Question

Giúp mình bài 7 câu 2 với ạ.

  

Answer 1

Bài 7:

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

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

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

\(=\dfrac{x-4\sqrt{x}+3-2x+5\sqrt{x}-2+x-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)

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

=>P-1<0

=>\(\dfrac{1}{\sqrt{x}-2}-1< 0\)

=>\(\dfrac{1-\sqrt{x}+2}{\sqrt{x}-2}< 0\)

=>\(\dfrac{3-\sqrt{x}}{\sqrt{x}-2}< 0\)

=>\(\dfrac{\sqrt{x}-3}{\sqrt{x}-2}>0\)

TH1: \(\left\{{}\begin{matrix}\sqrt{x}-3>0\\\sqrt{x}-2>0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\sqrt{x}>3\\\sqrt{x}>2\end{matrix}\right.\)

=>\(\sqrt{x}>3\)

=>x>9

TH2: \(\left\{{}\begin{matrix}\sqrt{x}-3< 0\\\sqrt{x}-2< 0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\sqrt{x}< 3\\\sqrt{x}< 2\end{matrix}\right.\)

=>\(\sqrt{x}< 2\)

=>0<=x<4

kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}0< =x< 4\\x\ne1\end{matrix}\right.\)