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a, \(\dfrac{\sqrt{15}-\sqrt{6}}{\sqrt{35}-\sqrt{14}}=\dfrac{\sqrt{3}.\sqrt{5}-\sqrt{3}.\sqrt{2}}{\sqrt{5}.\sqrt{7}-\sqrt{7}.\sqrt{2}}\)
\(=\dfrac{\sqrt{3}.\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{7}.\left(\sqrt{5}-\sqrt{2}\right)}=\dfrac{\sqrt{3}}{\sqrt{7}}\)
b, \(\dfrac{2\sqrt{15}-2\sqrt{10}+\sqrt{6}-3}{2\sqrt{5}-2\sqrt{10}-\sqrt{3}+\sqrt{6}}\)
\(=\dfrac{2.\sqrt{5}.\sqrt{3}-2.\sqrt{2}.\sqrt{5}-\sqrt{3}.\sqrt{3}+\sqrt{2}.\sqrt{3}}{2.\sqrt{5}-2.\sqrt{2}.\sqrt{5}-\sqrt{3}+\sqrt{2}.\sqrt{3}}\)
\(=\dfrac{2\sqrt{5}\left(\sqrt{3}-\sqrt{2}\right)-\sqrt{3}.\left(\sqrt{3}-\sqrt{2}\right)}{2\sqrt{5}.\left(1-\sqrt{2}\right)-\sqrt{3}.\left(1-\sqrt{2}\right)}\)
\(=\dfrac{\left(2\sqrt{5}+\sqrt{3}\right).\left(\sqrt{3}-\sqrt{2}\right)}{\left(2\sqrt{5}-\sqrt{3}\right).\left(1-\sqrt{2}\right)}=\dfrac{\sqrt{3}-\sqrt{2}}{1-\sqrt{2}}\)
c, \(\dfrac{x+\sqrt{xy}}{y+\sqrt{xy}}=\dfrac{\sqrt{x}.\sqrt{x}+\sqrt{x}.\sqrt{y}}{\sqrt{y}.\sqrt{y}+\sqrt{x}.\sqrt{y}}\)
\(=\dfrac{\sqrt{x}.\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{y}.\left(\sqrt{x}+\sqrt{y}\right)}=\dfrac{\sqrt{x}}{\sqrt{y}}\)
Chúc bạn học tốt!!!
d) \(\dfrac{\sqrt{a}+a\sqrt{b}-\sqrt{b}-b\sqrt{a}}{ab-1}\) = \(-\dfrac{\sqrt{a}\left(1+\sqrt{ab}\right)-\sqrt{b}\left(1+\sqrt{ab}\right)}{1-ab}\)
= \(-\dfrac{\left(\sqrt{a}-\sqrt{b}\right)\left(1+\sqrt{ab}\right)}{\left(1+\sqrt{ab}\right)\left(1-\sqrt{ab}\right)}\) = \(-\dfrac{\sqrt{a}-\sqrt{b}}{1-\sqrt{ab}}\) = \(\dfrac{\sqrt{b}-\sqrt{a}}{1-\sqrt{ab}}\)
Bài 1:
a: \(=\sqrt{32.4}=\dfrac{9}{5}\sqrt{10}\)
b: \(=\sqrt{5\cdot5\cdot7\cdot7\cdot11\cdot11}=5\cdot7\cdot11=385\)
c: \(=5-2\sqrt{6}\)
d: \(=18-1=17\)
e: \(=3\sqrt{2}-2\sqrt{3}+7\sqrt{3}-7\sqrt{2}=-4\sqrt{2}+5\sqrt{3}\)
a: \(=2\sqrt{5}-5\sqrt{5}-4\sqrt{5}+11\sqrt{5}=4\sqrt{5}\)
b: \(=2\sqrt{5}-2-2\sqrt{5}=-2\)
c: \(=3-\sqrt{6}+2\sqrt{6}-3=\sqrt{6}\)
d: \(=\dfrac{2\left(2\sqrt{2}-\sqrt{3}\right)}{\sqrt{6}\left(\sqrt{3}-2\sqrt{2}\right)}-\dfrac{1}{\sqrt{6}}\)
\(=\dfrac{-3}{\sqrt{6}}=-\dfrac{3\sqrt{6}}{6}=-\dfrac{\sqrt{6}}{2}\)
