Đưa một thừa số vào trong dấu căn.
a)$-\dfrac{2}{3} \sqrt{ab}$ với $a>0, b \geq 0 \text {; }$
b) $a \sqrt{\frac{3}{a}}$ với $a>0, b \geq 0 \text {; }$
c) $a\sqrt{7}$ với $\mathrm{a} \geq 0;$
d) $b \sqrt{3}$ với $b<0;$
e) $a b \sqrt{\dfrac{a}{b}}$ với $b \geq 0, a>0;$
f) $a b \sqrt{\dfrac{1}{a}+\dfrac{1}{b}}$ với $a>0 , b>0$.
a, \(-\frac{2}{3}\sqrt{ab}=-\sqrt{\frac{4ab}{9}}\)
b, \(a\sqrt{\frac{3}{a}}=\sqrt{\frac{3a^2}{a}}=\sqrt{3a}\)
c, \(a\sqrt{7}=\sqrt{7a^2}\)
d, \(b\sqrt{3}=\sqrt{3b^2}\)
e, \(ab\sqrt{\frac{a}{b}}=\sqrt{\frac{a^3b^2}{b}}=\sqrt{a^3b}\)
f, \(ab\sqrt{\frac{1}{a}+\frac{1}{b}}=\sqrt{\frac{a^2b^2}{a}+\frac{a^2b^2}{b}}=\sqrt{ab^2+a^2b}\)
a,\(-\sqrt{\dfrac{4}{9}ab}\)
d, trừ căn (3b^2)
b, căn 3a
c, căn 7a^2
e, căn của a^3b
f, căn của a^2b+ab^2
a) -√(4/9 *ab)
b) √3a
c) √7a²
d) √3b²
e) √a³b
f)√(ab²+a²b)
a) -√((4ab)/9)
b) √3a
c) √7a^2
d) -√3b^2
e) √a^3b
f) √(ab^2+a^2b)
a. \(-\sqrt{\dfrac{4ab}{9}}\)
b. \(\sqrt{3a}\)
c. \(\sqrt{7a^2}\)
d. \(\sqrt{3b^2}\)
e. \(\sqrt{a^3b}\)
f. \(\sqrt{ab^2+a^2b}\)
a) -căn 4ab/9
b) căn 3a
c) căn 7a^2
d) -căn 3b^2
a) - can 4ab/9
b) can 3a
c) can 7a^2
d)- can 3b^2
e) can a^3b
f) can ab^2+ a^2b
a,\(-\dfrac{2}{3}\sqrt{ab}=-\sqrt{\dfrac{4ab}{9}}\)
b,\(a\sqrt{\dfrac{3}{a}}=\sqrt{\dfrac{3a^2}{a}}=\sqrt{3a}\)
c,\(a\sqrt{7}=\sqrt{7a^2}\)
d,\(b\sqrt{3}=\sqrt{3b^2}\)
e,\(ab\sqrt{\dfrac{a}{b}}=\sqrt{\dfrac{a^3b^2}{b}}=\sqrt{a^3b}\)
f,\(ab\sqrt{\dfrac{1}{a}+\dfrac{1}{b}}=\sqrt{\dfrac{a^2b^2}{a}+\dfrac{a^2b^2}{b}}=\sqrt{ab^2+a^2b}\)
a, \(\dfrac{2}{3}\sqrt{\left(\dfrac{2}{3}\right)^2}nên\)
\(-\dfrac{2}{3}\sqrt{ab}=-\sqrt{\left(\dfrac{2}{3}\right)^2.\sqrt{ab}=-\sqrt{\dfrac{4ab}{9}}}\)
b, \(a\sqrt{a^2}\) do a>0 nên \(a\sqrt{\dfrac{3}{a}}=\sqrt{a^2}.\sqrt{\dfrac{3}{a}}=\sqrt{a^2.\dfrac{3}{a}}=\sqrt{3a}\)
c, \(a\sqrt{7}=\sqrt{7a^2}\\ \) ( vì a > hoặc = 0 )
d, \(b\sqrt{3}=-\left(-b\right)\sqrt{3}=-\sqrt{3b^2}\) ( vì b < 0 )
e, \(ab\sqrt{\dfrac{a}{b}}=\sqrt{\dfrac{a^2b^2a}{b}}=\sqrt{a^3b}\)
f, \(ab\sqrt{\dfrac{1}{a}+\dfrac{1}{b}}=\sqrt{a^2b^2\left(\dfrac{1}{a}+\dfrac{1}{b}\right)}=\sqrt{ab^2+a^2b}\)
A. -2/3Vab =-V4ab/9
B- aV3/a = V3a
C- aV7 =V7a^2
D- bV3 =-V3b^2
E- abVa/b = Va^3b
F- ab V1/a+1/b = Vab^2 + a^2b
a. \(-\dfrac{2}{3}\sqrt{ab}=-\sqrt{\dfrac{4ab}{9}}\)
b. \(a\sqrt{\dfrac{3}{a}}=\sqrt{\dfrac{3a^2}{a}}=\sqrt{3a}\)
