a)<=>(x^2+x-3)(x^2+x-2)-12=(x-2)(x+3)(x^2+x+1)

TH1:=>x-2=0

=>x=2

TH2:x+3=0

=>x=-3

dựa vô bệt thức ta thấy

D<0=> phương trình ko có nghiệm thực

=>x=-3 hoặc 2

nhớ tick nhé

a)x=-3 hoặc 2

câu 1 có sai đề ko z

\(1.6x\left(x-10\right)-2x+20=0\)

⇔\(6x\left(x-10\right)-2\left(x-10\right)=0\)

⇔ \(2\left(x-10\right)\left(3x-1\right)=0\)

⇔ x = 10 hoặc x = \(\dfrac{1}{3}\)

KL....

\(2.3x^2\left(x-3\right)+3\left(3-x\right)=0\)

⇔ \(3\left(x-3\right)\left(x^2-1\right)=0\)

⇔ \(x=+-1\) hoặc \(x=3\)

KL....

\(3.x^2-8x+16=2\left(x-4\right)\)

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

⇔ \(\left(x-4\right)\left(x-6\right)=0\)

⇔ \(x=4\) hoặc \(x=6\)

KL.....

\(4.x^2-16+7x\left(x+4\right)=0\)

\(\text{⇔}4\left(x+4\right)\left(2x-1\right)=0\)

⇔ \(x=-4hoacx=\dfrac{1}{2}\)

KL.....

\(5.x^2-13x-14=0\)

⇔ \(x^2+x-14x-14=0\)

\(\text{⇔}\left(x+1\right)\left(x-14\right)=0\)

\(\text{⇔}x=14hoacx=-1\)

KL......

Còn lại tương tự ( dài quá ~ )

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

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

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

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

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

\(=\dfrac{1}{x^2+x+1}\)

b) \(\dfrac{9}{x^3-9x}-\dfrac{-1}{x+3}\)

\(=\dfrac{9}{x\left(x-3\right)\left(x+3\right)}+\dfrac{1}{x+3}\)

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

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

c) \(\dfrac{x^3-8}{5x+10}.\dfrac{x^2+4x}{x^2+2x+4}\)

\(=\dfrac{x\left(x-2\right)\left(x^2+2x+4\right)\left(x+4\right)}{5\left(x+2\right)\left(x^2+2x+4\right)}\)

\(=\dfrac{x\left(x-2\right)\left(x+4\right)}{5\left(x+2\right)}\)

d) \(\dfrac{5x+10}{4x-8}.\dfrac{4-2x}{x+2}\)

\(=\dfrac{5\left(x+2\right)}{4\left(x-2\right)}.\dfrac{2\left(2-x\right)}{x+2}\)

\(=-\dfrac{10\left(x+2\right)\left(x-2\right)}{4\left(x-2\right)\left(x+2\right)}\)

\(=-\dfrac{5}{2}\)

e) \(\dfrac{\left(x-13\right)^2}{2x^5}.\dfrac{-3x^2}{x-13}\)

\(=\dfrac{x-13}{2x^3}.\dfrac{-3}{1}\)

\(=\dfrac{-3\left(x-13\right)}{2x^3}\)

g) \(\dfrac{x^2+6x+9}{1-x}.\dfrac{\left(x-1\right)^2}{2\left(x+3\right)^2}\)

\(=-\dfrac{\left(x+3\right)^2}{x-1}.\dfrac{\left(x-1\right)^2}{2\left(x+3\right)^2}\)

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

\(=-\dfrac{x-1}{2}\).