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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: \(a\ge0;a\ne1\)
\(\left(1+\frac{a+\sqrt{a}}{1+\sqrt{a}}\right).\left(1-\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)+a\)
\(=\left(1+\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{1+\sqrt{a}}\right).\left(1-\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)+a\)
\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)+a\)
\(=1-a+a=1\)
a. ĐK \(\hept{\begin{cases}x\ge0\\x\ne9\end{cases}}\)
b. \(Q=\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}+1}{\sqrt{x}-3}-\frac{3-11\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)-3+11\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{2x-6\sqrt{x}+x+4\sqrt{x}+3-3+11\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{3\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{3\sqrt{x}}{\sqrt{x}-3}\)
c. Để \(Q< 1\Rightarrow Q-1< 0\Leftrightarrow\frac{3\sqrt{x}-\sqrt{x}+3}{\sqrt{x}-3}< 0\Leftrightarrow\frac{2\sqrt{x}+3}{\sqrt{x}-3}< 0\)
\(\Rightarrow\sqrt{x}-3< 0\Rightarrow0\le x< 9\)
Vậy \(0\le x< 9\)thì \(Q< 1\)
Bn có thể qua hoc.24h.vn hỏi nha , ở đó có nhiều người biết đó
~ Hok tốt ~
#Gumball
a) \(=\sqrt{\frac{9}{2}}-\sqrt{16.2}+\sqrt{36.2}-\sqrt{81.2}\)
\(=\frac{3}{2}\sqrt{2}-4\sqrt{2}+6\sqrt{2}-9\sqrt{2}\)
\(=\left(\frac{3}{2}-4+6-9\right)\sqrt{2}=\frac{-11}{2}\sqrt{2}\)
b) \(=\frac{\sqrt{5}+3-\sqrt{5}+3}{\left(\sqrt{5}-3\right)\left(\sqrt{5}+3\right)}.\frac{\sqrt{3}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}\)
\(=\frac{6}{5-9}.\left(-\sqrt{3}\right)=\frac{3}{2}\sqrt{3}\)
c) \(=\left(\frac{a-1-4\sqrt{a}+\sqrt{a}+1}{a-1}\right):\frac{\sqrt{a}\left(\sqrt{a}-2\right)}{a-1}\)
\(=\frac{a-3\sqrt{a}}{a-1}.\frac{a-1}{\sqrt{a}\left(\sqrt{a}-2\right)}\)
\(=\frac{\sqrt{a}\left(\sqrt{a}-3\right)}{\sqrt{a}\left(\sqrt{a}-2\right)}=\frac{\sqrt{a}-3}{\sqrt{a}-2}\)
1. Câu hỏi của Nữ hoàng sến súa là ta - Toán lớp 9 - Học toán với OnlineMath
\(A=\frac{\sqrt{3}-\sqrt{6}}{1-\sqrt{2}}-\frac{2+\sqrt{8}}{1+\sqrt{2}}\\ A=\frac{\sqrt{3}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}-\frac{2\left(1+\sqrt{2}\right)}{1+\sqrt{2}}\\ A=\sqrt{3}-2\)
\(C=\left(\frac{a\sqrt{a}-1}{a-\sqrt{a}}-\frac{a\sqrt{a}+1}{a+\sqrt{a}}\right):\frac{a+2}{a-2}\\ C=\left(\frac{a\left(\sqrt{a}-1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}-\frac{a\left(\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}+1\right)}\right):\frac{a+2}{a-2}\\ C=\left(\frac{a}{\sqrt{a}}-\frac{a}{\sqrt{a}}\right):\frac{a+2}{a-2}\\ C=0\)
\(ĐKXĐ:\hept{\begin{cases}a\ge0\\a\ne\frac{1}{9}\end{cases}}\)
\(P=\left(\frac{\sqrt{a}-1}{3\sqrt{a}-1}-\frac{1}{1+3\sqrt{a}}+\frac{8\sqrt{a}}{9a-1}\right)\div\left(1-\frac{3\sqrt{a}-2}{3\sqrt{a}+1}\right)\)
\(\Leftrightarrow P=\frac{\left(\sqrt{a}-1\right)\left(1+3\sqrt{a}\right)-3\sqrt{a}+1+8\sqrt{a}}{9a-1}:\frac{3\sqrt{a}+1-3\sqrt{a}+2}{3\sqrt{a}+1}\)
\(\Leftrightarrow P=\frac{\sqrt{a}+3a-1-3\sqrt{a}-3\sqrt{a}+1+8\sqrt{a}}{\left(3\sqrt{a}-1\right)\left(3\sqrt{a}+1\right)}:\frac{3}{3\sqrt{a}+1}\)
\(\Leftrightarrow P=\frac{\left(3a+3\sqrt{a}\right)\left(3\sqrt{a}+1\right)}{3\left(3\sqrt{a}-1\right)\left(3\sqrt{a}+1\right)}\)
\(\Leftrightarrow P=\frac{a+\sqrt{a}}{3\sqrt{a}-1}\)