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3\(\sqrt{5}\)= \(\sqrt{3^2.5}\)=\(\sqrt{45}\)
-5\(\sqrt{2}\)= \(-\sqrt{5^2.2}\)= -\(\sqrt{50}\)
\(\dfrac{-2}{3}\sqrt{xy}\) = \(-\sqrt{\left(\dfrac{2}{3}\right)^2xy}\) = -\(\sqrt{\dfrac{4}{9}xy}\)
x\(\sqrt{\dfrac{2}{x}}\)= \(\sqrt{\dfrac{2x^2}{x}}=\sqrt{2x}\)
\(x\sqrt{\dfrac{2}{x}}=\sqrt{x^2\cdot\dfrac{2}{x}}=\sqrt{2x}\)
\(x\sqrt{\dfrac{2}{5}}=\sqrt{\dfrac{2}{5}\cdot x^2}=\sqrt{\dfrac{2x^2}{5}}\)
\(\left(x-5\right)\cdot\sqrt{\dfrac{x}{25-x^2}}=\sqrt{\left(x-5\right)^2\cdot\dfrac{x}{-\left(x-5\right)\left(x+5\right)}}=\sqrt{-\dfrac{x\left(x-5\right)}{x+5}}\)
\(x\sqrt{\dfrac{7}{x^2}}=\sqrt{x^2\cdot\dfrac{7}{x^2}}=\sqrt{7}\)
a) Ta có : Vì \(x\ge0\)và \(y\ge0\)nên \(x+y\ge0\)\(\Leftrightarrow\left|x+y\right|=x+y\)
\(\frac{2}{x^2-y^2}\sqrt{\frac{3\left(x+y\right)^2}{2}}\)
\(=\frac{2}{x^2-y^2}\sqrt{\frac{3}{2}.\left(x+y\right)^2}\)
\(=\frac{2}{x^2-y^2}.\sqrt{\frac{3}{2}}.\left|x+y\right|\)
\(=\frac{2}{\left(x-y\right)\left(x+y\right)}.\sqrt{\frac{3}{2}}.\left(x+y\right)\)
\(=\frac{2}{x-y}.\sqrt{\frac{3}{2}}\)
\(=\frac{1}{x-y}.2.\sqrt{\frac{3}{2}}\)
\(=\frac{1}{x-y}.\sqrt{\frac{2^2.3}{2}}\)
\(=\frac{1}{x-y}.\sqrt{6}=\frac{\sqrt{6}}{x-y}\)
a, \(\frac{2}{x^2-y^2}\sqrt{\frac{3\left(x+y\right)^2}{2}}=\frac{2}{x^2-y^2}\frac{\sqrt{3}\left|x+y\right|}{\sqrt{2}}=\frac{2\sqrt{3}\left(x+y\right)}{\left(x-y\right)\left(x+y\right)\sqrt{2}}\)
do \(x\ge0;y\ge0\)
\(=\frac{2\sqrt{3}}{\sqrt{2}\left(x-y\right)}=\frac{2\sqrt{6}}{2\left(x-y\right)}=\frac{\sqrt{6}}{x-y}\)
+ Ta có:
2√6−√5=2(√6+√5)(√6−√5)(√6+√5)26−5=2(6+5)(6−5)(6+5)
=2(√6+√5)(√6)2−(√5)2=2(√6+√5)6−5=2(6+5)(6)2−(5)2=2(6+5)6−5
=2(√6+√5)1=2(√6+√5)=2(6+5)1=2(6+5).
+ Ta có:
3√10+√7=3(√10−√7)(√10+√7)(√10−√7)310+7=3(10−7)(10+7)(10−7)
=3(√10−√7)(√10)2−(√7)2=3(10−7)(10)2−(7)2=3(√10−√7)10−7=3(10−7)10−7
=3(√10−√7)3=√10−√7=3(10−7)3=10−7.
+ Ta có:
1√x−√y=1.(√x+√y)(√x−√y)(√x+√y)1x−y=1.(x+y)(x−y)(x+y)
=√x+√y(√x)2−(√y)2=√x+√yx−y=x+y(x)2−(y)2=x+yx−y
+ Ta có:
2ab√a−√b=2ab(√a+√b)(√a−√b)(√a+√b)2aba−b=2ab(a+b)(a−b)(a+b)
=2ab(√a+√b)(√a)2−(√b)2=2ab(√a+√b)a−b=2ab(a+b)(a)2−(b)2=2ab(a+b)a−b.
