a) \(\dfrac{x}{3}=\dfrac{4}{12}\Rightarrow x=\dfrac{4}{12}\cdot3=\dfrac{12}{12}=1\)

b) \(\dfrac{x-1}{x-2}=\dfrac{3}{5}\) (Điều kiện : \(x\ne2\))

\(\Rightarrow5\left(x-1\right)=3\left(x-2\right)\)

\(\Leftrightarrow5x-5=3x-6\Leftrightarrow5x-3x=-6+5\Leftrightarrow2x=-1\Leftrightarrow x=-\dfrac{1}{2}\)

c) \(2x:6=\dfrac{1}{4}\Leftrightarrow2x=\dfrac{1}{4}\cdot6=\dfrac{6}{4}=\dfrac{3}{2}\Leftrightarrow x=\dfrac{3}{2}:2=\dfrac{3}{2}\cdot\dfrac{1}{2}=\dfrac{3}{4}\)

d) \(\dfrac{x^2+x}{2x^2+1}=\dfrac{1}{2}\)

\(\Rightarrow2\left(x^2+x\right)=2x^2+1\)

\(\Leftrightarrow2x^2+2x=2x^2+1\)

\(\Leftrightarrow2x^2+2x-2x^2=1\Leftrightarrow2x=1\Leftrightarrow x=\dfrac{1}{2}\).

\(a,\Leftrightarrow x^3-8-x\left(x^2-9\right)=1\\ \Leftrightarrow x^3-8-x^3+9x=1\\ \Leftrightarrow9x=9\Leftrightarrow x=1\\ b,\Leftrightarrow8x^3+12x^2+6x+1-8x^3 +12x^2-6x+1-24x^2+24x-1=0\Leftrightarrow1=0\Leftrightarrow x\in\varnothing\)

a) \(\Leftrightarrow x^3-8-x^3+9x=1\)

\(\Leftrightarrow9x=9\Leftrightarrow x=1\)

b) \(\Leftrightarrow8x^3+12x^2+6x+1-8x^3+12x^2-6x+1-24x^2+24x-6=5\)

\(\Leftrightarrow24x=9\Leftrightarrow x=\dfrac{3}{8}\)

Bài 1:
$2x(x+3)+(2x+3)(5-x)=2$

$\Leftrightarrow 2x^2+6x+(10x-2x^2+15-3x)=2$

$\Leftrightarrow 2x^2+6x+7x-2x^2+15=2$

$\Leftrightarrow 13x+15=2$

$\Leftrightarrow 13x=2-15=-13$

$\Leftrightarrow x=-13:13=-1$

Bài 2:

$x-y=4\Rightarrow x=y+4$. Thay vào $xy=5$ thì:

$(y+4)y=5$

$\Leftrightarrow y^2+4y-5=0$

$\Leftrightarrow (y-1)(y+5)=0$

$\Leftrightarrow y=1$ hoặc $y=-5$

Nếu $y=1$ thì $x=y+4=5$. Khi đó $x^3+y^3=5^3+1^3=126$

Nếu $y=-5$ thì $x=y+4=-1$. Khi đó: $x^3+y^3=(-1)^3+(-5)^3=-126$

1) \(\Rightarrow10x-16-12x+15=12x-16+11\)

\(\Rightarrow14x=4\Rightarrow x=\dfrac{2}{7}\)

2) \(\Rightarrow4x^2+4x+1-4x^2+13x-3-15=0\)

\(\Rightarrow17x=17\Rightarrow x=1\)

3) \(\Rightarrow\left(3x-1\right)\left(2x-7+6x-5\right)=0\)

\(\Rightarrow\left(2x-3\right)\left(3x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)

2: Ta có: \(\left(2x+1\right)^2-\left(4x-1\right)\left(x-3\right)-15=0\)

\(\Leftrightarrow4x^2+4x+1-4x^2+12x+x-3-15=0\)

\(\Leftrightarrow17x=17\)

hay x=1

a) 2x-5=3+2x-7x

2x-2x+7x=3+5

7x=8

  x=8/7

vậy x=8/7

a) 2x - 5 = 3 + 2x - 7x

=> 2x - 2x + 7x = 3 +5 

=> 7x = 8

=> x = 8/7

b) \(\left(2x-1\right)^2=\left(2x-1\right)^5\)

=> \(\left(2x-1\right)^2-\left(2x-1\right)^5=0\)

