Question

1)(3x²+2x+4)²=(x²-4)²

2) (2x²-3x-4)²=(x²-x)²

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

4) \(\dfrac{x}{x-3}=\dfrac{1}{x+2}\)

5) \(\dfrac{4}{x-2}+\dfrac{x}{x+1}=\dfrac{x^2-2}{x^2-x-2}\)

gúp em tl câu hỏi trên vs ạ em đag cần gấp  em c.ơn trước

Answer 1

\(5,\dfrac{4}{x-2}+\dfrac{x}{x+1}-\dfrac{x^2-2}{\left(x-2\right)\left(x+1\right)}=0\left(dkxd:x\ne2;-1\right)\)

\(\Rightarrow4\left(x+1\right)+x\left(x-2\right)-x^2-2=0\)

\(\Rightarrow4x+4+x^2-2x-x^2-2=0\)

\(\Rightarrow2x+2=0\)

\(\Rightarrow x=-1\left(loai\right)\)

Vậy \(S=\varnothing\)

Answer 2

\(4,\dfrac{x}{x-3}-\dfrac{1}{x+2}=0\left(dkxd:x\ne3;-2\right)\)

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

\(\Rightarrow x^2+3x-x+3=0\)

\(\Rightarrow x^2+2x+3=0\)

\(\Rightarrow S=\varnothing\)

Answer 3

giúp em tl những câu tl trên vs

Answer 4

1: \(\Leftrightarrow\left(3x^2+2x+4-x^2+4\right)\left(3x^2+2x+4+x^2-4\right)=0\)

\(\Leftrightarrow\left(2x^2+2x+8\right)\left(4x^2+2x\right)=0\)

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

=>x=0 hoặc x=-1/2

2: \(\Leftrightarrow\left(2x^2-3x-4-x^2+x\right)\left(2x^2-3x-4+x^2-x\right)=0\)

=>(x^2-2x-4)(3x^2-4x-4)=0

=>\(x\in\left\{1+\sqrt{5};1-\sqrt{5};2;-\dfrac{2}{3}\right\}\)

5: \(\Leftrightarrow4\cdot\left(x+1\right)+x\left(x-2\right)=x^2-2\)

=>4x+8+x^2-2x=x^2-2

=>2x+8=-2

=>2x=-10

=>x=-5(nhận)