\(2x^2-x-6=0\)

\(\Leftrightarrow2x^2-4x+3x-6=0\)

\(\Leftrightarrow2x\left(x-2\right)+3\left(x-2\right)=0\)

\(\Leftrightarrow\left(2x+3\right)\left(x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x+3=0\Rightarrow x=\frac{-3}{2}\\x-2=0\Rightarrow x=2\end{cases}}\)

\(2x^2-x-6=0\)

\(\Leftrightarrow2x^2+3x-4x-6=0\)

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

\(2x^2-x-6=0\)

\(\Leftrightarrow2x^2-x=6\)

\(\Leftrightarrow2x^2-x1=6\)

\(\Leftrightarrow x.\left(2x-1\right)=6\)

\(\Leftrightarrow x.\left(x+1\right)=6\)

\(\Rightarrow x=2\)

Ta có :
\(A=x^4-2x^3+3x^2-4x+5\)

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

\(=\left(x^2-x\right)^2+\left(\sqrt{2}x-\sqrt{2}\right)^2+3\)

Vì \(\hept{\begin{cases}\left(x^2-x\right)^2\ge0\forall x\\\left(\sqrt{2}x-\sqrt{2}\right)^2\ge0\forall x\end{cases}}\)

\(\Rightarrow\left(x^2-x\right)^2+\left(\sqrt{2}x-\sqrt{2}\right)^2\ge0\forall x\)

\(\Rightarrow\left(x^2-x\right)^2+\left(\sqrt{2}x-\sqrt{2}\right)^2+3\ge3\forall x\)

Dấu bằng xảy ra khi vì chỉ khi \(\hept{\begin{cases}\left(x^2-x\right)^2=0\\\left(\sqrt{2}x-\sqrt{2}=0\right)\end{cases}\Leftrightarrow\hept{\begin{cases}x^2-x=0\\\sqrt{2}x-\sqrt{2}=0\end{cases}\Leftrightarrow\hept{\begin{cases}x\left(x-1\right)=0\\\sqrt{2}\left(x-1\right)\end{cases}}}}\)

\(\Leftrightarrow x=1\)

Vậy \(A_{min}=3\)tại \(x=1\)

   x²-2xy-9z²+y²

= ( x²-2xy+y² ) -9z²

=(x-y)²-(3z)²

=(x-y-3z)(x-y+z)

Thay x=6;y=-4;z=30 ta có:

   (6+4-90)(6+4+90)

=(-80).100

=-8000

Hk tốt

a,  a(x-y) + bx - by = a(x - y ) +b.(x-y) = (x-y)(a-b)

b, ac+bc + a + b = c.(a+b) +(a+b) = (a+b)(c+1)

c, \(5a^2-5ax-7a+7x=5a.\left(a-x\right)-7.\left(a-x\right)=\left(a-x\right)\left(5a-7\right)\)

d, \(7z^2-7yz-4z+4y=7z.\left(z-y\right)-4.\left(z-y\right)=\left(z-y\right)\left(7z-4\right)\)

e, \(x^3+3x^2+3x+9=x^2.\left(x+3\right)+3\left(x+3\right)=\left(x+3\right)\left(x^2+3\right)\)

g, \(pq-p^2-5\left(p-q\right)=p.\left(q-p\right)+5\left(q-p\right)=\left(q-p\right)\left(p+5\right)\)

xin lỗi bài này mình không biết

( a + b + c )² = a² + b² + c² + 2ab + 2ac + 2bc

( a - b + c )² = a² + b² + c² - 2ab + 2ac - 2bc

( a - b - c )² = a² + b² + c² - 2ab - 2ac + 2bc

sao nữa