=> \(\left(2x-1\right)^2\left[1-\left(2x-1\right)^3\right]=0\)

=> \(\orbr{\begin{cases}\left(2x-1\right)^2=0\\1-\left(2x-1\right)^3=0\end{cases}}\)

=> \(\orbr{\begin{cases}2x-1=0\\\left(2x-1\right)^3=1\end{cases}}\)

=> \(\orbr{\begin{cases}2x=1\\2x-1=1\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{1}{2}\\2x=2\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{1}{2}\\x=1\end{cases}}\)

\(-2x-\frac{3}{4}=x-\frac{3}{5}\)

\(3x=\frac{3}{5}-\frac{3}{4}=-\frac{3}{20}\)

\(x=-\frac{1}{20}\)

b) \(\left|\frac{x}{2}-\frac{1}{3}\right|=\frac{1}{4}+\frac{3}{2}=\frac{7}{4}\)

=> \(\orbr{\begin{cases}\frac{x}{2}-\frac{1}{3}=\frac{7}{4}\\\frac{x}{2}-\frac{1}{3}=-\frac{7}{4}\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{25}{6}\\x=-\frac{17}{6}\end{cases}}}\)

a, \(-2x-\frac{3}{4}=x-\frac{3}{5}\)

\(-2x-\frac{3}{4}-x+\frac{3}{5}=0\)

\(-3x-\frac{3}{20}=0\)

\(\frac{3}{20}=-3x\Leftrightarrow x=\frac{1}{20}\)

\(a,-2x+\frac{-3}{4}=x-\frac{3}{5}\)

\(-2x-x=\frac{3}{4}-\frac{3}{5}\)

\(-3x=\frac{15}{20}-\frac{12}{20}\)

\(-3x=\frac{3}{20}\)

\(x=\frac{3}{20}\div\left(-3\right)\)

\(x=\frac{3}{20}.\frac{-1}{3}\)

\(x=-\frac{1}{20}\)

Vậy \(x=-\frac{1}{20}\).

\(b,\left|\frac{1}{2}x-\frac{1}{3}\right|-\frac{3}{2}=\frac{1}{4}\)

\(\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{1}{4}+\frac{3}{2}\)

\(\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{1}{4}+\frac{6}{4}\)

\(\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{7}{4}\)

\(\Rightarrow\orbr{\begin{cases}\frac{1}{2}x-\frac{1}{3}=\frac{7}{4}\\\frac{1}{2}x-\frac{1}{3}=-\frac{7}{4}\end{cases}}\Rightarrow\orbr{\begin{cases}\frac{1}{2}x=\frac{25}{12}\\\frac{1}{2}x=-\frac{17}{12}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{25}{6}\\x=-\frac{17}{6}\end{cases}}\)

Vậy \(x\in\left\{\frac{25}{6};-\frac{17}{6}\right\}\).

1: \(\left(2x-2\right)\left(3x+1\right)-\left(3x-2\right)\left(2x-3\right)=5\)

=>\(6x^2+2x-6x-2-\left(6x^2-9x-4x+6\right)=5\)

=>\(6x^2-4x-2-6x^2+13x-6=5\)

=>9x-8=5

=>9x=13

=>\(x=\frac{13}{9}\)

2: \(\left(1-3x\right)\left(3x-5\right)-\left(2x-4\right)\left(2-3x\right)=x-6\)

=>\(3x-5-9x^2+15x+\left(2x-4\right)\left(3x-2\right)=x-6\)

=>\(-9x^2+18x-5+6x^2-4x-12x+8=x-6\)

=>\(-3x^2+2x+3-x+6=0\)

=>\(-3x^2+x+9=0\)

=>\(3x^2-x-9=0\)

=>\(x^2-\frac13x-3=0\)

=>\(x^2-2\cdot x\cdot\frac16+\frac{1}{36}-\frac{109}{36}=0\)

=>\(\left(x-\frac16\right)^2=\frac{109}{36}\)

=>\(x-\frac16=\pm\frac{\sqrt{109}}{6}\)

=>\(x=\frac16\pm\frac{\sqrt{109}}{6}\)

3: \(\left(2x-1\right)\left(4x^2+2x+1\right)-\left(2x+1\right)\left(4x^2-2x+1\right)=5x+6\)

=>\(8x^3-1-8x^3-1=5x+6\)

=>5x+6=-2

=>5x=-8

=>\(x=-\frac85\)