=>\(\frac{6\left(x-1\right)}{18}+\frac{9\left(3x-5\right)}{18}+\frac{2\left(2x\right)}{18}+\frac{2\left(-5x+3\right)}{18}=\frac{1}{2}\)

=>\(\frac{6x-6}{18}+\frac{27x-45}{18}+\frac{4x}{18}+\frac{-10x+6}{18}=\frac{1}{2}\)

=>\(\frac{\left(6x-6\right)+\left(27x-45\right)+4x+\left(-10x+6\right)}{18}=\frac{1}{2}\)

=>\(\frac{27x-45}{18}=\frac{1}{2}\)

=>54x-90=18

54x =108

=>x=2

quy đồng lên rồi nhân chéo rồi giải

x=2

x=2 nhớ k  cho mình nha

a) x= 23

b) x=8

\(\frac{x-3}{x+5}=\frac{5}{7}\)

=> 7 ( x - 3 ) = 5 ( x + 5 )

7x - 21 = 5x + 25

7x - 5x = 25 + 21

2x = 46

x = 23

\(\frac{x-3}{x+5}=\frac{5}{7}\)

=>   \(7\left(x-3\right)=5\left(x+5\right)\)

<=>  \(7x-21=5x+25\)

<=>  \(2x=46\)

<=>  \(x=23\)

Vậy...

\(\frac{7}{x-1}=\frac{x+1}{9}\)

<=>  \(\left(x-1\right)\left(x+1\right)=63\)

<=>  \(x^2-1=63\)

<=>  \(x^2=64\)

<=>  \(x=\pm8\)

Vậy...

(x - 2/3)³ = -1/27

=> (x - 2/3)³ = (-1/3)³

=> x - 2/3 = -1/3

=> x = -1/3 + 2/3

=> x = 1/3

Từ bài ra ta có \(\left(x-\frac{2}{3}\right)^3=\left(\frac{-1}{3}\right)^3\)

\(\Rightarrow x-\frac{2}{3}=\frac{-1}{3}\)

\(\Rightarrow x=\frac{-1}{3}+\frac{2}{3}\)

\(\Rightarrow x=\frac{1}{3}\)

Vậy ... nếu đúng thì k nha

\(C=\frac{2x^2-3x-9}{x^2-9}=0\)

\(\Leftrightarrow2x^2-3x-9=0\)

Ta có : \(\Delta=\left(-3\right)^2-4.2.\left(-9\right)=81>0\)

\(\Rightarrow x1=\frac{3-\sqrt{81}}{2.2}=\frac{3-9}{4}=-\frac{3}{2}\)

\(\Rightarrow x2=\frac{3+\sqrt{81}}{2.2}=\frac{3+9}{4}=3\)