Trục căn thức ở mẫu :
f) \(\dfrac{2}{\sqrt{6}-\sqrt{5}}\)
l) \(\dfrac{3}{\sqrt{10}+\sqrt{7}}\)
m) \(\dfrac{1}{\sqrt{x}-\sqrt{y}}\) (\(x>0;y>0;x\ne y\))
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Đề : Trục căn thức ở mẫuf) dfrac{2}{sqrt{6}-sqrt{5}} l) dfrac{3}{sqrt{10}+sqrt{7}} m) dfrac{1}{sqrt{x}-sqrt{y}} ( x0 ,y0,xne y )o) dfrac{2ab}{sqrt{a}-sqrt{b}} (age0,bge0,ane b) P) dfrac{P}{2sqrt{P}-1} (Pge0 , Pnedfrac{1}{4})
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Đề : Trục căn thức ở mẫu
f) \(\dfrac{2}{\sqrt{6}-\sqrt{5}}\) l) \(\dfrac{3}{\sqrt{10}+\sqrt{7}}\) m) \(\dfrac{1}{\sqrt{x}-\sqrt{y}}\) ( x>0 ,y>0,\(x\ne y\) )
o) \(\dfrac{2ab}{\sqrt{a}-\sqrt{b}}\) (\(a\ge0,b\ge0,a\ne b\))
P) \(\dfrac{P}{2\sqrt{P}-1}\) (\(P\ge0\) , \(P\ne\dfrac{1}{4}\))
f: \(\dfrac{2}{\sqrt{6}-\sqrt{5}}=2\sqrt{6}+2\sqrt{5}\)
l: \(\dfrac{3}{\sqrt{10}+\sqrt{7}}=\sqrt{10}-\sqrt{7}\)
m: \(\dfrac{1}{\sqrt{x}-\sqrt{y}}=\dfrac{\sqrt{x}+\sqrt{y}}{x-y}\)
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Đề : Trục căn thức ở mẫua) dfrac{5}{sqrt{10}} b) dfrac{5}{2sqrt{5}} c) dfrac{1}{3sqrt{20}}d) dfrac{2sqrt{2}+2}{5sqrt{2}} e) dfrac{y+bsqrt{y}}{bsqrt{y}} (với bge0 vàbne0 )
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Đề : Trục căn thức ở mẫu
a) \(\dfrac{5}{\sqrt{10}}\) b) \(\dfrac{5}{2\sqrt{5}}\) c) \(\dfrac{1}{3\sqrt{20}}\)
d) \(\dfrac{2\sqrt{2}+2}{5\sqrt{2}}\) e) \(\dfrac{y+b\sqrt{y}}{b\sqrt{y}}\) (với \(b\ge0\) và\(b\ne0\) )
a: \(\dfrac{5}{\sqrt{10}}=\dfrac{5\sqrt{10}}{10}=\dfrac{\sqrt{10}}{2}\)
b: \(\dfrac{5}{2\sqrt{5}}=\dfrac{\sqrt{5}}{2}\)
c: \(\dfrac{1}{3\sqrt{20}}=\dfrac{\sqrt{5}}{30}\)
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a)\(\dfrac{5}{\sqrt{10}}=\dfrac{5\sqrt{10}}{10}=\dfrac{\sqrt{10}}{2}\)
b)\(\dfrac{5}{2\sqrt{5}}=\dfrac{5\sqrt{5}}{2.5}=\dfrac{\sqrt{5}}{2}\)
c)\(\dfrac{1}{3\sqrt{20}}=\dfrac{\sqrt{20}}{3.20}=\dfrac{\sqrt{20}}{60}=\dfrac{\sqrt{5}}{30}\)
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d)\(\dfrac{2\sqrt{2}+2}{5\sqrt{2}}=\dfrac{2.2+2\sqrt{2}}{5.2}=\dfrac{2+\sqrt{2}}{5}\)
e)\(\dfrac{y+b\sqrt{y}}{b\sqrt{y}}=\dfrac{y\sqrt{y}+by}{by}=\dfrac{\sqrt{y}+b}{b}\)
