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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. ĐKXĐ \(x\ge0\)và \(x\ne9\)
Ta có \(K=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3\sqrt{x}-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\frac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)
\(=\frac{2x-6\sqrt{x}+x+3\sqrt{x}-3\sqrt{x}-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(=\frac{3x-6\sqrt{x}-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}+1}=\frac{3\left(x-2\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\frac{3\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}+1}=\frac{3\left(\sqrt{x}-3\right)}{\sqrt{x}+3}\)
b. Để \(K< -1\Rightarrow\frac{3\sqrt{x}-9+\sqrt{x}+3}{\sqrt{x}+3}< 0\Rightarrow\frac{4\sqrt{x}-6}{\sqrt{x}+3}< 0\Rightarrow4\sqrt{x}-6< 0\)vì \(\sqrt{x}+3\ge3\)
\(\Rightarrow0\le x< \frac{9}{4}\left(tm\right)\)
Vậy với \(0\le x< \frac{9}{4}\)thì K<-1
c. \(K=\frac{3\sqrt{x}-9}{\sqrt{x}+3}=3+\frac{-18}{\sqrt{x}+3}\)
Ta có \(\sqrt{x}+3\ge3\Rightarrow\frac{1}{\sqrt{x}+3}\le\frac{1}{3}\Rightarrow-\frac{18}{\sqrt{x}+3}\ge-6\Rightarrow3+\frac{-18}{\sqrt{x}+3}\ge-3\)
\(\Rightarrow K\ge-3\)
Vậy \(MinK=-3\Leftrightarrow\sqrt{x}=0\Leftrightarrow x=0\)
C = \(\left(1-\frac{x-3\sqrt{x}}{x-9}\right):\)\(\left(\frac{-\sqrt{x}+3}{\sqrt{x}-2}+\frac{\sqrt{x}-2}{\sqrt{x}+3}-\frac{9-x}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\right)\)( \(x\ge0\) , \(x\ne9;4\))
= \(\frac{x-9-x+3\sqrt{x}}{x-9}\): \(\frac{9-x+\left(\sqrt{x}-2\right)^2-9+x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
= \(\frac{3\sqrt{x}-9}{x-9}\): \(\frac{\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
= \(\frac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)\(:\frac{\sqrt{x}-2}{\sqrt{x}+3}\)
= \(\frac{3}{\sqrt{x}+3}.\frac{\sqrt{x}+3}{\sqrt{x}-2}\)
= \(\frac{3}{\sqrt{x}-2}\)
#mã mã#
a/
\(=\left(\frac{1}{\sqrt{x}+3}+\frac{3}{\sqrt{x}\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right):\left(\frac{\sqrt{x}}{\sqrt{x}+3}-\frac{3\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+3\right)}\right)\)
\(=\left(\frac{x-3\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right):\left(\frac{\sqrt{x}-3}{\sqrt{x}+3}\right)\)
\(=\left(\frac{x-3\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right).\frac{\sqrt{x}+3}{\sqrt{x}-3}\)
\(=\frac{x-3\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)^2}\)
\(=\frac{x-3\sqrt{x}+3}{x\sqrt{x}-6\text{x}+9\sqrt{x}}\)
\(=\frac{x-3\sqrt{x}+3}{x\sqrt{x}-6\text{x}+9\sqrt{x}}\)
b/ Vậy để P>1 khi BT trên>1
Ta có phương trình tương đương
\(x-3\sqrt{x}+3-x\sqrt{x}+6\text{x}-9>0\)
\(-x\sqrt{x}+7\text{x}-3\sqrt{x}-6>0\)
Giải pt rồi suy ra
tick cho mình nha
\(a)\)\(R=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3\left(\sqrt{x}+3\right)}{x-9}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(R=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right):\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(R=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3}{\sqrt{x-3}}\right):\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(R=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}-3}{\sqrt{x}-3}\right):\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(R=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+1\right):\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(R=\frac{3\sqrt{x}+3}{\sqrt{x}+3}.\frac{\sqrt{x}-3}{\sqrt{x+1}}\)
\(R=\frac{3\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}\)
\(R=\frac{3\left(\sqrt{x}-3\right)}{\sqrt{x}+3}\)
\(b)\) 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 ~
Hoàng Anh Tú gọi là quy đồng cho nhanh