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\(P=B:A\)
\(P=\dfrac{\sqrt{x}-2}{\sqrt{x}-1}:\dfrac{\sqrt{x}+3}{\sqrt{x}-1}=\dfrac{\sqrt{x}-2}{\sqrt{x}-1}.\dfrac{\sqrt{x}-1}{\sqrt{x}+3}=\dfrac{\sqrt{x}-2}{\sqrt{x}+3}\)
\(P=\dfrac{1}{3}\Leftrightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}+3}=\dfrac{1}{3}\Leftrightarrow3\sqrt{x}-6=\sqrt{x}+3\)
\(\Leftrightarrow2\sqrt{x}=9\Leftrightarrow\sqrt{x}=4,5\Leftrightarrow x=\dfrac{81}{4}\)
b. \(P=\dfrac{\sqrt{x}-2}{\sqrt{x}+3}=\dfrac{\sqrt{x}+3-5}{\sqrt{x}+3}=1-\dfrac{5}{\sqrt{x}+3}\)
Ta có: \(-\dfrac{5}{\sqrt{x}+3}\ge-\dfrac{5}{\sqrt{0}+3}=-\dfrac{5}{3}\)
\(\Rightarrow1-\dfrac{5}{\sqrt{x}+3}\ge1-\dfrac{5}{3}=-\dfrac{2}{3}\)
Suy ra: \(P\ge-\dfrac{2}{3}\) khi \(x=0\)
\(A=\left(\sqrt{x}-\frac{x+2}{\sqrt{x}+1}\right):\left(\frac{\sqrt{x}}{\sqrt{x}+1}-\frac{\sqrt{x}-4}{1-x}\right)\) \(ĐKXĐ:x\ge0;x\ne1;x\ne4\)
\(A=\left[\frac{\sqrt{x}\left(\sqrt{x}+1\right)-x-2}{\sqrt{x}+1}\right]:\left[\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\frac{\sqrt{x}-4}{x-1}\right]\)
\(A=\frac{x+\sqrt{x}-x-2}{\sqrt{x}+1}:\left[\frac{x-\sqrt{x}+\sqrt{x}-4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right]\)
\(A=\frac{\sqrt{x}-2}{\sqrt{x}+1}:\frac{x-4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(A=\frac{\sqrt{x}-2}{\sqrt{x}+1}.\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(A=\frac{\sqrt{x}-1}{\sqrt{x}+2}\)
vậy \(A=\frac{\sqrt{x}-1}{\sqrt{x}+2}\)
b)theo bài ra: \(A=\frac{1}{\sqrt{x}}\)
\(\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}+2}=\frac{1}{\sqrt{x}}\)
\(\Leftrightarrow\left(\sqrt{x}-1\right).\sqrt{x}=\sqrt{x}+2\)
\(\Leftrightarrow x-\sqrt{x}-\sqrt{x}-2=0\)
\(\Leftrightarrow x-2\sqrt{x}-2=0\)
\(\Leftrightarrow x-2\sqrt{x}+1-3=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)^2-\left(\sqrt{3}\right)^2=0\)
\(\Leftrightarrow\left(\sqrt{x}-1-\sqrt{3}\right)\left(\sqrt{x}-1+\sqrt{3}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}-1-\sqrt{3}=0\\\sqrt{x}-1+\sqrt{3}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}\sqrt{x}=\sqrt{3}+1\\\sqrt{x}=1-\sqrt{3}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\left(\sqrt{3}+1\right)^2\\x=\left(1-\sqrt{3}\right)^2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=3+2\sqrt{3}+1\\x=3-2\sqrt{3}+1\end{cases}}\)
vậy......
\( a)A = \dfrac{{a - \sqrt a - 6}}{{4 - a}} - \dfrac{1}{{\sqrt a - 2}}\\ A = \dfrac{{a + 2\sqrt a - 3\sqrt a - 6}}{{\left( {2 - \sqrt a } \right)\left( {2 + \sqrt a } \right)}} - \dfrac{1}{{\sqrt a - 2}}\\ A = \dfrac{{\left( {\sqrt a + 2} \right)\left( {\sqrt a - 3} \right)}}{{\left( {2 - \sqrt a } \right)\left( {2 + \sqrt a } \right)}} - \dfrac{1}{{\sqrt a - 2}}\\ A = - \dfrac{{\sqrt a - 3}}{{\sqrt a - 2}} - \dfrac{1}{{\sqrt a - 2}}\\ A = - \dfrac{{\sqrt a - 2}}{{\sqrt a - 2}} = - 1 \)
\( b)B = \dfrac{1}{{\sqrt x - 1}} + \dfrac{1}{{\sqrt x + 1}} - \dfrac{2}{{x - 1}}\\ B = \dfrac{1}{{\sqrt x - 1}} + \dfrac{1}{{\sqrt x + 1}} - \dfrac{2}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\ B = \dfrac{{\sqrt x + 1 + \sqrt x - 1 - 2}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\ B = \dfrac{{2\sqrt x - 2}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\ B = \dfrac{{2\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}} = \dfrac{2}{{\sqrt x + 1}} \)
Mới đc câu a ak, thog cảm nha, trih độ mih thấp lắm:
