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Câu 1:
a: \(P=\dfrac{x+\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
b: Để \(2P=2\sqrt{5}+5\) thì \(P=\dfrac{2\sqrt{5}+5}{2}\)
\(\Leftrightarrow\sqrt{x}\left(2\sqrt{5}+5\right)=2\left(\sqrt{x}+1\right)\)
\(\Leftrightarrow\sqrt{x}\left(2\sqrt{5}+3\right)=2\)
hay \(x=\dfrac{4}{29+12\sqrt{5}}=\dfrac{4\left(29-12\sqrt{5}\right)}{121}\)
Bài 2:
a: \(A=\left(5+\sqrt{5}\right)\left(\sqrt{5}-2\right)+\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{4}-\dfrac{3\sqrt{5}\left(3-\sqrt{5}\right)}{4}\)
\(=-5+3\sqrt{5}+\dfrac{5+\sqrt{5}-9\sqrt{5}+15}{4}\)
\(=-5+3\sqrt{5}+5-2\sqrt{5}=\sqrt{5}\)
b: \(B=\left(\dfrac{x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+3\right)}\right):\dfrac{x+3\sqrt{x}-2\left(\sqrt{x}+3\right)+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{x+3\sqrt{x}+6-2\sqrt{x}-6}=1\)
\(a,P=\dfrac{3\left(x+2\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\left(dk:x\ge0,x\ne1\right)\)
\(=\dfrac{3\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\\ =\dfrac{3\sqrt{x}}{\sqrt{x}-1}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\\ =\dfrac{3\sqrt{x}-\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\\ =\dfrac{2\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\\ =\dfrac{2\left(\sqrt{x}-1\right)}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\\ =\dfrac{2\left(\sqrt{x}+2\right)-\left(\sqrt{x}+1\right)}{\sqrt{x}+2}\\ =\dfrac{2\sqrt{x}+4-\sqrt{x}-1}{\sqrt{x}+2}\\ =\dfrac{\sqrt{x}+3}{\sqrt{x}+2}\)
\(b,x=6-2\sqrt{5}=\left(\sqrt{5}-1\right)^2\)
\(\Rightarrow P=\dfrac{\sqrt{\left(\sqrt{5}-1\right)^2}+3}{\sqrt{\left(\sqrt{5}-1\right)^2}+2}=\dfrac{\left|\sqrt{5}-1\right|+3}{\left|\sqrt{5}-1\right|+2}=\dfrac{\sqrt{5}-1+3}{\sqrt{5}-1+2}=\dfrac{\sqrt{5}+2}{\sqrt{5}+1}\)
đk : \(x\ge0,x\ne1\)
\(=>P=\left[\dfrac{2\left(\sqrt{x}+2\right)-5}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right]:\left[\dfrac{x+\sqrt{x}-2+3-x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right]\)
\(P=\left[\dfrac{2\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right].\left[\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{\sqrt{x}+1}\right]\)
\(P=\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}\)
b,\(x=6-2\sqrt{5}=\left(\sqrt{5}-1\right)^2\) thay vào P
\(=>P=\dfrac{2\sqrt{\left(\sqrt{5}-1\right)^2}-1}{\sqrt{\left(\sqrt{5}-1\right)^2}+1}=\dfrac{2\sqrt{5}-3}{\sqrt{5}}\)
c,\(=>\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}=\dfrac{1}{\sqrt{x}}=>2x-\sqrt{x}=\sqrt{x}+1\)
\(=>2x-2\sqrt{x}-1=0< =>2\left(x-\sqrt{x}-\dfrac{1}{2}\right)=0\)
\(=>x-\sqrt{x}-\dfrac{1}{2}=>\Delta=1-4\left(-\dfrac{1}{2}\right)=3>0=>\left[{}\begin{matrix}x1=\dfrac{1+\sqrt{3}}{2}\\x2=\dfrac{1-\sqrt{3}}{2}\end{matrix}\right.\)
đối chiếu đk loại x2 còn x1 thỏa
\(A=2\left|2-\sqrt{5}\right|-\dfrac{8\left(3+\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\)
\(=2\left(\sqrt{5}-2\right)-\dfrac{8\left(3+\sqrt{5}\right)}{4}=2\sqrt{5}-4-2\left(3+\sqrt{5}\right)\)
