\(x^2-xy-12y^2=0\)

\(\Leftrightarrow\left(x^2+3xy\right)-\left(4xy-12y^2\right)=0\)

\(\Leftrightarrow x\left(x+3y\right)-4y\left(x+3y\right)=0\)

\(\Leftrightarrow\left(x+3y\right)\left(x-4y\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=-3y\\x=4y\end{cases}}\)

TH1:\(x=-3y\)

\(A=\frac{3\cdot\left(-3y\right)+2y}{3\left(-3y\right)-2y}=\frac{-9y+2y}{-9y-2y}=\frac{-7y}{-11y}=\frac{7}{11}\)

TH2:\(x=4y\)

\(A=\frac{3\cdot4y+2y}{3\cdot4y-2y}=\frac{12y+2y}{12y-2y}=\frac{14y}{10y}=\frac{7}{5}\)

\(A=x^2-7xy+12y^2\)

\(A=x^2-3xy-4xy+12y^2\)

\(A=x\left(x-3y\right)-4y\left(x-3y\right)\)

\(A=\left(x-4y\right)\left(x-3y\right)\)

\(B=x^2-3xy-4y^2\)

\(B=x^2+xy-4xy-4y^2\)

\(B=x\left(x+y\right)-4y\left(x+y\right)\)

\(B=\left(x-4y\right)\left(x+y\right)\)

\(A=x^2-7xy+12y^2\)

\(=x^2-3xy-4xy+12y^2\)

\(=x\left(x-3y\right)-4y\left(x-3y\right)\)

\(=\left(x-4y\right)\left(x-3y\right)\)

4/3xy.(x - 7xy)

= 4/3xy.x + 4/3xy.(-7xy)

= 4/3x²y - 28/3x²y²

\(4/3xy . ( x - 7xy )\)

\(\iff 4/3xy . x + 4/3xy . ( -7xy )\)

\(\iff 4/3x^2y - 28/3x^2y^2\)

_Hok_tốt

\(x^2-x-1=0\)

\(\Leftrightarrow\left(x^2-2\cdot\frac{1}{2}\cdot x+\frac{1}{4}\right)-\frac{5}{4}=0\)

\(\Leftrightarrow\left(x-\frac{1}{2}\right)^2-\frac{5}{4}=0\)

\(\Leftrightarrow\left(x-\frac{1}{2}\right)^2=\frac{5}{4}\)

\(\Leftrightarrow x=\frac{\sqrt{5}}{2}+\frac{1}{2};x=\frac{-\sqrt{5}}{2}+\frac{1}{2}\)

\(x^2-2x-1=0\)

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

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

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

\(\Leftrightarrow x=\sqrt{2}+1;x=-\sqrt{2}+1\)

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

<=> x² - 4 = 0

<=> x² = 4

<=> x² = 2²;(-2)²

=> x = 2; -2

Mình phân tích thành nhân tử đó do ko rõ đề

\(\left(a^3-b^3\right)+\left(a-b\right)^2\)

\(=\left(a-b\right)\left(a^2+ab+b^2\right)+\left(a-b\right)^2\)

\(=\left(a-b\right)\left(a^2+ab+b^2+a-b\right)\)

1,\(\left(x-3\right)^3-5\left(x-2\right)+5=0\)

\(\Rightarrow\left(x-3\right)^3-5\left(x-3\right)=0\)

\(\Rightarrow\left(x-3\right)\left[\left(x-3\right)^2-5\right]=0\)

\(\Rightarrow\orbr{\begin{cases}x-3=0\\\left(x-3\right)^2-5=0\end{cases}}\)

\(\Rightarrow x=3\) hoặc \(x=\sqrt{5}+3\) hoặc \(x=-\sqrt{5}+3\) 

Vậy........

1) \(2x^2-3x-2\)

\(=2x^2-4x+x-2\)

\(=2x\left(x-2\right)+x-2\)

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

2) \(3x^2-7x-10\)

\(=3x^2+3x-10x-10\)

\(=3x\left(x+1\right)-10\left(x+1\right)\)

\(=\left(3x-10\right)\left(x+1\right)\)