\(5-7x^2=\left(\sqrt{5}\right)^2-\left(x\sqrt{7}\right)^2\)

\(=\left(\sqrt{5}-x\sqrt{7}\right)\left(\sqrt{5}+x\sqrt{7}\right)\)

\(3+4x=\left(\sqrt{3}\right)^2-\left(2\sqrt{x}\right)^2\) ( do x<0 )

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

\(x^4-14x^2-7x+30=\left(x^4+x^3-3x^2\right)+\left(-x^3-x^2+3x\right)+\left(-10x^2-10x+30\right)\)

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

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

x^4-14*x^2-7*x+30=(x^2-x-10)*(x^2+x-3)

đề sai ko? ~0~

\(x^4-14x^2-7x+30\)

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

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

Nhớ chọn cho tớ nhe!

\(=\left(x+2\right)\left(x+5\right)\)

\(x^2+7x+10=\left(x^2+5x\right)+\left(2x+10\right)=x\left(x+5\right)+2\left(x+5\right)=\left(x+2\right)\left(x+5\right)\)

=8x^2+12x=4x(2x+3)

\(x^2+7x^2+12x=8x^2+12x=4x\left(2x+3\right)\)

x³ - 7x² + 10x

=x³ - 2x² - 5x² + 10x

=(x³ -2x²) - (5x² - 10x)

= x²( x - 2) - 5x( x - 2)

= (x - 2) (x² - 5x)

Gõ trên máy tính nên hơi lâu 

Cảm ơn mình đê

x²-7x+12

=x²-4x-3x+12

=x(x-4)-3(x-4)

=(x-3)(x-4)

        \(x^2-7x+12\)

\(=\left(x^2-4x\right)-\left(3x-12\right)\)

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

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