e: \(=\dfrac{8}{3}\sqrt{3}-\dfrac{1}{3}\sqrt{3}-\dfrac{4}{5}\sqrt{3}=\dfrac{23}{15}\sqrt{3}\)
b: \(=\left(\sqrt{ab}+\dfrac{2\sqrt{ab}}{a}-\sqrt{\dfrac{a^2+1}{ab}}\right)\cdot\sqrt{ab}\)
\(=ab+\dfrac{2ab}{a}-\sqrt{a^2+1}=ab+2b-\sqrt{a^2+1}\)
c: \(=2\sqrt{6b}-6\sqrt{18}+10\sqrt{12}-\sqrt{48}\)
\(=2\sqrt{6b}-18\sqrt{2}+20\sqrt{3}-4\sqrt{3}\)
\(=2\sqrt{6n}-18\sqrt{2}+16\sqrt{3}\)
d: \(=\dfrac{\sqrt{3}\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{7}\left(\sqrt{5}-\sqrt{2}\right)}=\dfrac{\sqrt{21}}{7}\)
a) \(\dfrac{2\sqrt{3}+2}{4\sqrt{3}+4}=\dfrac{2\left(\sqrt{3}+1\right)}{4\left(\sqrt{3}+1\right)}=\dfrac{1}{2}\)
b) \(\dfrac{\sqrt{10}+\sqrt{15}}{\sqrt{8}+\sqrt{12}}=\dfrac{\sqrt{5}\left(\sqrt{2}+\sqrt{3}\right)}{\sqrt{4}\left(\sqrt{2}+\sqrt{3}\right)}=\dfrac{\sqrt{5}}{2}\)
c) \(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{4}+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\\ =\dfrac{\left(1+\sqrt{2}\right)\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}=1+\sqrt{2}\)
d) \(\sqrt{9+\sqrt{17}}.\sqrt{9-\sqrt{17}}=\sqrt{\left(9+\sqrt{17}\right)\left(9-\sqrt{17}\right)}\\ =\sqrt{81-17}=\sqrt{64}=8\)
\(a.\dfrac{2\sqrt{3}+2}{4\sqrt{3}+4}=\dfrac{2\left(\sqrt{3}+1\right)}{4\left(\sqrt{3}+1\right)}=\dfrac{2}{4}=\dfrac{1}{2}\)
\(b.\dfrac{\sqrt{10}+\sqrt{15}}{\sqrt{8}+\sqrt{12}}=\dfrac{\sqrt{5}\left(\sqrt{2}+\sqrt{3}\right)}{2\left(\sqrt{2}+\sqrt{3}\right)}=\dfrac{\sqrt{5}}{2}\)
\(c.\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\dfrac{\sqrt{2}+\sqrt{3}+2+2+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+2}=\dfrac{\sqrt{2}+\sqrt{3}+2}{\sqrt{2}+\sqrt{3}+2}+\dfrac{\sqrt{2}\left(\sqrt{2}+\sqrt{3}+2\right)}{\sqrt{2}+\sqrt{3}+2}=1+\sqrt{2}\)
\(d.\sqrt{9+\sqrt{17}}.\sqrt{9-\sqrt{17}}=\sqrt{\left(9+\sqrt{17}\right)\left(9-\sqrt{17}\right)}=\sqrt{81-17}=8\)
\(a,\dfrac{x+2\sqrt{x}-3}{\sqrt{x}-1}\)
\(\Leftrightarrow\dfrac{x+3\sqrt{x}-\sqrt{x}-3}{\sqrt{x}-1}\)
\(\Leftrightarrow\dfrac{\sqrt{x}.\left(\sqrt{x}+3\right)-\left(\sqrt{x}+3\right)}{\sqrt{x}-1}\)
\(\Leftrightarrow\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\)
\(\Rightarrow\sqrt{x}+3\)
\(b,\dfrac{4y+3\sqrt{y}-7}{4\sqrt{y}+7}\)
