12 : 3x 4=
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a) \(\Rightarrow72-20x-36x+84=30x-240-6x-84\)
\(\Rightarrow80x=480\Rightarrow x=6\)
b) \(\Rightarrow15x+25-8x+12=5x+6x+36+1\)
\(\Rightarrow4x=0\Rightarrow x=0\)
c) \(\Rightarrow10x-16-12x+15=12x-16+11\)
\(\Rightarrow14x=4\Rightarrow x=\dfrac{2}{7}\)

\(a,\frac{7}{3}x+5=\frac{9}{6}x-7\)
\(\frac{7}{3}x-\frac{9}{6}x=-7-5\)
\(\left(\frac{7}{3}-\frac{9}{6}\right).x=-12\)
\(\frac{5}{6}.x=-12\)
\(x=\left(-12\right):\frac{5}{6}\)
\(x=-\frac{72}{5}\)
\(b,3x+\frac{6}{4}=5-\frac{7x}{6}\)
\(3x+\frac{7x}{6}=5-\frac{6}{4}\)
\(\left(3+\frac{7}{6}\right).x=\frac{7}{2}\)
\(\frac{25}{6}.x=\frac{7}{2}\)
\(x=\frac{7}{2}:\frac{25}{6}\)
\(x=\frac{21}{25}\)
\(c,\frac{12}{5}x+6=\frac{17}{3}x+12\)
\(\frac{12}{5}x-\frac{17}{3}x=12-6\)
\(-\frac{49}{15}x=6\)
\(x=6:\left(-\frac{49}{15}\right)\)
\(x=-\frac{90}{49}\)
\(d,6\left(3x+7\right)=12\left(2x-4\right)\)
\(18x+42=24x-48\)
\(18x-24x=-48-42\)
\(-6x=-90\)
\(x=15\)
\(e,13\left(2x-9\right)=8\left(3x-13\right)\)
\(26x-117=24x-104\)
\(26x-24x=-104+117\)
\(2x=13\)
\(x=\frac{13}{2}\)

a) Ta có: \(\dfrac{x+5}{3x-6}-\dfrac{1}{2}=\dfrac{2x-3}{2x-4}\)
\(\Leftrightarrow\dfrac{2\left(x+5\right)}{6\left(x-2\right)}-\dfrac{3\left(x-2\right)}{6\left(x-2\right)}=\dfrac{3\left(2x-3\right)}{6\left(x-2\right)}\)
Suy ra: \(2x+5-3x+6=6x-9\)
\(\Leftrightarrow-x+11-6x+9=0\)
\(\Leftrightarrow20-7x=0\)
\(\Leftrightarrow7x=20\)
hay \(x=\dfrac{20}{7}\)(thỏa ĐK)
Vậy: \(S=\left\{\dfrac{20}{7}\right\}\)
Tim x, bt:
a) 4.(18- 5x) - 12.( 3x-7) =15.(2x-16) - 6.(x+14)
b) 5.(3x+5) - 4.(2x-3) =5x + 3x(2x-12) +1

a) 4.(18- 5x) - 12.( 3x-7) =15.(2x-16) - 6.(x+14)
72 - 20x - 36x + 84 = 30x - 240 - 6x - 84
-20x - 36x - 30x + 6x = -240 - 84 - 72 -84
-80x = -480
x= 6
4.(18 - 5x) - 12.(3x - 7) = 15.(2x - 16) - 6.(x+14)
4.18 - 4.5x - 12.3x + 12.7 = 15.2x - 15.16 - 6x - 6.14
72 - 20x - 36x + 84 = 30x - 240 - 6x - 84
72 + 84 - 20x - 36x = 30x - 6x - 240 - 84
156 - 56x = 24x - 324
156 = 24x - 324 + 56x
156 = 80x - 324
80x - 324 = 156
80x = 156 + 324
80x = 480
x = 480:80
x = 6
câu b giải tương tự

a) ĐKXĐ: \(x\notin\left\{\dfrac{1}{3};-\dfrac{1}{3}\right\}\)
Ta có: \(\dfrac{12}{1-9x^2}=\dfrac{1-3x}{1+3x}-\dfrac{1+3x}{1-3x}\)
\(\Leftrightarrow\dfrac{\left(1-3x\right)^2}{\left(1+3x\right)\left(1-3x\right)}-\dfrac{\left(1+3x\right)^2}{\left(1-3x\right)\left(1+3x\right)}=\dfrac{12}{\left(1-3x\right)\left(1+3x\right)}\)
Suy ra: \(9x^2-6x+1-9x^2-6x-1=12\)
\(\Leftrightarrow-12x=12\)
hay x=-1(thỏa ĐK)
Vậy: S={-1}
12:3x4=16
12 : 3 x 4
= 4 x 4
= 16