\(a,\Rightarrow3x\left(x-5\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\\ b,\Rightarrow\left(x-3\right)\left(2x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{2}\end{matrix}\right.\\ c,Đề.sai\\ d,Sửa:\left(x-2\right)^2-16\left(5-2x\right)^2=0\\ \Rightarrow\left[x-2-4\left(5-2x\right)\right]\left[x-2+4\left(5-2x\right)\right]=0\\ \Rightarrow\left(x-2-20+8x\right)\left(x-2+20-8x\right)=0\\ \Rightarrow\left(9x-22\right)\left(18-7x\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{22}{9}\\x=\dfrac{18}{7}\end{matrix}\right.\)

5-x-16=40+x

-x-x=40+16-5

(-x).2=51

-x=51/2=>x=51/2

1.x=40

2.x=10

3.x=

15-x = 4x -5 

15 + 5 = 4x + x 

5x = 20

x= 4

x - 15 = 6 + 4x 

-15 - 6 = 4x -x 

3x = -21 

x= -7

a: =>6(2x-5)-3x(2x-5)=0

=>(2x-5)(6-3x)=0

=>x=5/2 hoặc x=2

b: \(\Leftrightarrow\left(x+2\right)^2=x^2+5x+4+x^2\)

\(\Leftrightarrow2x^2+5x+4=x^2+4x+4\)

=>x²+x=0

=>x(x+1)=0

=>x=-1

\(\Rightarrow4x^2-12x+9-x^2+9=3x^2+5x-2-15x\\ \Rightarrow-2x=-20\Rightarrow x=10\)

\(\Leftrightarrow4x^2-12x+9-x^2+9=3x^2+5x-2-15x\)

\(\Leftrightarrow-2x=-20\Leftrightarrow x=10\)

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

\(\Leftrightarrow4x^2-12x+9-x^2+9=3x^2+6x-x-2-15x\)

\(\Leftrightarrow-12x+18=-10x-2\)

\(\Leftrightarrow x=10\)

a: =>6(2x-5)-3x(2x-5)=0

=>(2x-5)(6-3x)=0

=>x=5/2 hoặc x=2

b: \(\Leftrightarrow x^2+5x+4+x^2=x^2+4x+4\)

=>x²+x=0

=>x(x+1)=0

=>x=0(loại) hoặc x=-1(nhận)

Ta có 4( 18 - 5x ) - 12( 3x - 7 ) = 15( 2x - 16 ) - 6( x + 14 )

⇔ 4.18 - 4.5x - 12.3x - 12.( - 7 ) = 15.2x - 15.16 - 6.x - 6.14

⇔ 72 - 20x - 36x + 84 = 30x - 240 - 6x - 84

⇔ 156 - 56x = 24x - 324 ⇔ 56x + 24x = 156 + 324

⇔ 80x = 480 ⇔ x = 6.

Vậy giá trị x cần tìm là x = 6.