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TRục căn thức ở mẫu A =\(\frac{1}{\sqrt{2}+\sqrt{3}-\sqrt{6}}\)
\(\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{6}\right)\left(2\sqrt{6}+1\right)}{23}\)
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Khử mẫu của biểu thức dưới dấu căn bậc hai
a) \(\sqrt{\dfrac{5x^3}{49y}}\)
với x ≥ 0, y >0
b) 7xy\(\sqrt{\dfrac{-3}{xy}}\)
với x<0, y>0
\(\sqrt{18}-\frac{1}{3}\sqrt{72}-\sqrt{8}+\frac{2-3\sqrt{2}}{3-\sqrt{2}}\)
Trục căn thức ở mẫu
Ta có: \(\sqrt{18}-\frac{1}{3}\sqrt{72}-\sqrt{8}+\frac{2-3\sqrt{2}}{3-\sqrt{2}}\)
\(=3\sqrt{2}-\frac{6\sqrt{2}}{3}-2\sqrt{2}+\frac{\left(3+\sqrt{2}\right)\left(2-3\sqrt{2}\right)}{9-2}\)
\(=3\sqrt{2}-2\sqrt{2}-2\sqrt{2}-\sqrt{2}\)
\(=-2\sqrt{2}\)
Rút gọn các biểu thức sau
a) $M=\sqrt{\dfrac{3 a}{7}}-2 \sqrt{\dfrac{7 a}{3}}+\sqrt{21 a};$
b) $N=\sqrt{\dfrac{8 x}{3}}-\sqrt{\dfrac{27 x}{2}}+\sqrt{6 x};$
c) $P=2 \sqrt{\dfrac{8 y}{5}}+\sqrt{\dfrac{45 y}{2}}-\sqrt{10 y}$.
a)\(\sqrt{\frac{3a}{7}}-2\sqrt{\frac{7a}{3}}+\sqrt{21a}\) =\(\sqrt{\frac{3}{7}.\frac{1}{21}.21a}\) - \(2\sqrt{\frac{7}{3}.\frac{1}{21}.21a}\)+ \(\sqrt{21}\)
=\(\sqrt{\frac{1}{49}.21a}\) - \(2\sqrt{\frac{1}{9}.21a}\)+\(\sqrt{21}\)
=\(\sqrt{\frac{1}{49}}.\sqrt{21a}\) - \(2.\sqrt{\frac{1}{9}}.\sqrt{21a}\)+ \(\sqrt{21a}\)
=\(\frac{1}{7}\sqrt{21a}\) - \(\frac{2}{3}\sqrt{21a}\) + \(\sqrt{21a}\)
=\(\frac{-10}{21}\sqrt{21a}\)
b)
N=\(\sqrt{\frac{8x}{3}}\) - \(\sqrt{\frac{27x}{2}}\) + \(\sqrt{6x}\)
=\(\sqrt{\frac{8}{3}.\frac{1}{6}.6x}\) - \(\sqrt{\frac{27}{2}.\frac{1}{6}.6x}\)+ \(\sqrt{6x}\)
=\(\frac{2}{3}\sqrt{6x}-\frac{3}{2}.\sqrt{6x}+\sqrt{6x}\)
=\(\frac{1}{6}\sqrt{6x}\)
em lớp 8 nene làm theo cách hiểu thôi ạ
c)P=\(2\sqrt{\frac{8y}{5}}\) + \(\sqrt{\frac{45y}{2}}\) - \(\sqrt{10y}\)
=\(2\sqrt{\frac{8}{5}.\frac{1}{10}.10y}\) + \(\sqrt{\frac{45}{2}.\frac{1}{10}.10y}\) - \(\sqrt{10y}\)
=\(2\sqrt{\frac{4}{25}.10y}\) + \(\sqrt{\frac{9}{4}.10y}\) - \(\sqrt{10y}\)
=\(2\).\(\sqrt{\frac{4}{25}}\) \(.\sqrt{10y}\) + \(\sqrt{\frac{9}{4}}.\sqrt{10y}\) - \(\sqrt{10y}\)
=\(\frac{4}{5}\sqrt{10y}\) + \(\frac{3}{2}\sqrt{10y}\) - \(\sqrt{10y}\)
=\(\frac{13}{10}\sqrt{10y}\)
Trục căn thức ở mẫu và giả thiết các biểu thức đều có nghĩa:
