12+x trên 42=5/6
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Bài 2:
a: =>x-35=-23
=>x=12
b: =>|x-8|=13
=>x-8=13 hoặc x-8=-13
=>x=21 hoặc x=-5
Bài 1:
a: =42-98-42+12-12=-98
b: =10x4x3x(-25)=40x(-25)x3=-1000x3=-3000
a, 2 x 31 x 12 + 4 x 6 x 42 + 8 x 27 x 3
= 2 x 12 x 31 + 4 x 6 x 42 + 8 x 3 x 27
= 24 x 31 + 24 x 42 + 24 x 27
= 24 x ( 31 + 24 + 27 )
= 24 x 82
= 1968
b, 2 x 53 x 12 + 4 x 6 x 87 - 3 x 8 x 40
= 24 x 53 + 24 x 87 - 24 x 40
= 24 x ( 53 + 87 - 40 )
= 24 x 100
= 2400
c, Tương tự
\(a,12+\frac{x}{42}=\frac{5}{6}\)
\(\Rightarrow\frac{x}{42}=\frac{5}{6}-12\)
\(\Rightarrow\frac{x}{42}=\frac{-67}{6}\)
\(\Leftrightarrow6x=42\times(-67)\)
\(\Leftrightarrow6x=-2814\)
\(\Leftrightarrow x=-2814\div6=-469\)
Vậy x = -469
\(b,25-\frac{x}{40}=\frac{3}{8}\)
\(\frac{x}{40}=25-\frac{3}{8}\)
\(\frac{x}{40}=\frac{197}{8}\)
\(\Leftrightarrow8x=40\times197\)
\(\Leftrightarrow8x=7880\)
\(\Leftrightarrow x=7880\div8=985\)
Vậy x = 985
Chúc bạn học tốt
a, \(12+\frac{x}{42}=\frac{5}{6}\Leftrightarrow\frac{x}{42}=-\frac{67}{6}\Leftrightarrow\frac{x}{42}=\frac{-469}{42}\)
Khử mẫu đi ta đc : \(x=-469\)
b, \(25-\frac{x}{40}=\frac{3}{8}\Leftrightarrow\frac{x}{40}=\frac{197}{8}\Leftrightarrow\frac{x}{40}=\frac{985}{40}\)
Khử mẫu đi ta đc : \(x=985\)
6 x 4 = 24 12 : 6 = 2 7 x 3 = 21 63 : 7 = 9
7 x 5 = 35 42 : 7 = 6 6 x 8 = 48 48 : 6 = 8
6 x 6 = 36 28 : 7 = 4 7 x 8 = 56 49 : 7 = 7
6 x 4 = 24 12 : 6 = 02 7 x 3 = 21 63 : 7 = 09 7 x 5 = 35 42 : 7 = 06
6 x 8 = 48 48 :6 = 8 6 x 6 = 36 28 : 7 = 04 7 x 8 = 56 49 : 7 = 07
a)=>x=13+3/5=68/5.
Vậy x = 68/5
b)=>x/6=5/7+-23/42
=>x/6=1/6
=>x=1
Vậy x=1
còn lại tự làm nha😆😆😆😆😆
a) x-3/5=13
suy ra x=13+3/5
suy ra x=68/5
suy ra x=13.6
b)x/6=5/7+23/-42
suy ra x/6=1/6
suy ra x=1
d) x+11/12=2/5
suy ra x=2/5-11/12
suy ra x=-31/60
`#3107.101107`
a)
\(x+x+\dfrac{1}{2}\times\dfrac{2}{5}+x+\dfrac{8}{10}=121\\3x+\dfrac{1}{5}+\dfrac{4}{5}=121\\ 3x+1=121\\ 3x=121-1\\ 3x=120\\ x=40 \)
Vậy, `x = 40`
b)
\(\dfrac{12+x}{42}=\dfrac{5}{6}\\ \dfrac{12+x}{42}=\dfrac{35}{42}\\ \dfrac{12+x}{42}-\dfrac{35}{42}=0\\ \dfrac{12+x-35}{42}=0\\ \dfrac{x-\left(35-12\right)}{42}=0\\ \dfrac{x-23}{42}=0\\ x-23=0\\ x=23\)
Vậy,` x = 23.`
a: \(x+x+\dfrac{1}{2}\cdot\dfrac{2}{5}+x+\dfrac{8}{10}=121\)
=>\(3x+\dfrac{1}{5}+\dfrac{4}{5}=121\)
=>3x+1=121
=>3x=120
=>x=40
b: \(\dfrac{x+12}{42}=\dfrac{5}{6}\)
=>\(x+12=42\cdot\dfrac{5}{6}=35\)
=>x=35-12=23
a) \(⇔x^2-9x+20=12 \)
\(⇔x^2-9x+8=0\)
\(⇔x^2-x-8x+8=0\)
\(⇔(x-1)(x-8)=0\)
\(⇔\left[\begin{array}{} x-1=0\\ x-8=0 \end{array}\right.⇔\left[\begin{array}{} x=1\\ x=8 \end{array}\right.\)
b) \(⇔4x^2-12x+8=3\)
\(⇔4x^2-12x+5=0\)
\(⇔(2x-1)(2x-5)=0\)
\(⇔\left[\begin{array}{} 2x-1=0\\ 2x-5=0 \end{array}\right.⇔\left[\begin{array}{} x=\frac{1}{2}\\ x=\frac{5}{2} \end{array}\right.\)
c) \(⇔x^2+x-30=42\)
\(⇔x^2+x-72=0\)
\(⇔(x-9)(x+8)=0\)
\(⇔\left[\begin{array}{} x-9=0\\ x+8=0 \end{array}\right.⇔\left[\begin{array}{} x=9\\ x=-8 \end{array}\right.\)
d) \(⇔2x^2+5x-3=-6\)
\(⇔2x^2+5x+3=0\)
\(⇔(x+1)(2x+3)=0\)
\(⇔\left[\begin{array}{} x+1=0\\ 2x+3=0 \end{array}\right.⇔\left[\begin{array}{} x=-1\\ x=-\frac{3}{2} \end{array}\right.\)
a, x=1; y=2 => 12
x=2; y=1 => 21
b, x=1; y=5 => 15
x=5; y=1 => 51
c, x=1; y=6 => 16
x=6;y=1 => 61
x=2; y=3=> 23
x=3; y=2 => 32
d, x=1; y=8 => 18
x=2; y=4 => 24
x=4; y=2 => 42
x=8; y=1 => 81
\(\frac{12+x}{42}=\frac{5}{6}\)
=> \(\frac{12+x-35}{42}=0\)
=> \(\frac{x-23}{42}=0\)
=> x - 23 = 0
-> x = 23
\(\frac{12+x}{42}\)\(=\frac{5}{6}\)
\(=>12+x.6=42.5\)
\(12+x.6=210\)
\(12+x=210:6=35\)
\(x=35-12=23\)
\(=>x=23\)