g: Ta có: \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)=0\)

\(\Leftrightarrow3\left(6x^2-5x+1\right)-\left(18x^2-29x+3\right)=0\)

\(\Leftrightarrow18x^2-15x+3-18x^2+29x-3=0\)

\(\Leftrightarrow14x=0\)

hay x=0

w

\(3.\left(2x-1\right).\left(3x-1\right)-\left(2x-3\right).\left(9x-1\right)=0\) 0

\(\Leftrightarrow18x^2-15x+1-18x^2+29x-3=0\)

\(\Leftrightarrow14x-2=0\)

\(\Leftrightarrow14x=2\)

\(\Leftrightarrow x=\frac{2}{14}\)

Vậy \(x=\frac{1}{7}\)

Xin lỗi chị, em thật sự ko biết giai bài này vì em lớp 5. chị kb vơi em nha

a) Ta có: \(6x\left(x-5\right)+3x\left(7-2x\right)=18\)

\(\Leftrightarrow6x^2-30x+21x-6x^2=18\)

\(\Leftrightarrow-9x=18\)

hay x=-2

Vậy: S={-2}

b) Ta có: \(2x\left(3x+1\right)+\left(4-2x\right)\cdot3x=7\)

\(\Leftrightarrow6x^2+2x+12x-6x^2=7\)

\(\Leftrightarrow14x=7\)

hay \(x=\dfrac{1}{2}\)

Vậy: \(S=\left\{\dfrac{1}{2}\right\}\)

c) Ta có: \(0.5x\left(0.4-4x\right)+\left(2x+5\right)\cdot x=-6.5\)

\(\Leftrightarrow0.2x-2x^2+2x^2+5x=-6.5\)

\(\Leftrightarrow5.2x=-6.5\)

hay \(x=-\dfrac{5}{4}\)

Vậy: \(S=\left\{-\dfrac{5}{4}\right\}\)

d) Ta có: \(\left(x+3\right)\left(x+2\right)-\left(x-2\right)\left(x+5\right)=6\)

\(\Leftrightarrow x^2+5x+6-\left(x^2+3x-10\right)=6\)

\(\Leftrightarrow x^2+5x+6-x^2-3x+10=6\)

\(\Leftrightarrow2x+16=6\)

\(\Leftrightarrow2x=-10\)

hay x=-5

Vậy: S={-5}

e) Ta có: \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)=0\)

\(\Leftrightarrow3\left(6x^2-5x+1\right)-\left(18x^2-29x+3\right)=0\)

\(\Leftrightarrow18x^2-15x+3-18x^2+29x-3=0\)

\(\Leftrightarrow14x=0\)

hay x=0

Vậy: S={0}

giúp mình với ;-;

ghi này chả hiểu j bn ak

ghi rõ ra coi

tui chịu,???

a: Ta có: \(\left(x+2\right)\left(x+3\right)-\left(x-2\right)\left(x+5\right)=6\)

\(\Leftrightarrow x^2+5x+6-x^2-3x+10=6\)

\(\Leftrightarrow2x=-10\)

hay x=-5

b: Ta có: \(\left(3x+2\right)\left(2x+9\right)-\left(x+2\right)\left(6x+1\right)=\left(x+1\right)-\left(x-6\right)\)

\(\Leftrightarrow6x^2+27x+4x+18-6x^2-x-12x-2=x+1-x+6\)

\(\Leftrightarrow18x+16=7\)

hay \(x=-\dfrac{1}{2}\)

c: Ta có: \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)=0\)

\(\Leftrightarrow3\left(6x^2-2x-3x+1\right)-\left(18x^2-2x-27x+3\right)=0\)

\(\Leftrightarrow18x^2-15x+3-18x^2+27x-3=0\)

hay x=0

a)  -3x(2x-5)-2x(2-3x)=7

=> -6x² + 15 - 4x + 6x² = 7

=> -6x² + 6x² + 15 -4x =7

=> 15 - 4x =7

=> 4x = 15-7 =8

=> x= 8:4 = 2

b) \(\left(9x-12x+4\right)\left(2-5x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}9x-12x+4=0\\2-5x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x\left(9-12\right)=-4\\5x=2\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}-3x=-4\\x=\frac{2}{5}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{4}{3}\\x=\frac{2}{5}\end{cases}}\)

Vay...

c)  (4-3x) = (5+2x)

=> 4-3x=5+2x

=> -3x - 2x = 5-4

=> x(-3-2) = 1

=> -5x = 1

=> x= \(\frac{-1}{5}\)

d) (2x-1)-3(2x-1)=0

=> 2x-1 - 6x + 3 =0

=> 2x - 6x = 1 -3

=> x(2-6)=-2

=> -4x= -2

=> x = \(\frac{1}{2}\)

d)(2x-1)-3(2x-1)

=>1(2x-1)-3(2x-1)=0

=>(1-3).(2x-1)=0

=>-2(2x-1)=0

=>2x-1=0

=>2x=-1

=>x=-0,5 

vay x =-0,5