\(\frac{\sqrt{a}}{\sqrt{a}-\sqrt{b}}-\frac{\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\frac{2b}{a-b}\)
=\(\frac{a+\sqrt{ab}-\sqrt{ab}+b}{a-b}-\frac{2b}{a-b}\)
=\(\frac{a+b-2b}{a-b}=\frac{a-b}{a-b}=1\)
\(B=\frac{x-1-4\sqrt{x}+\sqrt{x}+1}{x-1}.\frac{x-1}{x-2\sqrt{x}}\)
\(=\frac{x-3\sqrt{x}}{x-2\sqrt{x}}\)
\(=\frac{\sqrt{x}-3}{\sqrt{x}-2}\)
a.Ta co:
\(\frac{\sqrt{x}-3}{\sqrt{x}-2}< 1\left(x\ge0,x\ne4\right)\)
\(\Leftrightarrow\sqrt{x}-3< \sqrt{x}-2\)
\(\Leftrightarrow3>2\)
Vay \(B< 1\left(\forall x\ge0,x\ne4\right)\)
Lát mình giải 2 câu kia,di ăn com cái
b.Ta co:
\(\frac{\sqrt{x}-3}{\sqrt{x}-2}< \frac{3}{2}\)
\(\Leftrightarrow2\sqrt{x}-6< 3\sqrt{x}-6\)
\(\Leftrightarrow x>0\)
Vay \(B< \frac{3}{2}\left(\forall x>0,x\ne4\right)\)
c.Ta co:
\(\frac{\sqrt{x}-3}{\sqrt{x}-2}>\sqrt{x}-1\)
\(\Leftrightarrow\sqrt{x}-3>x-3\sqrt{x}+2\)
\(\Leftrightarrow x-4\sqrt{x}+5< 0\)
\(\Leftrightarrow\left(\sqrt{x}-2\right)^2+1< 0\) (vo ly)
Vay khong co gia tri nao cua x thoa man \(B>\sqrt{x}-1\)
\(P=\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{2\sqrt{x}-1}{\sqrt{x}-1}+\frac{x-2}{x-3\sqrt{x}+2}\)
ĐK : \(\hept{\begin{cases}x\ge0\\x\ne1\\x\ne4\end{cases}}\)
\(=\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{2\sqrt{x}-1}{\sqrt{x}-1}+\frac{x-2}{x-\sqrt{x}-2\sqrt{x}+2}\)
\(=\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{2\sqrt{x}-1}{\sqrt{x}-1}+\frac{x-2}{\sqrt{x}\left(\sqrt{x}-1\right)-2\left(\sqrt{x}-1\right)}\)
\(=\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{2\sqrt{x}-1}{\sqrt{x}-1}+\frac{x-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}-\frac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}+\frac{x-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{x-4\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}-\frac{2x-5\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}+\frac{x-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{x-4\sqrt{x}+3-2x+5\sqrt{x}-2+x-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}=\frac{1}{\sqrt{x}-2}\)
b) Để P < 1
=> \(\frac{1}{\sqrt{x}-2}< 1\)
<=> \(\frac{1}{\sqrt{x}-2}-1< 0\)
<=> \(\frac{1}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}-2}< 0\)
<=> \(\frac{1-\sqrt{x}+2}{\sqrt{x}-2}< 0\)
<=> \(\frac{3-\sqrt{x}}{\sqrt{x}-2}< 0\)
Xét hai trường hợp :
1. \(\hept{\begin{cases}3-\sqrt{x}>0\\\sqrt{x}-2< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}-\sqrt{x}>-3\\\sqrt{x}< 2\end{cases}}\Leftrightarrow\hept{\begin{cases}\sqrt{x}< 3\\\sqrt{x}< 2\end{cases}}\Leftrightarrow\hept{\begin{cases}x< 9\\x< 4\end{cases}}\Leftrightarrow x< 4\)
2. \(\hept{\begin{cases}3-\sqrt{x}< 0\\\sqrt{x}-2>0\end{cases}}\Leftrightarrow\hept{\begin{cases}-\sqrt{x}< -3\\\sqrt{x}>2\end{cases}}\Leftrightarrow\hept{\begin{cases}\sqrt{x}>3\\\sqrt{x}>2\end{cases}}\Leftrightarrow\hept{\begin{cases}x>9\\x>4\end{cases}}\Leftrightarrow x>9\)
Kết hợp với ĐK => Với \(\orbr{\begin{cases}x\in\left\{0;2;3\right\}\\x>9\end{cases}}\)thì thỏa mãn đề bài
\(ĐK:\)\(x\ge0;x\ne1;x\ne4\)
\(P=B:A=\frac{\sqrt{x}-2}{\sqrt{x}-1}:\frac{\sqrt{x}+3}{\sqrt{x}-1}\)
\(=\frac{\sqrt{x}-2}{\sqrt{x}+3}\)
\(P=\frac{1}{3}\)\(\Rightarrow\)\(\frac{\sqrt{x}-2}{\sqrt{x}+3}=\frac{1}{3}\)
\(\Rightarrow\)\(3\left(\sqrt{x}-2\right)=\sqrt{x}+3\)
\(\Leftrightarrow\)\(2\sqrt{x}-9=0\)
\(\Leftrightarrow\)\(2\sqrt{x}=9\)
\(\Leftrightarrow\)\(\sqrt{x}=\frac{9}{2}\)
\(\Leftrightarrow\)\(x=\frac{81}{4}\)