Question

  làm bài 2 và 3 ạ, em cảm ơn ạa

Answer 1

Bài 22:

\(a,P=\dfrac{2}{\sqrt{x}-2}:\left(\dfrac{\sqrt{x}}{x-4}+\dfrac{1}{\sqrt{x}-2}\right)\\ =\dfrac{2}{\sqrt{x}-2}:\left[\dfrac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right]\\ =\dfrac{2}{\sqrt{x}-2}:\dfrac{2\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\\ =\dfrac{2}{\sqrt{x}-2}\cdot\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{2\left(\sqrt{x}+1\right)}\\ =\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\) 

b) Ta có: 

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

Để P nguyên thì 1 chia hết cho \(\sqrt{x}+1\) 

Mà: \(\sqrt{x}+1\ge1\)

\(=>\sqrt{x}+1=1\\ =>\sqrt{x}=0\\ =>x=0\left(tm\right)\)

Answer 2

Bài 23:

a:

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

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

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

\(=x-\sqrt{x}+1\)

b: Để P=7 thì \(x-\sqrt{x}+1=7\)

=>\(x-\sqrt{x}-6=0\)

=>\(\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)=0\)

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

=>x=9(nhận)

Bài 21:

a: \(A=\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{2+5\sqrt{x}}{4-x}\)

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

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

\(=\dfrac{3x-6\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{3\sqrt{x}}{\sqrt{x}+2}\)

b: Để A=2 thì \(3\sqrt{x}=2\left(\sqrt{x}+2\right)\)

=>\(3\sqrt{x}=2\sqrt{x}+4\)

=>\(\sqrt{x}=4\)

=>x=16(nhận)