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a: \(A=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\sqrt{a}-\sqrt{b}-\sqrt{a}-\sqrt{b}=-2\sqrt{b}\)
b: \(B=\dfrac{2\sqrt{x}-x-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{x-1}\)
\(=\dfrac{-2x+\sqrt{x}-1}{\sqrt{x}-1}\cdot\dfrac{1}{x-1}\)
c: \(C=\dfrac{x-9-x+3\sqrt{x}}{x-9}:\left(\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}+3}+\dfrac{x-9}{x+\sqrt{x}-6}\right)\)
\(=\dfrac{3\left(\sqrt{x}-3\right)}{x-9}:\dfrac{9-x+x-4\sqrt{x}+4+x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{x-4\sqrt{x}+4}\)
\(=\dfrac{3}{\sqrt{x}-2}\)
\(a)\)\(R=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3\left(\sqrt{x}+3\right)}{x-9}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(R=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right):\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(R=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3}{\sqrt{x-3}}\right):\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(R=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}-3}{\sqrt{x}-3}\right):\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(R=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+1\right):\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(R=\frac{3\sqrt{x}+3}{\sqrt{x}+3}.\frac{\sqrt{x}-3}{\sqrt{x+1}}\)
\(R=\frac{3\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}\)
\(R=\frac{3\left(\sqrt{x}-3\right)}{\sqrt{x}+3}\)
\(b)\) Ta có : \(R< -1\)
\(\Leftrightarrow\)\(\frac{3\left(\sqrt{x}-3\right)}{\sqrt{x}+3}< -1\)
\(\Leftrightarrow\)\(\frac{\sqrt{x}-3}{\sqrt{x}+3}< \frac{-1}{3}\)
\(\Leftrightarrow\)\(3\sqrt{x}-9< -\sqrt{x}-3\)
\(\Leftrightarrow\)\(4\sqrt{x}< 6\)
\(\Leftrightarrow\)\(\sqrt{x}< \frac{3}{2}\)
\(\Leftrightarrow\)\(x< \frac{9}{4}\)
Chúc bạn học tốt ~
a) ĐKXĐ: \(x\ge0;x\ne9\)
mk chỉnh lại đề bài nhé, chắc có lẽ bn ghi nhầm:
\(P=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3x+3}{x-9}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(=\left(\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-\frac{\sqrt{x}-3}{\sqrt{x}-3}\right)\)
\(=\frac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\frac{2\sqrt{x}-\sqrt{x}+1}{\sqrt{x}-3}\)
\(=\frac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\frac{-3}{\sqrt{x}+3}\)
P/s : sửa đề
ĐKXĐ : \(\hept{\begin{cases}x\ge0\\x\ne9\end{cases}}\)
a) \(P=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3x+3}{x-9}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(P=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\frac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)
\(P=\frac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{x-9}.\frac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(P=\frac{-3\sqrt{x}-3x}{x-9}.\frac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(P=\frac{-3\sqrt{x}\left(1+\sqrt{x}\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}\)
\(P=\frac{-3\sqrt{x}}{\sqrt{x}+3}\)
b) \(P< -\frac{1}{2}\)
\(\Leftrightarrow\frac{-3\sqrt{x}}{\sqrt{x}+3}+\frac{1}{2}< 0\)
\(\Leftrightarrow\frac{-6\sqrt{x}+\sqrt{x}+3}{2\left(\sqrt{x}+3\right)}< 0\)
\(\Leftrightarrow\frac{-5\sqrt{x}+3}{2\left(\sqrt{x}+3\right)}< 0\)
Mà \(2\left(\sqrt{x}+3\right)>0\)
\(\Rightarrow-5\sqrt{x}+3< 0\)
\(\Leftrightarrow-5\sqrt{x}< -3\)
\(\Leftrightarrow\sqrt{x}>\frac{3}{5}\)
\(\Leftrightarrow x>\frac{9}{25}\)
Vấy .................
c) \(P.\left(\sqrt{x}+3\right)+2\sqrt{x}-2+x=2\)
\(\Leftrightarrow\frac{-3\sqrt{x}}{\sqrt{x}+3}\left(\sqrt{x}+3\right)+2\sqrt{x}-2+x=2\)
\(\Leftrightarrow-3\sqrt{x}+2\sqrt{x}-2-2+x=0\)
\(\Leftrightarrow-\sqrt{x}-4+x=0\)
\(\Leftrightarrow-\sqrt{x}\left(1-\sqrt{x}\right)=4\)
Còn lại lập bảng tự tìm giá trị của x là ra .( Chú ý : đối chiếu ĐKXĐ )
d)
\(P.\left(\sqrt{x}+3\right)+x\left(\sqrt{x}-m\right)=x-\sqrt{x}\left(3+m\right)\)
\(\Leftrightarrow\frac{-3\sqrt{x}}{\sqrt{x}+3}\left(\sqrt{x}+3\right)+x\sqrt{x}-xm=x-3\sqrt{x}-m\sqrt{x}\)
\(\Leftrightarrow-3\sqrt{x}+x\sqrt{x}-xm-x+3\sqrt{x}+m\sqrt{x}=0\)
\(\Leftrightarrow\sqrt{x}\left(x+m\right)-x\left(m+1\right)=0\)
\(\Leftrightarrow\sqrt{x}\left[x+m-m\sqrt{x}-\sqrt{x}\right]=0\)
\(\Leftrightarrow\sqrt{x}\left[m\left(1-\sqrt{x}\right)-\sqrt{x}\left(1-\sqrt{x}\right)\right]=0\)
\(\Leftrightarrow\sqrt{x}=0;m-\sqrt{x}=0;1-\sqrt{x}=0\)
+) \(\sqrt{x}=0\Leftrightarrow x=0\left(TM\right)\)
+) \(1-\sqrt{x}=0\)
\(\Leftrightarrow x=1\left(TM\right)\)
+) \(m-\sqrt{x}=0\)
\(\Leftrightarrow\orbr{\begin{cases}m-\sqrt{0}=0\\m-\sqrt{1}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}m=0\\m=1\end{cases}}}\)
Vậy ..................
