Bài 21 :
a) \(A=\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{2+5\sqrt{x}}{4-x}\left(x\ge0;x\ne4\right)\)
\(A=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)+2\sqrt{x}\left(\sqrt{x}-2\right)-2-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(A=\dfrac{x+3\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(A=\dfrac{3x-6\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{3\sqrt{x}}{\sqrt{x}+2}\)
b) \(A=2\Leftrightarrow\dfrac{3\sqrt{x}}{\sqrt{x}+2}=2\)
\(\Leftrightarrow3\sqrt{x}=2\sqrt{x}+4\)
\(\Leftrightarrow\sqrt{x}=4\)
\(\Leftrightarrow x=16\)
Bài 23:
a: \(P=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\dfrac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\)
\(=\sqrt{x}\left(\sqrt{x}-1\right)-2\sqrt{x}-1+2\left(\sqrt{x}+1\right)\)
\(=x-\sqrt{x}-2\sqrt{x}-1+2\sqrt{x}+2=x-\sqrt{x}+1\)
b: P=7
=>P-7=0
=>\(x-\sqrt{x}-6=0\)
=>\(\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)=0\)
mà \(\sqrt{x}+2>=2>0\forall x\) thỏa mãn ĐKXĐ
nên \(\sqrt{x}-3=0\)
=>x=9(nhận)
Bài 24:
a: Thay x=9 vào A, ta được:
\(A=\dfrac{\sqrt{9}}{\sqrt{9}+1}=\dfrac{3}{3+1}=\dfrac{3}{4}\)
b: A+B
\(=\dfrac{\sqrt{x}}{\sqrt{x}+1}+\dfrac{\sqrt{x}}{\sqrt{x}+1}+\dfrac{\sqrt{x}+2}{\sqrt{x}}-\dfrac{2}{x+\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}+1}+\dfrac{\sqrt{x}+2}{\sqrt{x}}-\dfrac{2}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{2x+\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)-2}{\sqrt{x}\left(\sqrt{x}+1\right)}=\dfrac{2x+x+3\sqrt{x}+2-2}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{3x+3\sqrt{x}}{x+\sqrt{x}}=3\)