\(\frac{2}{\sqrt{6}-\sqrt{5}}=\frac{2\left(\sqrt{6}+\sqrt{5}\right)}{\left(\sqrt{6}-\sqrt{5}\right)\left(\sqrt{6}+\sqrt{5}\right)}=\frac{2\left(\sqrt{6}+\sqrt{5}\right)}{6-5}=2\left(\sqrt{6}+\sqrt{5}\right)\)
\(\frac{3}{\sqrt{10}+\sqrt{7}}=\frac{3\left(\sqrt{10}-\sqrt{7}\right)}{\left(\sqrt{10}-\sqrt{7}\right)\left(\sqrt{10}+\sqrt{7}\right)}=\frac{3\left(\sqrt{10}-\sqrt{7}\right)}{10-7}=\sqrt{10}-\sqrt{7}\)
\(\frac{1}{\sqrt{x}-\sqrt{y}}=\frac{\sqrt{x}+\sqrt{y}}{x-y}\)
\(\frac{2ab}{\sqrt{a}-\sqrt{b}}=\frac{2ab\left(\sqrt{a}+\sqrt{b}\right)}{a-b}\)
\(3\sqrt{5}=\sqrt{45}\)
\(-5\sqrt{2}=-\sqrt{25}.\sqrt{2}=-\sqrt{50}\)
\(\dfrac{-2}{3}\sqrt{xy}=-\sqrt{\dfrac{4}{9}}.\sqrt{xy}=-\sqrt{\dfrac{4}{9}xy}\left(xy\ge0\right)\)
\(x\sqrt{\dfrac{2}{x}}=\sqrt{x^2}.\sqrt{\dfrac{2}{x}}=\sqrt{\dfrac{2x^2}{x}}=\sqrt{2x}\left(x>0\right)\)
\(\sqrt{48\cdot45}=12\sqrt{15}\\ \sqrt{225\cdot17}=15\sqrt{17}\\ \sqrt{a^3b^7}=\left|ab^3\right|\sqrt{ab}=ab^3\sqrt{ab}\\ \sqrt{x^5\left(x-3\right)^2}=\left|x^2\left(x-3\right)\right|\sqrt{x}=x^2\left(x-3\right)\sqrt{x}\)
\(\sqrt{48\cdot45}=4\sqrt{3}\cdot3\sqrt{5}=12\sqrt{15}\)
\(\sqrt{225\cdot17}=15\sqrt{17}\)
a) a) Biến đổi vế trái thành và làm tiếp.
b) Biến đổi vế trái thành và làm tiếp
(do xy > 0 (gt) nên đưa thừa số xy vào trong căn để khử mẫu)
#Học tốt!!!
\(ab\cdot\sqrt{\dfrac{a}{b}}=a\cdot\sqrt{ab}\)
\(\dfrac{a}{b}\cdot\sqrt{\dfrac{b}{a}}=\dfrac{\sqrt{a\cdot b}}{b}\)
\(\sqrt{\dfrac{1}{b}+\dfrac{1}{b^2}}=\dfrac{\sqrt{b+1}}{b}\)
\(\sqrt{\dfrac{9\cdot a^3}{36\cdot b}}=\dfrac{\sqrt{a^3\cdot b}}{2\cdot b}\)
\(3\cdot x\cdot y\cdot\sqrt{\dfrac{2}{x\cdot y}}=3\cdot\sqrt{2\cdot x\cdot y}\)
a: \(a^2\cdot\sqrt{\dfrac{2}{3a}}=a^2\cdot\dfrac{\sqrt{2}}{\sqrt{3}\cdot\sqrt{a}}=\dfrac{a\sqrt{2}}{\sqrt{3}}=\dfrac{a\sqrt{6}}{3}\)
b: \(\dfrac{x-3}{x}\cdot\sqrt{\dfrac{x^3}{9-x^2}}\)
\(=\dfrac{x-3}{x}\cdot\dfrac{x\sqrt{x}}{\sqrt{x-3}\cdot\sqrt{x+3}}\)
\(=\dfrac{\sqrt{x}\cdot\sqrt{x-3}}{\sqrt{x+3}}\)
Ta có:
+) 3√5=√32.5=√9.5=√45.35=32.5=9.5=45.
+) −5√2=−√52.2=−√25.2=−√50.−52=−52.2=−25.2=−50.
+) Với xy>0xy>0 thì √xyxy có nghĩa nên ta có:
−23√xy=−√(23)2.xy=−√49xy.−23xy=−(23)2.xy=−49xy.
+) Với x>0x>0 thì √2x2x có nghĩa nên ta có:
x√2x=√x2.2x=√x2.2xx2x=x2.2x=x2.2x=√2x.xx=√2x.
a, \(3\sqrt{5}=\sqrt{9.5}=\sqrt{45}\)
b, \(-5\sqrt{2}=-\sqrt{25.2}=-\sqrt{50}\)
c, \(-\frac{2}{3}\sqrt{xy}=-\sqrt{\frac{4}{9}xy}\)
d, \(x\sqrt{\frac{2}{x}}=\sqrt{\frac{2x^2}{x}}=\sqrt{2x}\)