2:

a: A(x)=0

=>5x-10-2x-6=0

=>3x-16=0

=>x=16/3

b: B(x)=0

=>5x^2-125=0

=>x^2-25=0

=>x=5 hoặc x=-5

c: C(x)=0

=>2x^2-x-3=0

=>2x^2-3x+2x-3=0

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

=>x=3/2 hoặc x=-1

a: \(=\dfrac{5}{3}x^2-x+\dfrac{1}{3}\)

b: \(=-5y-9+xy\)

a: \(=x^2-4-x^2+x+8=x+4\)

a) (x-2)(x+2)-x(x-1)+8
= x²-4-x²+x+8
= (x²-x²)+(-4+8)+x
= 4+x
b) bn viết lại đề đi:v
đọc khó quá.

b: Ta có: \(5\left(x-1\right)^2-\left(1-x\right)\)

\(=5\left(x-1\right)^2+\left(x-1\right)\)

\(=\left(x-1\right)\left(5x-5+1\right)\)

\(=\left(x-1\right)\left(5x-4\right)\)

a: Ta có: \(5x^2-4xy-x^2y\)

\(=x\left(5x-4y-xy\right)\)

\(\text{a. (x – 2y)^2=x^2- 2x2y+ (2y)^2 =x^2-4xy+4y^2 }\)

\(\text{b. (2x^2 +3)^2= (2x^2)^2+2.2x.3+3^2=16x^4+12x+9}\)

\(\text{c. (x – 2)(x^2 + 2x + 4)=x^3-2^3=x^3-6 }\)

\(\text{d. (2x – 1)^3= 2x^3-3.(2x)^2.1+3.2x.1^2-1^3=8x^3-12x^2+6x-1}\)

c: \(=x^2+6xy+9y^2\)

e: \(=x^4-4y^2\)

\(1,\\ a,=2x^2+2x\\ b,=x^2+4x+3-4=x^2+4x-1\\ c,=x^2+4x+4+3x-5=x^2+7x-1\\ 2,\\ a,=3\left(x+y\right)\\ b,=\left(x-3\right)^2\\ c,=7\left(x+y\right)\\ 3,\\ \Leftrightarrow\left(x-1\right)\left(3x-5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{5}{3}\end{matrix}\right.\)

a) 2x²+2x

a: \(=\dfrac{2x-2x+y}{2\left(2x-y\right)}=\dfrac{y}{2\left(2x-y\right)}\)

b: \(=\dfrac{3x+1}{\left(x-1\right)\left(x+1\right)}-\dfrac{x}{2\left(x-1\right)}\)

\(=\dfrac{6x+2-x^2-x}{2\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{-x^2+5x+2}{2\left(x-1\right)\left(x+1\right)}\)

c: \(=\dfrac{1}{x+2}+\dfrac{x+8}{3x\left(x+2\right)}\)

\(=\dfrac{3x+x+8}{3x\left(x+2\right)}=\dfrac{4x+8}{3x\left(x+2\right)}=\dfrac{4}{3x}\)

d: \(=\dfrac{4x+6-2x^2+3x+2x+1}{\left(2x-3\right)\left(2x+3\right)}\)

\(=\dfrac{-2x^2+9x+7}{\left(2x-3\right)\left(2x+3\right)}\)