\(\Leftrightarrow\dfrac{4y+7\sqrt{y}-4\sqrt{y}-7}{4\sqrt{y}+7}\)
\(\Leftrightarrow\dfrac{\sqrt{y}.\left(4\sqrt{y}\right)-\left(4\sqrt{y}+7\right)}{4\sqrt{y}+7}\)
\(\Leftrightarrow\dfrac{\left(4\sqrt{y}+7\right).\left(\sqrt{y}-1\right)}{4\sqrt{y}+7}\)
\(\Rightarrow\sqrt{y}-1\)
\(c,\dfrac{x\sqrt{y}-y\sqrt{x}}{\sqrt{x}-\sqrt{y}}\)
\(\Leftrightarrow\dfrac{\sqrt{xy}.\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}\)
\(\Rightarrow\sqrt{xy}\)
\(d,\dfrac{x-3\sqrt{x}-4}{x-\sqrt{x}-12}\)
\(\Leftrightarrow\dfrac{x+\sqrt{x}-4\sqrt{x}-4}{x+3\sqrt{x}-4\sqrt{x}-12}\)
\(\Leftrightarrow\dfrac{\sqrt{x}.\left(\sqrt{x}+1\right)-4\left(\sqrt{x}+1\right)}{\sqrt{x}.\left(x+3\right)-4\left(\sqrt{x}+3\right)}\)
\(\Leftrightarrow\dfrac{\left(\sqrt{x}+1\right).\left(\sqrt{x}-4\right)}{\left(\sqrt{x}+3\right).\left(\sqrt{x}-4\right)}\)
\(\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)
\(\Rightarrow\dfrac{x-2\sqrt{x}-3}{x-9}\)
\(e,\dfrac{1+\sqrt{x}+\sqrt{y}+\sqrt{xy}}{1+\sqrt{4}}\)
\(\Leftrightarrow\dfrac{1+\sqrt{x}+\sqrt{y}+\sqrt{xy}}{1+2}\)
\(\Rightarrow\dfrac{1+\sqrt{x}+\sqrt{y}+\sqrt{xy}}{3}\)
a) \(\dfrac{\sqrt{15}-\sqrt{6}}{\sqrt{35}-\sqrt{14}}=\dfrac{\left(\sqrt{15}-\sqrt{6}\right)\left(\sqrt{35}+\sqrt{14}\right)}{21}\)
\(=\dfrac{\sqrt{525}+\sqrt{210}-\sqrt{210}-\sqrt{84}}{21}=\dfrac{5\sqrt{21}-2\sqrt{21}}{21}\)
\(=\dfrac{3\sqrt{21}}{21}=\dfrac{\sqrt{21}}{7}\)
b) \(\dfrac{\sqrt{10}+\sqrt{15}}{\sqrt{8}+\sqrt{12}}=\dfrac{\sqrt{10}+\sqrt{15}}{2\sqrt{2}+2\sqrt{3}}\)
\(=\dfrac{\left(\sqrt{10}+\sqrt{15}\right)\left(2\sqrt{2}-2\sqrt{3}\right)}{-4}=\dfrac{\left(\sqrt{10}+\sqrt{15}\right)\left(\sqrt{2}-\sqrt{3}\right)}{-2}\)
\(=\dfrac{\left(\sqrt{10}+\sqrt{15}\right)\left(\sqrt{2}-\sqrt{3}\right)}{-2}=\dfrac{\sqrt{20}-\sqrt{30}+\sqrt{30}-\sqrt{45}}{-2}\)
\(=\dfrac{2\sqrt{5}-3\sqrt{5}}{-2}=\dfrac{-\sqrt{5}}{-2}=\dfrac{\sqrt{5}}{2}\)
c) \(\dfrac{2\sqrt{15}-2\sqrt{10}+\sqrt{6}-3}{2\sqrt{5}-2\sqrt{10}-\sqrt{3}+\sqrt{6}}\) có sai k nhỉ
d) \(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\) (tự làm đc kq là \(1+\sqrt{2}\))
e,f) xem lại đề
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