Câu hỏi của vũ nguyễn gia linh - Toán lớp 8 - Học toán với OnlineMath

Bn kham khảo nhé ! 

a, \(3x\left(x-2\right)-2x+4=3x\left(x-2\right)-2\left(x-2\right)=\left(3x-2\right)\left(x-2\right)\)

b, \(x^2+2xy+y^2-9z^2=\left(x+y\right)^2-\left(3z\right)^2=\left(x+y-3z\right)\left(x+y+3z\right)\)

c, \(x^2+xy-3x-3y=x\left(x+y\right)-3\left(x+y\right)=\left(x-3\right)\left(x+y\right)\)

d, \(x^2-4xy-9+4y^2=\left(x+2y\right)^2-3^2=\left(x+2y-3\right)\left(x+2y+3\right)\)

e, \(x^2-5x-xy+5y=x\left(x-y\right)-5\left(x-y\right)=\left(x-5\right)\left(x-y\right)\)

g, \(x^2+2xy-9+y^2=\left(x+y\right)^2-3^2=\left(x+y-3\right)\left(x+y+3\right)\)

h, \(7x\left(x-2021\right)-x+2021=7x\left(x-2021\right)-\left(x-2021\right)=\left(7x-1\right)\left(x-2021\right)\)

i, \(\left(3x+1\right)^2-\left(2x-3\right)^2=\left(3x+1-2x+3\right)\left(3x+1+2x-3\right)=\left(x+4\right)\left(5x-2\right)\)

c) \(3x\left(x-2\right)-2x+4\)

\(3x\left(x-2\right)-2\left(x-2\right)\)

\(\left(x-2\right)\left(3x-2\right)\)

d) \(x^2+2xy+y^2-9z^2\)

\(\left(x+y\right)^2-\left(3z\right)^2\)

\(\left(x-y-3z\right)\left(x-y+3z\right)\)

e)\(x^2+xy-3x-3y\)

\(x\left(x+y\right)-3\left(x+y\right)\)

\(\left(x+y\right)\left(x-3\right)\)

f) \(x^2-4xy-9+4y^2\)

\(\left(x-2y\right)^2-3^2\)

\(\left(x-2y-3\right)\left(x-2y+3\right)\)

g) \(x^2-5x-xy+5y\)

\(x\left(x-5\right)-y\left(x-5\right)\)

\(\left(x-5\right)\left(x-y\right)\)

h) \(x^2+2xy+y^2-9\)

\(\left(x+y\right)^2-3^2\)

\(\left(x-y-3\right)\left(x-y+3\right)\)

i) \(7x\left(x-2021\right)-x+2021\)

\(\left(x-2021\right)\left(7x-1\right)\)

j) \(\left(3x+1\right)^2-\left(2x-3\right)^2\)

\(\left(3x+1-2x+3\right)\left(3x+1+2x-3\right)\)

\(\left(x+4\right)\left(5x-2\right)\)

\(x^2-2xy+y^2-9z^2=\left(x-y\right)^2-\left(3z\right)^2=\left(x-y-3z\right)\left(x-y+3z\right)\)

\(x^2+2xy+y^2-9z^2=\left(x^2+2xy+y^2\right)-\left(3z\right)^2\)

\(=\left(x+y\right)^2-\left(3z\right)^2=\left(x+y+3z\right)\left(x+y-3z\right)\)

đề bài là gì?-..-

phân tích các đa thức sau thnahf nhân tử

\(x^2-2x+1=4\left(x-1\right)\)

\(\Leftrightarrow\left(x-1\right)^2-4\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1-4\right)\left(x-1\right)=0\Leftrightarrow x=5;1\)

\(\frac{20x^2-45}{\left(2x+3\right)^2}=\frac{5\left(4x^2-9\right)}{\left(2x+3\right)^2}=\frac{5\left(2x-3\right)\left(2x+3\right)}{\left(2x+3\right)^2}=\frac{5\left(2x-3\right)}{2x+3}\)( ĐKXĐ : x ≠ -3/2 )

\(\frac{5x^2-10xy}{2\left(2y-x\right)^3}=\frac{-\left(10xy-5x^2\right)}{2\left(2y-x\right)^3}=\frac{-5x\left(2y-x\right)}{2\left(2y-x\right)^3}=\frac{-5x}{2\left(2y-x\right)^2}\)( ĐKXĐ : y ≠ x/2 )

\(\frac{80x^3-125x}{3\left(x-3\right)-\left(x-3\right)\left(8-4x\right)}=\frac{5x\left(16x^2-25\right)}{\left(x-3\right)\left(3-8+4x\right)}=\frac{5x\left(4x-5\right)\left(4x+5\right)}{\left(x-3\right)\left(4x-5\right)}=\frac{5x\left(4x+5\right)}{x-3}\)

( ĐKXĐ : x ≠ 3 ; x ≠ 5/4 )