\(A=x^4+6x^2+8x^3-2x-3\)

\(B=3x^2+x^4+4x^3-3x+5\)\(\Rightarrow2B=6x^2+2x^4+8x^3-6x+10\)

\(\Rightarrow A-2B=x^4+6x^2+8x^3-2x-3\)\(-6x^2-2x^4-8x^3+6x-10\)

\(=-x^4+4x-13\)

Ta có 

\(A=x^4+8x^3+6x^2-2x-3\)

\(B=x^4+4x^3+3x^2-3x+5\Rightarrow2B=2x^4+8x^3+6x^2-6x+10\)

\(A-2B=x^4+8x^3+6x^2-2x-3\)\(-2x^4-8x^3-6x^2+6x-10\)

\(A-2B=-x^4+4x-13\)

len google di ban

mk chua hoc bai nay

a.

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

\(\Rightarrow6x^5-3x^3+15x^2=6x^5-3x^3+15x^2\)

\(=6x^5-3x^2+15x^2-6x^5-3x^3+15x^2\)

= 0

b.

\(\left(4xy+3y-5x\right)x^2y=4x^3y^2+3x^2y^2-5x^3y\)

\(\Rightarrow4x^3y^2+3x^2y^2-5x^3y=4x^3y^2+3x^2y^2-5x^3y\)

= 0

\(\frac{2x-3}{\left(7-6x\right)^2}+\frac{x-2}{\left(7-6x\right)^2}=\frac{6x-3}{\left(3x-5\right)^2}-\frac{12x-10}{\left(3x-5\right)^2}\)

\(\Leftrightarrow\frac{2x-3+x-2}{\left(7-6x\right)^2}=\frac{6x-3-12x+10}{\left(3x-5\right)^2}\)

\(\Leftrightarrow\frac{3x-5}{\left(7-6x\right)^2}=\frac{7-6x}{\left(3x-5\right)^2}\)

\(\Leftrightarrow\left(7-6x\right)^3=\left(3x-5\right)^3\)

\(\Leftrightarrow7-6x=3x-5\)

\(\Leftrightarrow7+5=3x+6x\)

\(\Leftrightarrow12=9x\)

\(\Leftrightarrow x=\frac{4}{3}\)

Vậy \(x=\frac{4}{3}\)