Question

câu 6

Answer 1

Câu 5: 

a: Để (d)//(d') thì \(\left\{{}\begin{matrix}m^2-2=-1\\m+2< >3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m^2=1\\m< >1\end{matrix}\right.\Leftrightarrow m=-1\)

b: \(P=\left(\dfrac{x-\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}}{\sqrt{x}-2}\right)\cdot\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\)

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

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

Answer 2

Câu 5: 

a: Để (d)//(d') thì {m2−2=−1m+2<>3⇔{m2=1m<>1⇔m=−1{m2−2=−1m+2<>3⇔{m2=1m<>1⇔m=−1

b: =x−√x+2−x−√x(√x−2)(√x+1)⋅√x−2√x−1=x−x+2−x−x(x−2)(x+1)⋅x−2x−1