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Mới đc câu a ak, thog cảm nha, trih độ mih thấp lắm:
\(\frac{\sqrt{a}}{\sqrt{a}-\sqrt{b}}-\frac{\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\frac{2b}{a-b}\)
=\(\frac{a+\sqrt{ab}-\sqrt{ab}+b}{a-b}-\frac{2b}{a-b}\)
=\(\frac{a+b-2b}{a-b}=\frac{a-b}{a-b}=1\)
a) \(A=\frac{4}{\sqrt{x}+3}+\frac{2x-\sqrt{x}-13}{x-9}-\frac{\sqrt{x}}{\sqrt{x}-3}\)
\(=\frac{4\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{2x-\sqrt{x}-13}{x-9}-\frac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{4\sqrt{x}-12}{x-9}+\frac{2x-\sqrt{x}-13}{x-9}-\frac{x+3\sqrt{x}}{x-9}\)
\(=\frac{4\sqrt{x}-12+2x-\sqrt{x}-13-x-3\sqrt{x}}{x-9}\)
\(=\frac{x-25}{x-9}\)
b) \(P=\frac{A}{B}=\frac{\frac{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}}{\frac{\sqrt{x}+5}{\sqrt{x}-3}}\)
\(=\frac{\sqrt{x}-5}{\sqrt{x}+3}\)
\(\sqrt{P}< \frac{1}{3}\Rightarrow\sqrt{\frac{\sqrt{x}-5}{\sqrt{x}+3}}< \frac{1}{3}\)
\(\Rightarrow\frac{\sqrt{x}-5}{\sqrt{x}+3}< \frac{1}{9}\Leftrightarrow9\sqrt{x}-45< \sqrt{x}+3\)
\(\Leftrightarrow8\sqrt{x}< 48\Leftrightarrow\sqrt{x}< 6\Rightarrow0\le x< 36\)
\(a,\)\(A=\frac{4}{\sqrt{x}+3}+\frac{2x-\sqrt{x}-13}{x-9}=\frac{4\left(\sqrt{x}-3\right)+2x-\sqrt{x}-13}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{4\sqrt{x}-12+2x-\sqrt{x}-13}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)\(=\frac{2x+3\sqrt{x}-1}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(b,P=\frac{A}{B}=\frac{2x+3\sqrt{x}-1}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\frac{\sqrt{x}+5}{\sqrt{x}-3}\)
\(=\frac{2x+3\sqrt{x}-1}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\frac{\left(\sqrt{x}-3\right)}{\sqrt{x}+5}=\frac{2x+3\sqrt{x}-1}{\sqrt{x}+5}\)
Để \(\sqrt{P}< \frac{1}{3}\Rightarrow\frac{2x+3\sqrt{x}-1}{\sqrt{x}+5}< \frac{1}{3}\)
\(\Rightarrow\frac{2x+3\sqrt{x}-1}{\sqrt{x}+5}-\frac{1}{3}< 0\)
\(\Rightarrow\frac{3\left(2x+3\sqrt{x}-1\right)-\sqrt{x}-5}{3\left(\sqrt{x}+5\right)}< 0\)
\(\Rightarrow6x+9\sqrt{x}-3-\sqrt{x}-5< 0\)( do \(3\left(\sqrt{x}+5\right)>0\))
\(\Rightarrow6x-8\sqrt{x}-8< 0\Rightarrow3x-4\sqrt{x}-4< 0\)
\(\Rightarrow3x-6\sqrt{x}+2\sqrt{x}-4< 0\)
\(\Rightarrow3\sqrt{x}\left(\sqrt{x}-2\right)+2\left(\sqrt{x}-2\right)< 0\)
\(\Rightarrow\left(\sqrt{x}-2\right)\left(3\sqrt{x}+2\right)< 0\)
Vì \(3\sqrt{x}+2>0\Rightarrow\sqrt{x}-2< 0\)
\(\Rightarrow\sqrt{x}< 2\Rightarrow x< 4\)
Vậy để \(\sqrt{P}< \frac{1}{3}\)thì \(0\le x< 4\)
a/ \(P=\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}+1}{\sqrt{x}-3}-\frac{3-11\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)-\left(3-11\sqrt{x}\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{2x-6\sqrt{x}+x+4\sqrt{x}+3-3+11\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{3x+9\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{3\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{3\sqrt{x}}{\sqrt{x}-3}\)
b/ \(P< 1\Rightarrow\frac{3\sqrt{x}}{\sqrt{x}-3}< 1\Rightarrow\frac{2\sqrt{x}+3}{\sqrt{x}-3}< 0\)
Xét 2 trường hợp:
- \(\hept{\begin{cases}2\sqrt{x}+3>0\\\sqrt{x}-3< 0\end{cases}\Rightarrow\hept{\begin{cases}2\sqrt{x}>-3\\\sqrt{x}< 3\end{cases}\Rightarrow}\hept{\begin{cases}\sqrt{x}>-\frac{3}{2}\\\sqrt{x}< 3\end{cases}}\Rightarrow-\frac{3}{2}< \sqrt{x}< 3}\)
\(\Rightarrow-\frac{9}{4}< x< 9\)
- \(\hept{\begin{cases}2\sqrt{x}+3< 0\\\sqrt{x}>3\end{cases}\Rightarrow\hept{\begin{cases}\sqrt{x}< -\frac{3}{2}\\\sqrt{x}>3\end{cases}}}\) (vô lí)