a) \(ĐKXĐ:\hept{\begin{cases}x>0\\x\ne9\end{cases}}\)
\(C=\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{x+9}{9-x}\right):\left(\frac{3\sqrt{x}+1}{x-3\sqrt{x}}-\frac{1}{\sqrt{x}}\right)\)
\(\Leftrightarrow C=\frac{\sqrt{x}\left(3-\sqrt{x}\right)+x+9}{9-x}:\frac{3\sqrt{x}+1-\sqrt{x}+3}{x-3\sqrt{x}}\)
\(\Leftrightarrow C=\frac{3\sqrt{x}+9}{9-x}:\frac{2\sqrt{x}+4}{x-3\sqrt{x}}\)
\(\Leftrightarrow C=\frac{3}{3-\sqrt{x}}\cdot\frac{x-3\sqrt{x}}{2\sqrt{x}+4}\)
\(\Leftrightarrow C=\frac{-3}{2\sqrt{x}+4}\)
b) Để \(-\frac{3}{2\sqrt{x}+4}< -1\)
\(\Leftrightarrow\frac{1+2\sqrt{x}}{2\sqrt{x}+4}< 0\)
Vì \(\hept{\begin{cases}1+2\sqrt{x}>0\\2\sqrt{x}+4>0\end{cases}\Leftrightarrow C>0}\)
Vậy để C <-1 <=> \(x\in\varnothing\)
c) \(A=\frac{1}{\sqrt{3}-\sqrt{2}}=\sqrt{3}+\sqrt{2}\)
\(\Leftrightarrow A^2=3+2+2\sqrt{5}=5+2\sqrt{5}\)
\(B=\sqrt{5}+1\)
\(\Leftrightarrow B^2=5+1+2\sqrt{5}=6+2\sqrt{5}\)
Vì \(5+2\sqrt{5}< 6+2\sqrt{5}\)
\(\Leftrightarrow A^2< B^2\)
\(\Leftrightarrow A< B\)
Vậy \(\frac{1}{\sqrt{3}-\sqrt{2}}< \sqrt{5}+1\)
câu 2
\(...=\sqrt{\left(2-\sqrt{5}\right)^2}-\sqrt{\left(2+\sqrt{5}\right)^2}=\left|2-\sqrt{5}\right|-\left|2+\sqrt{5}\right|=-4\)
câu 1
\(P=\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{x+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\right):\left(\frac{3\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\frac{1}{\sqrt{x}}\right)\)
\(=\left(\frac{\sqrt{x}\left(3-\sqrt{x}\right)+x+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}\right):\left(\frac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\right)\)
\(=\frac{3\sqrt{x}+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}:\frac{2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\frac{3}{\left(3-\sqrt{x}\right)}.\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\sqrt{x}+4}=\frac{-3\sqrt{x}}{2\sqrt{x}+4}\)
\(P< -1\Leftrightarrow\frac{-3\sqrt{x}}{2\sqrt{x}+4}+1< 0\Leftrightarrow-\sqrt{x}+4< 0\Leftrightarrow\sqrt{x}>4\Leftrightarrow x>16\)
\(ĐKXĐ:\)
\(\hept{\begin{cases}x-9\ne0\\\sqrt{x}-2\ne0\\\sqrt{x}+3\ne0;x\ge0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x\ne9\\x\ne4\\x\ge0\end{cases}}\)
Vậy...................................................
\(A=\left(\frac{x-3\sqrt{x}}{x-9}-1\right):\left(\frac{9-x}{x+\sqrt{x}-6}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}+3}\right)\)
\(=\left(\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-1\right):\left(\frac{9-x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}+3}\right)\)
\(=\frac{\sqrt{x}-\sqrt{x}-3}{\left(\sqrt{x}+3\right)}:\left(\frac{9-x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}+\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\right)\)
\(=\frac{-3}{\sqrt{x}+3}:\left(\frac{9-x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}+\frac{x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}-\frac{x-4}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\right)\)
\(=\frac{-3}{\sqrt{x}+3}:\frac{9-x+x-9-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{-3}{\sqrt{x}+3}:\frac{-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{-3}{\sqrt{x}+3}.\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{4-x}\)
\(=\frac{3\left(2-\sqrt{x}\right)}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\)
\(=\frac{3}{\left(2+\sqrt{x}\right)}\)
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