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: \(\sqrt{x}+1\ge1\)
\(=>\sqrt{x}+1=1\\ =>\sqrt{x}=0\\ =>x=0\left(tm\right)\)
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)
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