\(=2\sqrt{5}-4-6-2\sqrt{5}=-10\)
\(B=\left(\dfrac{2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\left(\dfrac{\sqrt{x}-2+2}{\sqrt{x}-2}\right)\)
\(=\left(\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}\right)\)
\(=\dfrac{1}{\sqrt{x}-2}.\dfrac{\sqrt{x}-2}{\sqrt{x}}=\dfrac{1}{\sqrt{x}}\)
Lời giải:
ĐK: $x\geq 0; x\neq 4; x\neq 9$
\(P=\frac{1}{\sqrt{x}+1}:\left[\frac{(\sqrt{x}+3)(\sqrt{x}-3)}{(\sqrt{x}-2)(\sqrt{x}-3)}-\frac{(\sqrt{x}+2)(\sqrt{x}-2)}{(\sqrt{x}-3)(\sqrt{x}-2)}+\frac{\sqrt{x}+2}{(\sqrt{x}-2)(\sqrt{x}-3)}\right]\)
\(=\frac{1}{\sqrt{x}+1}:\frac{x-9-(x-4)+\sqrt{x}+2}{(\sqrt{x}-2)(\sqrt{x}-3)}=\frac{1}{\sqrt{x}+1}:\frac{\sqrt{x}-3}{(\sqrt{x}-2)(\sqrt{x}-3)}=\frac{\sqrt{x}-2}{\sqrt{x}+1}\)
Để $P>0\Leftrightarrow \frac{\sqrt{x}-2}{\sqrt{x}+1}>0$
$\Leftrightarrow \sqrt{x}-2>0$ (do $\sqrt{x}+1>0$)
$\Leftrightarrow x>4$
Kết hợp với ĐKXĐ suy ra $x>4; x\neq 9$
a, \(P=\left(1-\dfrac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{\sqrt{x}+2}{3-\sqrt{x}}+\dfrac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right)\)
\(P=\left(\dfrac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}+1}\right):\left(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-3}+\dfrac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right)\)
\(P=\dfrac{1}{\sqrt{x}+1}:\left[\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\right]\)
\(P=\dfrac{1}{\sqrt{x}+1}:\left[\dfrac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right]\)
\(P=\dfrac{1}{\sqrt{x}+1}:\dfrac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(P=\dfrac{1}{\sqrt{x}+1}.\sqrt{x}-2=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\)
1. ĐKXĐ: $x>0; x\neq 9$
\(A=\frac{\sqrt{x}+3+\sqrt{x}-3}{(\sqrt{x}-3)(\sqrt{x}+3)}.\frac{\sqrt{x}-3}{\sqrt{x}}=\frac{2\sqrt{x}}{(\sqrt{x}-3)(\sqrt{x}+3)}.\frac{\sqrt{x}-3}{\sqrt{x}}=\frac{2}{\sqrt{x}+3}\)
2. ĐKXĐ: $x\geq 0; x\neq 4$
\(B=\left[\frac{\sqrt{x}(\sqrt{x}+2)+\sqrt{x}-2}{(\sqrt{x}-2)(\sqrt{x}+2)}+\frac{6-7\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}\right](\sqrt{x}+2)\)
\(=\frac{x+3\sqrt{x}-2+6-7\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}.(\sqrt{x}+2)=\frac{x-4\sqrt{x}+4}{\sqrt{x}-2}=\frac{(\sqrt{x}-2)^2}{\sqrt{x}-2}=\sqrt{x}-2\)
b: Ta có: \(B=\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right)\cdot\left(\dfrac{x\sqrt{x}-1}{\sqrt{x}-1}+\dfrac{x+\sqrt{x}}{\sqrt{x}+1}\right)\)
\(=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}\cdot\left(x+\sqrt{x}+1+\sqrt{x}\right)\)
\(=\dfrac{x+\sqrt{x}-2-x+\sqrt{x}+2}{\sqrt{x}-1}\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}-1}\)
Câu 1:
a: \(P=\dfrac{x+\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
b: Để \(2P=2\sqrt{5}+5\) thì \(P=\dfrac{2\sqrt{5}+5}{2}\)
\(\Leftrightarrow\sqrt{x}\left(2\sqrt{5}+5\right)=2\left(\sqrt{x}+1\right)\)
\(\Leftrightarrow\sqrt{x}\left(2\sqrt{5}+3\right)=2\)
hay \(x=\dfrac{4}{29+12\sqrt{5}}=\dfrac{4\left(29-12\sqrt{5}\right)}{121}\)