Question

A=\(\left(\dfrac{\sqrt{x}-1}{x-4}-\dfrac{\sqrt{x}+1}{x-4\sqrt{x}+4}\right).\dfrac{x\sqrt{x}-2x-4\sqrt{x}+8}{6\sqrt{x}-18}\)

1,\(\dfrac{x+4}{x^2-9}-\dfrac{2}{x+3}=\dfrac{4x}{3x-x^2}\)

2,x(x+1)(x² +x+1)=42

3, Trong mặt phẳng tọa độ Oxy, cho parabol(P) y=-x² và đường thẳng(d)y=mx-m-2. CMR khi m thay đổi, (d) cắt (P) tại 2 điểm có hoành độ phân biệt.

Answer 1

Bài 1 :

\(\dfrac{x+4}{x^2-9}-\dfrac{2}{x+3}=\dfrac{4x}{3x-x^2}\) ( ĐK : \(\left\{{}\begin{matrix}x\ne0\\x\ne-3\\x\ne3\end{matrix}\right.\) )

\(\Leftrightarrow\dfrac{x\left(x+4\right)}{x\left(x-3\right)\left(x+3\right)}-\dfrac{2x\left(x-3\right)}{x\left(x-3\right)\left(x+3\right)}=\dfrac{-4x\left(x+3\right)}{x\left(x-3\right)\left(x+3\right)}\)

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

\(\Leftrightarrow x^2+4x-2x^2+6x+4x^2+12x=0\)

\(\Leftrightarrow3x^2+22x=0\)

\(\Leftrightarrow x\left(3x+22\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\3x+22=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(L\right)\\x=-\dfrac{22}{3}\left(N\right)\end{matrix}\right.\)

Vậy \(x=-\dfrac{22}{3}\)

Answer 2

Bài 2 : \(x\left(x+1\right)\left(x^2+x+1\right)=42\)

\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x+1\right)=42\)

Đặt \(x^2+x=t\) . Phương trình trở thành :

\(t\left(t+1\right)=42\)

\(\Leftrightarrow t^2+t-42=0\)

\(\Leftrightarrow\left(t-6\right)\left(t+7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t-6=0\\t+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}t=6\\t=-7\end{matrix}\right.\)

Với \(t=6\)

\(\Leftrightarrow x^2+x=6\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)

Với \(t=-7\)

\(\Leftrightarrow x^2+x=-7\)

\(\Leftrightarrow x^2+x+7=0\)

---> Phương trình vô nghiệm !

Vậy \(x=-3;x=2\)