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\(C=\frac{2\sqrt{a}\left(\sqrt{a}-3\right)+\sqrt{a}\left(\sqrt{a}+3\right)-\left(3a+3\right)}{a-9}:\frac{2\sqrt{a}-2-\left(\sqrt{a}-3\right)}{\sqrt{a}-3}\)
\(C=\frac{2a-6\sqrt{a}+a+3\sqrt{a}-3a-3}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}.\frac{\sqrt{a}-3}{2\sqrt{a}-2-\sqrt{a}+3}\)
\(C=\frac{-3\sqrt{a}-3}{\sqrt{a}+3}.\frac{1}{\sqrt{a}+1}\)
\(C=\frac{-3}{\sqrt{a}+3}\)
Thay a = \(21-12\sqrt{3}\) vào C , ta có
\(C=\frac{-3}{\sqrt{21-12\sqrt{3}}+3}\)
\(C=\frac{-3}{\sqrt{\left(2\sqrt{3}-3\right)^2}+3}\)
\(C=\frac{-3}{2\sqrt{3}-3+3}=\frac{-3}{2\sqrt{3}}=\frac{-\sqrt{3}}{2}\)
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: \(C=\dfrac{x-9-x+3\sqrt{x}}{x-9}:\left(\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}+3}+\dfrac{x-9}{x+\sqrt{x}-6}\right)\)
\(=\dfrac{3\left(\sqrt{x}-3\right)}{x-9}:\dfrac{9-x+x-4\sqrt{x}+4+x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{x-4\sqrt{x}+4}\)
\(=\dfrac{3}{\sqrt{x}-2}\)
Bài 1:
a) \(\frac{2}{\sqrt{3}-1}-\frac{2}{\sqrt{3}+1}\)
\(=\frac{2\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}-\frac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\)
\(=\frac{2\left(\sqrt{3}+1\right)}{2}-\frac{2\left(\sqrt{3}-1\right)}{2}\)
\(=\sqrt{3}+1-\left(\sqrt{3}-1\right)=2\)
b) \(\frac{2}{5-\sqrt{3}}+\frac{3}{\sqrt{6}+\sqrt{3}}\)
\(=\frac{2\left(5+\sqrt{3}\right)}{\left(5-\sqrt{3}\right)\left(5+\sqrt{3}\right)}+\frac{3\left(\sqrt{6}-\sqrt{3}\right)}{\left(\sqrt{6}+\sqrt{3}\right)\left(\sqrt{6}-\sqrt{3}\right)}\)
\(=\frac{2\left(5+\sqrt{3}\right)}{2}+\frac{3\left(\sqrt{6}-\sqrt{3}\right)}{3}\)
\(=5+\sqrt{3}+\sqrt{6}-\sqrt{3}=5+\sqrt{6}\)
c) ĐK: \(a\ge0;a\ne1\)
\(\left(1+\frac{a+\sqrt{a}}{1+\sqrt{a}}\right).\left(1-\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)+a\)
\(=\left(1+\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{1+\sqrt{a}}\right).\left(1-\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)+a\)
\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)+a\)
\(=1-a+a=1\)
Bài 1:
Ta có:
\(\left(a-b+c\right)^3=a^3-b^3+c^3-3a^2b+3a^2c+3ab^2+3b^2c+3ac^2-3bc^2-6abc\)
\(\Rightarrow\left(\sqrt[3]{\frac{1}{9}}-\sqrt[3]{\frac{2}{9}}+\sqrt[3]{\frac{4}{9}}\right)^3=\frac{1}{9}-\frac{2}{9}+\frac{4}{9}-\frac{1}{3}.\sqrt[3]{2}+\frac{1}{3}.\sqrt[3]{4}+\frac{1}{3}.\sqrt[3]{4}+\frac{2}{3}.\sqrt[3]{2}\)
\(+\frac{2}{3}.\sqrt[3]{2}-\frac{2}{3}.\sqrt[3]{4}-\frac{4}{3}=\sqrt[3]{2}-1\)
\(\Rightarrow\sqrt[3]{\sqrt[3]{2}-1}=\sqrt[3]{\frac{1}{9}}-\sqrt[3]{\frac{2}{9}}+\sqrt[3]{\frac{4}{9}}\)
Câu 3 :
\(ĐKXĐ:x>0\)
\(P=\left(\frac{2}{\sqrt{x}}+\frac{\sqrt{x}}{\sqrt{x}+2}\right):\frac{2\sqrt{x}}{x+2\sqrt{x}}\)
\(\Leftrightarrow P=\frac{2\sqrt{x}+4+x}{x+2\sqrt{x}}\cdot\frac{x+2\sqrt{x}}{2\sqrt{x}}\)
