Tìm x bieets:
a) x + 17 \(⋮\) (x - 12)
b) 3x + 6 \(⋮\) (x - 3)
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\(a,0,5x-\frac{2}{3}x=\frac{7}{12}\Rightarrow\frac{1}{2}x-\frac{2}{3}x=\frac{7}{12}\)
\(\Rightarrow x\left(\frac{1}{2}-\frac{2}{3}\right)=\frac{7}{12}\Rightarrow x\cdot\left(\frac{3}{6}-\frac{4}{6}\right)=\frac{7}{12}\)
\(\Rightarrow x\cdot\left(-1\right)=\frac{7}{12}\Rightarrow x=\frac{7}{12}:\left(-1\right)=\frac{7}{-12}\)
\(c,\frac{\left(x-5\right)}{12}\cdot\frac{9}{29}=\frac{-6}{29}\Rightarrow\frac{\left(x-5\right)}{12}=\frac{-6}{29}:\frac{9}{26}\)
\(\frac{\Rightarrow\left(x-5\right)}{12}=\frac{-6}{9}=\frac{-2}{3}\Rightarrow x-5=-\frac{2}{3}\cdot12\)
\(\Rightarrow x-5=\frac{-24}{3}=-8\Rightarrow x=-8+5=-3\)
\(a,0,5x-\frac{2}{3}x=\frac{7}{12}\)
\(\Rightarrow\frac{1}{2}x-\frac{2}{3}x=\frac{7}{12}\)
\(\Rightarrow-\frac{1}{6}x=\frac{7}{12}\)
\(\Rightarrow x=-\frac{7}{2}\)
\(c,\frac{x-5}{12}\cdot\frac{9}{29}=-\frac{6}{29}\)
\(\Rightarrow\frac{x-5}{12}=-\frac{2}{3}\)
\(\Rightarrow x-5=12.\left(-\frac{2}{3}\right)\)
\(\Rightarrow x-5=-8\)
\(\Rightarrow x=-3\)
Tìm số nguyên x, biết:
1) -16 + 23 + x = - 16
7+x=-16
x=-16-7
x=-23
2) 2x – 35 = 15
2x=15+35
2x=50
x=50:2
x=25
3) 3x + 17 = 12
3x=12-17
3x=-5
x=-5/3
4) (2x – 5) + 17 = 6
2x-5=6-17
2x-5=-11
2x=-11+5
2x=-6
x=-6:2
x=-3
5) 10 – 2(4 – 3x) = -4
2(4-3x)=10-(-4)
2(4-3x)=14
4-3x=14:2
4-3x=7
3x=4-7
3x=-3
x=-3:3
x=-1
6) - 12 + 3(-x + 7) = -18
3(-x+7)=-18-(-12)
3(x+7)=-6
x+7=-6:3
x+7=-2
x=-2-7
x=-9
a) x/7 = 6/21
x/7 = 2/7
=>x=2
b)2/3x -1/2=1/10
2/3x = 1/10+1/2
2/3x = 3/5
x=3/5:2/3
x=9/10
c)1/4+1/3 : x = -5
d)3x+17=2
đ)2/3x +1/4 =7/12
2x+3/10= 11/6 . 6/11
x : 4 1/3 = -2,5
a) -16 + 23 + x = -16
=> 7 + x = -16
=> x = -16 - 7
=> x = -23
b) 3x + 17 = 12
=> 3x = 12 - 17
=> 3x = -5
=> x = -5/3
vì x là số nguyên => x ko có gtri thõa mãn đề bài
c) (2x - 5) + 17 = 16
=> 2x - 5 = 16 - 17
=> 2x - 5 = -1
=> 2x = -1 + 5
=> 2x = 4
=> x = 4 : 2
=> x = 2
\(-16+23+x=-16\)
\(\Rightarrow x=-16-23+16\)
\(\Rightarrow x=-23\)
\(3x+17=12\)
\(3x=-5\)
\(x=-\frac{5}{3}\)
\(24:\left(3x-2\right)=-3\)
\(\Rightarrow3x-2=-8\)
\(3x=-6\)
\(x=-2\)
\(\left(2x-5\right)+17=16\)
\(2x=16-17+5\)
\(x=2\)
\(-6< x< 3\)
\(\Rightarrow x\in\left\{-5;-4;-3;-2;-1;0;1;2\right\}\)
"Tự tính tổng "
Bài 1:
a: \(x=\dfrac{2}{3}:\dfrac{3}{5}=\dfrac{2}{3}\cdot\dfrac{5}{3}=\dfrac{10}{9}\)
b: \(x=\dfrac{17}{8}:\dfrac{7}{17}=\dfrac{17}{8}\cdot\dfrac{17}{7}=\dfrac{289}{56}\)
c: \(x=-\dfrac{3}{4}:\dfrac{7}{12}=\dfrac{-3}{4}\cdot\dfrac{12}{7}=\dfrac{-63}{28}=-\dfrac{9}{4}\)
d: \(\Leftrightarrow x\cdot\dfrac{1}{6}=\dfrac{3}{8}-\dfrac{1}{4}=\dfrac{1}{4}\)
hay \(x=\dfrac{1}{4}:\dfrac{1}{6}=\dfrac{3}{2}\)
e: \(\Leftrightarrow\dfrac{1}{2}:x=-4-\dfrac{1}{3}=-\dfrac{17}{3}\)
hay \(x=-\dfrac{1}{2}:\dfrac{17}{3}=\dfrac{-3}{34}\)
a/ \(\frac{x+17}{x-12}=\frac{x-12+29}{x-12}=1+\frac{29}{x-12}\)
Để x+17 chia hết cho (x-12) thì (x-12) là ước của 29
Bạn tự liệt kê ra.
b/ ta có \(\frac{3x+6}{x-3}=\frac{3\left(x-3\right)+33}{x-3}=3+\frac{33}{x-3}\)
Đến đây giải tương tự như trên.