x=1/9:7/3=1/9*3/7=3/63=1/21
j: 1/4+5/12=8/3:x
=>8/3:x=3/12+5/12=8/12=2/3
=>x=4
h: =>7/4:x=1/5+1/2=7/10
=>x=7/4:7/10=10/4=5/2
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9/12 . x + 1/4 .x = 3/4
(9/12 + 1/4) .x = 3/4
1 .x = 3/4
x = 3/4 : 1
x = 3/4
Trả lời:
\(\frac{9}{12}\cdot x+\frac{1}{4}\cdot x=\frac{3}{4}\)
\(\left(\frac{9}{12}+\frac{1}{4}\right)\cdot x=\frac{3}{4}\)
\(1\cdot x=\frac{3}{4}\)
\(x=\frac{3}{4}:1\)
\(x=\frac{3}{4}\)
Trả lời :
a, \(\frac{3}{4}-\left(\frac{1}{2}\div x+\frac{1}{2}\right)=\frac{3}{5}\)
=> \(\frac{1}{2}\div x+\frac{1}{2}=\frac{3}{20}\)
=> \(\frac{1}{2}\div x=\frac{-7}{20}\)
=> \(x=\frac{-10}{7}\)
b, (4 - x) . (2x + 3) = 0
=> \(\orbr{\begin{cases}4-x=0\\2x+3=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=4\\x=\frac{-3}{2}\end{cases}}\)
c, \(\frac{4}{-3}=\frac{-12}{x}\)
=> 4x = 36
=> x = 9
d, \(\frac{4x}{-3}=\frac{12}{-x}\)
=> \(-4x^2=-36\)
=> 4x2 = 36
=> x2 = 9
=> x = \(\pm3\)
\(x+\dfrac{1}{4}=12\)
\(x=12-\dfrac{1}{4}\)
\(x=\dfrac{47}{4}\)
\(x+\dfrac{1}{4}=12\)
\(x=12-\dfrac{1}{4}\)
\(x=\dfrac{47}{4}\)
Vậy \(x=\dfrac{47}{4}\)