3xX=72:4

3xX=18

X=18:3

X=6

a)Ta có:

\(x^2-7x+12=0\)

\(\Leftrightarrow x^2-3x-4x+12=0\)

\(\Leftrightarrow x\left(x-3\right)-4\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=3\end{matrix}\right.\)

b) Ta có:

\(x\left(x-4\right)-3\left(4-x\right)=0\)

\(\Leftrightarrow x\left(x-4\right)+3\left(x-4\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=4\end{matrix}\right.\)

\(a,\dfrac{3}{8}=\dfrac{6}{x}\\ \Rightarrow x=6:\dfrac{3}{8}\\ \Rightarrow x=16\\ b,\dfrac{1}{9}=\dfrac{x}{27}\\ \Rightarrow x=\dfrac{1}{9}.27\\ \Rightarrow x=3\\ c,\dfrac{4}{x}=\dfrac{8}{6}\\ \Rightarrow x=4:\dfrac{4}{3}\\ \Rightarrow x=3\\ d,\dfrac{3}{x-5}=\dfrac{-4}{x+2}\\ \Rightarrow3\left(x+2\right)=-4\left(x-5\right)\\ \Rightarrow3x+6=-4x+20\\ \Rightarrow3x+6+4x-20=0\\ \Rightarrow7x-14=0\\ \Rightarrow7x=14\\ \Rightarrow x=2\)

a: =>6/x=3/8

hay x=16

b: =>x/27=1/9

nên x=3

c: =>4/x=4/3

nên x=3

d: =>3/x-5=-4/x+2

=>3x+2=-4x+20

=>7x=18

hay x=18/7

a: Ta có: \(4\left(2x+7\right)^2-9\left(x+3\right)^2=0\)

\(\Leftrightarrow\left(4x+14-3x-9\right)\left(4x+14+3x+9\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(7x+23\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\dfrac{23}{7}\end{matrix}\right.\)

c: Ta có: \(\left(x-3\right)^2-4=0\)

\(\Leftrightarrow\left(x-5\right)\cdot\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)

b. 

PT $\Leftrightarrow (5x^2-2x+10)^2-(3x^2+10x-8)^2=0$

$\Leftrightarrow (5x^2-2x+10-3x^2-10x+8)(5x^2-2x+10+3x^2+10x-8)=0$

$\Leftrightarrow (2x^2-12x+18)(8x^2+8x+2)=0$

$\Leftrightarrow (x^2-6x+9)(4x^2+4x+1)=0$

$\Leftrightarrow (x-3)^2(2x+1)^2=0$

$\Leftrightarrow (x-3)(2x+1)=0$

$\Leftrightarrow x-3=0$ hoặc $2x+1=0$

$\Leftrightarrow x=3$ hoặc $x=-\frac{1}{2}$

d.

$x^2-2x=24$

$\Leftrightarrow x^2-2x-24=0$

$\Leftrightarrow (x+4)(x-6)=0$
$\Leftrightarrow x+4=0$ hoặc $x-6=0$

$\Leftrightarrow x=-4$ hoặc $x=6$

b: Ta có: \(\left(2x+7\right)^2=9\left(x+2\right)^2\)

\(\Leftrightarrow\left(3x+4-2x-7\right)\left(3x+4+2x+7\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(5x+11\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{11}{5}\end{matrix}\right.\)

c: ta có: \(\left(x+2\right)^2=9\left(x^2-4x+4\right)\)

\(\Leftrightarrow\left(3x-6\right)^2-\left(x+2\right)^2=0\)

\(\Leftrightarrow\left(3x-6-x-2\right)\left(3x-6+x+2\right)=0\)

\(\Leftrightarrow\left(2x-8\right)\left(4x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=1\end{matrix}\right.\)

\(2^x+2^{x+3}=72\)

\(\Rightarrow2^x\cdot\left(1+2^3\right)=72\)

\(\Rightarrow2^x\cdot9=72\)

\(\Rightarrow2^x=72:9\)

\(\Rightarrow2^x=8\)

\(\Rightarrow2^x=2^3\)

\(\Rightarrow x=3\)

tìm x? `(x+3)/(x-2)=0`?

Bài 1:

a. 

$\frac{2}{3}\times \frac{x}{y}=\frac{8}{15}$

$\frac{x}{y}=\frac{8}{15}: \frac{2}{3}=\frac{4}{5}$

b.

$\frac{x}{y}: \frac{3}{4}=\frac{2}{5}$

$\frac{x}{y}=\frac{3}{4}\times \frac{2}{5}=\frac{3}{10}$

c.

$\frac{3}{5}: \frac{x}{y}=\frac{4}{7}$

$\frac{x}{y}=\frac{3}{5}: \frac{4}{7}=\frac{21}{20}$

Bài 2:

Chiều dài hình chữ nhật là:

$\frac{3}{5}: \frac{3}{4}=\frac{4}{5}$ (m)

Chu vi hình chữ nhật:

$2\times (\frac{3}{4}+\frac{4}{5})=\frac{31}{10}$ (m)