Vậy -9/4 < x < 9
\(B=\frac{x-1-4\sqrt{x}+\sqrt{x}+1}{x-1}.\frac{x-1}{x-2\sqrt{x}}\)
\(=\frac{x-3\sqrt{x}}{x-2\sqrt{x}}\)
\(=\frac{\sqrt{x}-3}{\sqrt{x}-2}\)
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 \(B>\sqrt{x}-1\)
a) Ta có: \(\frac{a-b}{\sqrt{a}-\sqrt{b}}-\frac{\sqrt{a^3}-\sqrt{b^3}}{a-b}\)
\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}-\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\sqrt{a}+\sqrt{b}}-\frac{a+\sqrt{ab}+b}{\sqrt{a}+\sqrt{b}}\)
\(=\frac{a+2\sqrt{ab}+b-a-\sqrt{ab}-b}{\sqrt{a}+\sqrt{b}}\)
\(=\frac{\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\)
b)Sửa đề: \(\frac{\left(\sqrt{a}+\sqrt{b}\right)^2-4\sqrt{ab}}{\sqrt{a}-\sqrt{b}}-\frac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}\)
Ta có: \(\frac{\left(\sqrt{a}+\sqrt{b}\right)^2-4\sqrt{ab}}{\sqrt{a}-\sqrt{b}}-\frac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}\)
\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(\sqrt{a}-\sqrt{b}\right)}-\frac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\sqrt{a}-\sqrt{b}-\sqrt{a}-\sqrt{b}\)
\(=-2\sqrt{b}\)
c) Ta có: \(\left(\frac{1}{\sqrt{a}-1}-\frac{1}{\sqrt{a}}\right):\left(\frac{\sqrt{a}+1}{\sqrt{a}-2}-\frac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(=\left(\frac{\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}-\frac{\sqrt{a}-1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}-\frac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\right)\)
\(=\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\frac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\frac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}\)
\(=\frac{\sqrt{a}-2}{3\sqrt{a}}\)
d) Ta có: \(\left(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right)\left(\frac{\sqrt{a}+\sqrt{b}}{a-b}\right)^2\)
\(=\left(\frac{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}{\left(\sqrt{a}+\sqrt{b}\right)}-\sqrt{ab}\right)\left(\frac{\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\right)^2\)
\(=\left(a-\sqrt{ab}+b-\sqrt{ab}\right)\cdot\left(\frac{1}{\sqrt{a}-\sqrt{b}}\right)^2\)
\(=\left(a-2\sqrt{ab}+b\right)\cdot\frac{1}{\left(\sqrt{a}-\sqrt{b}\right)^2}\)
\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(\sqrt{a}-\sqrt{b}\right)^2}=1\)
e) Ta có: \(\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{x+9}{9-x}\right):\left(\frac{3\sqrt{x}+1}{x-3\sqrt{x}}-\frac{1}{\sqrt{x}}\right)\)
\(=\left(\frac{\sqrt{x}\left(3-\sqrt{x}\right)}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}+\frac{x+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}\right):\left(\frac{3\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\frac{\sqrt{x}-3}{\sqrt{x}\left(\sqrt{x}-3\right)}\right)\)
\(=\frac{3\sqrt{x}+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}:\frac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\frac{3\left(\sqrt{x}+3\right)}{-\left(\sqrt{x}-3\right)\cdot\left(\sqrt{x}+3\right)}\cdot\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\left(\sqrt{x}+2\right)}\)
\(=\frac{-3\sqrt{x}}{2\sqrt{x}+4}\)
Với x >= 0 ; \(x\ne\frac{1}{9}\)
\(A=\frac{x+\sqrt{x}}{3\sqrt{x}-1}< 1\Leftrightarrow\frac{x+\sqrt{x}}{3\sqrt{x}-1}-1< 0\)
\(\Leftrightarrow\frac{x+\sqrt{x}-3\sqrt{x}+1}{3\sqrt{x}-1}< 0\Leftrightarrow\frac{x-2\sqrt{x}+1}{3\sqrt{x}-1}< 0\)
\(\Leftrightarrow\frac{\left(\sqrt{x}-1\right)^2}{3\sqrt{x}-1}< 0\Rightarrow3\sqrt{x}-1< 0\Leftrightarrow x< \frac{1}{9}\)
Kết hợp với đk vậy : 0 =< x < 1/9