\(\dfrac{2}{\sqrt{6}-\sqrt{5}};\dfrac{3}{\sqrt{10}+\sqrt{7}};\dfrac{1}{\sqrt{x}-\sqrt{y}};\dfrac{2ab}{\sqrt{a}-\sqrt{b}}.\)
\(\dfrac{2ab}{\sqrt{a}-\sqrt{b}}=\dfrac{2ab\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}=\dfrac{2ab\left(\sqrt{a}+\sqrt{b}\right)}{a-b}\)
\(\dfrac{1}{\sqrt{x}-\sqrt{y}}=\dfrac{\sqrt{x}+\sqrt{y}}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}=\dfrac{\sqrt{x}+\sqrt{y}}{x-y}\)
\(\dfrac{3}{\sqrt{10}+\sqrt{7}}=\dfrac{3\left(\sqrt{10}-\sqrt{7}\right)}{\left(\sqrt{10}+\sqrt{7}\right)\left(\sqrt{10}-\sqrt{7}\right)}=\dfrac{3\left(\sqrt{10}-\sqrt{7}\right)}{10-7}=\dfrac{3\left(\sqrt{10}-\sqrt{7}\right)}{3}=\sqrt{10}-\sqrt{7}\)
\(\dfrac{2}{\sqrt{6}-\sqrt{5}}=\dfrac{2\left(\sqrt{6}+\sqrt{5}\right)}{\left(\sqrt{6}-\sqrt{5}\right)\left(\sqrt{6}+\sqrt{5}\right)}=\dfrac{2\left(\sqrt{6}+\sqrt{5}\right)}{6-5}=2\left(\sqrt{6}+\sqrt{5}\right)\)
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Trục căn thức ở mẫu : \(\frac{1}{\sqrt{10}+\sqrt{15}+\sqrt{14}+\sqrt{21}}\)
bài tập 1 rút gọn
a) \(\dfrac{\sqrt{7}-5}{2}-\dfrac{6-\sqrt{7}}{4}+\dfrac{6}{4-\sqrt{7}}-\dfrac{5}{4+\sqrt{7}}\)
b) \(\dfrac{x\sqrt{y}+y\sqrt{y}}{\sqrt{xy}}:\dfrac{x+y}{\sqrt{x}-\sqrt{y}}\left(x,y>0\right)\)
c) (\(\dfrac{\sqrt{x}-1}{3\sqrt{x}-1}+\dfrac{1}{3\sqrt{x}+1}\)) : \(\dfrac{3\sqrt{x}-5}{3\sqrt{x}-1}\)
a: \(=\dfrac{2\sqrt{7}-10-6+\sqrt{7}}{4}+\dfrac{24+6\sqrt{7}-20+5\sqrt{7}}{9}\)
\(=\dfrac{3\sqrt{7}-16}{4}+\dfrac{4+11\sqrt{7}}{9}\)
\(=\dfrac{27\sqrt{7}-144+16+44\sqrt{7}}{36}=\dfrac{71\sqrt{7}-128}{36}\)
b: \(=\dfrac{\sqrt{y}\left(x+y\right)}{\sqrt{xy}}\cdot\dfrac{\sqrt{x}-\sqrt{y}}{x+y}\)
\(=\dfrac{\sqrt{x}-\sqrt{y}}{\sqrt{x}}\)
c: \(=\left(\dfrac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)+3\sqrt{x}-1}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}\right)\cdot\dfrac{3\sqrt{x}-1}{3\sqrt{x}-5}\)
\(=\dfrac{3x+\sqrt{x}-3\sqrt{x}-1+3\sqrt{x}-1}{3\sqrt{x}+1}\cdot\dfrac{1}{3\sqrt{x}-5}\)
\(=\dfrac{3x+\sqrt{x}-2}{\left(3\sqrt{x}+1\right)}\cdot\dfrac{1}{3\sqrt{x}-5}\)
\(=\dfrac{3x+\sqrt{x}-2}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-5\right)}\)
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