\(\Leftrightarrow P=\frac{2\sqrt{x}+4+x}{2\sqrt{x}}\)
b) Để P = 3
\(\Leftrightarrow\frac{2\sqrt{x}+4+x}{x+2\sqrt{x}}=3\)
\(\Leftrightarrow2\sqrt{x}+4+x=6\sqrt{x}\)
\(\Leftrightarrow x-4\sqrt{x}+4=0\)
\(\Leftrightarrow\left(\sqrt{x}-2\right)^2=0\)
\(\Leftrightarrow\sqrt{x}-2=0\)
\(\Leftrightarrow\sqrt{x}=2\)
\(\Leftrightarrow x=4\)(tm)
Vậy để \(P=3\Leftrightarrow x=4\)
Câu 1 : Hình như sai đề !! Mik sửa :
\(ĐKXĐ:\hept{\begin{cases}x\ge0\\x\ne4\end{cases}}\)
\(A=\left(\frac{x}{x\sqrt{x}-4\sqrt{x}}-\frac{6}{3\sqrt{x}-6}+\frac{1}{\sqrt{x}+2}\right):\left(\sqrt{x}-2+\frac{10-x}{\sqrt{x}+2}\right)\)
\(\Leftrightarrow A=\left(\frac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{2}{\sqrt{x}-2}+\frac{1}{\sqrt{x}+2}\right):\left(\frac{x-4+10-x}{\sqrt{x}+2}\right)\)
\(\Leftrightarrow A=\frac{\sqrt{x}-2\sqrt{x}-4+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}:\frac{6}{\sqrt{x}+2}\)
\(\Leftrightarrow A=\frac{-6\left(\sqrt{x}+2\right)}{6\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(\Leftrightarrow A=-\frac{1}{\sqrt{x}-2}\)
b) Để A < 2
\(\Leftrightarrow-\frac{1}{\sqrt{x}-2}< 2\)
\(\Leftrightarrow-1< 2\sqrt{x}-4\)
\(\Leftrightarrow2\sqrt{x}>3\)
\(\Leftrightarrow\sqrt{x}>1,5\)
\(\Leftrightarrow x>2,25\)
Vậy để \(A< 2\Leftrightarrow x>2,25\)
ĐKXĐ:
\(\left\{{}\begin{matrix}a\ge0\\\sqrt{a}\ne3\\a\ne9\\\frac{2\sqrt{a}-2}{\sqrt{a}-3}-1\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a\ge0\\a\ne9\\\sqrt{a}+1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a\ge0\\a\ne9\end{matrix}\right.\)
a,
\(Q=\left(\frac{2\sqrt{a}}{\sqrt{a}+3}-\frac{\sqrt{a}}{\sqrt{a}-3}-\frac{3a+3}{a-9}\right):\left(\frac{2\sqrt{a}-2}{\sqrt{a}-3}-1\right)\)
\(=\frac{2\sqrt{a}\left(\sqrt{a}-3\right)-\sqrt{a}\left(\sqrt{a}+3\right)-3a-3}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}:\frac{2\sqrt{a}-2-\sqrt{a}+3}{\sqrt{a}-3}\)
\(=\frac{2a-6\sqrt{a}-a-3\sqrt{a}-3a-3}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}\cdot\frac{\sqrt{a}-3}{\sqrt{a}+1}\)
\(=\frac{-2a-9\sqrt{a}-3}{\left(\sqrt{a}+1\right)\left(\sqrt{a}+3\right)}=\frac{-2a-9\sqrt{a}-3}{a+4\sqrt{a}+3}\)
b,
\(Q< -\frac{1}{2}\Leftrightarrow\frac{-2a-9\sqrt{a}-3}{a+4\sqrt{a}+3}< -\frac{1}{2}\)
\(\Leftrightarrow\frac{2a+9\sqrt{a}+3}{a+4\sqrt{a}+3}>\frac{1}{2}\)
\(\Leftrightarrow4a+18\sqrt{a}+6>a+4\sqrt{a}+3\)
\(\Leftrightarrow3a+14\sqrt{a}+3>0\)
Vậy với mọi thỏa ĐKXĐ thì \(Q< -\frac{1}{2}\)
c,
\(Q=\frac{-2a-9\sqrt{a}-3}{a+4\sqrt{a}+3}=-\frac{\left(a+4\sqrt{a}+3\right)+a+5\sqrt{a}}{a+4\sqrt{a}+3}=-1-\frac{a+5\sqrt{a}}{a+4\sqrt{a}+3}\)
mình nghx đề có vấn đề, số xấu quá
mk sửa đề lại xíu nha
\(Q=\left(\frac{2\sqrt{a}}{\sqrt{a}+3}+\frac{\sqrt{a}}{\sqrt{a}-3}-\frac{3a+3}{a-9}\right):\left(\frac{2\sqrt{a}-2}{\sqrt{a}-3}-1\right)\)