c. \(a\sqrt{7}=\sqrt{7a^2}\)
d. \(b\sqrt{3}=-\sqrt{3b^2}\)
e. \(ab\sqrt{\dfrac{a}{b}}=\sqrt{a^3b}\)
f. \(ab\sqrt{\dfrac{1}{a}+\dfrac{1}{b}}=\sqrt{ab^2+a^2b}\)
a,\(-\dfrac{2}{3}\sqrt{ab}=-\dfrac{49ab}{9}\)
b,\(a\sqrt{\dfrac{3}{a}}=\sqrt{\dfrac{3a^2}{a}}=\sqrt{3a}\)
c,\(a\sqrt{7}=\sqrt{7a^2}\)
d,\(b\sqrt{3}=\sqrt{3b^2}\)
e,\(ab\sqrt{\dfrac{a}{b}}=\sqrt{\dfrac{a^3b^2}{b}}=\sqrt{a^3b}\)
f,\(ab\sqrt{\dfrac{1}{a}+\dfrac{1}{b}}=\sqrt{\dfrac{a^2b^2}{a}+\dfrac{a^2b^2}{b}}=\sqrt{ab^2+a^2b}\)
a) -\(\dfrac{2}{3}\) . \(\sqrt{ab}\)
-\(\dfrac{2}{3}\) . \(\sqrt{ab}\)= \(\sqrt{\left(\dfrac{2}{3}\right)^2}\) . \(\sqrt{ab}\)= -\(\sqrt{\dfrac{4ab}{9}}\)
b) a\(\sqrt{\dfrac{3}{a}}\)
a.\(\sqrt{a^2}\) với a>0
nên a\(\sqrt{\dfrac{3}{a}}\) = \(\sqrt{a^2}\). \(\sqrt{\dfrac{3}{a}}\) = \(\sqrt{a^2\dfrac{3}{a}}\) = \(\sqrt{3a}\)
c) \(a\sqrt{7}\) với a>= 0
a\(\sqrt{7}\) = \(\sqrt{7a^2}\) ( vì a>=0 )
d) \(b\sqrt{3}\) = -(-b).\(\sqrt{3}\) = -\(\sqrt{3b^2}\) ( vì b<0 )
e) ab.\(\sqrt{\dfrac{a}{b}}\) với b>=0,a>0
ab.\(\sqrt{\dfrac{a}{b}}\) = \(\sqrt{\dfrac{a^2b^2a}{b}}\) = \(\sqrt{a^3b}\)
f) ab.\(\sqrt{\dfrac{1}{a}+\dfrac{1}{b}}\) với a>0,b>0
ab.\(\sqrt{\dfrac{1}{a}+\dfrac{1}{b}}\) = \(\sqrt{a^2b^2\left(\dfrac{1}{a}+\dfrac{1}{b}\right)}\) = \(\sqrt{ab^2+a^2b}\)
a, \(\dfrac{2}{3}\sqrt{\left(\dfrac{2}{3}\right)^2}\) nên \(-\dfrac{2}{3}\sqrt{ab}\) =\(-\sqrt{\left(\dfrac{2}{3}\right)^2}.\sqrt{ab}=-\sqrt{\dfrac{4ab}{9}}\)
b, \(a\sqrt{a^2}\) do a > 0 nên \(a\sqrt{\dfrac{3}{a}}=\sqrt{a^2}.\sqrt{\dfrac{3}{a}}=\sqrt{a^2.\dfrac{3}{a}}=\sqrt{3a}\)
c, \(a\sqrt{7}=\sqrt{7a^2}\) ( vì a > 0 )
d, \(b\sqrt{3}=-\left(-b\right)\sqrt{3}=-\sqrt{3b^2}\) ( vì b < 0 )
e, \(ab\sqrt{\dfrac{a}{b}}=\sqrt{\dfrac{a^2b^2a}{b}}=\sqrt{a^3b}\)
f, \(ab\sqrt{\dfrac{1}{a}+\dfrac{1}{b}}=\sqrt{a^2b^2\left(\dfrac{1}{a}+\dfrac{1}{b}\right)}=\sqrt{ab^2+a^2b}\)
a) \(\dfrac{2}{3}\sqrt{\left(\dfrac{2}{3}\right)^2}\) nên -
-\(\dfrac{2}{3}\sqrt{ab}=-\sqrt{\left(\dfrac{2}{3}\right)^2}.\sqrt{ab}=-\sqrt{\dfrac{4ab}{9}}\)
b) \(a\sqrt{a^2}\) do \(a>0\) nên \(a\sqrt{\dfrac{3}{a}}=\sqrt{a^2}.\sqrt{\dfrac{3}{a}}=\sqrt{a^2.\dfrac{3}{a}}=\sqrt{3a}\)
c) \(a\sqrt{7}=\sqrt{7a^2}\) ( vì b < 0 )
d) \(b\sqrt{3}=-\left(-b\right)\sqrt{3}=-\sqrt{3b^2}\) ( vì b < 0 )
e) \(ab\sqrt{\dfrac{a}{b}}=\sqrt{\dfrac{a^2b^2a}{b}}=\sqrt{a^3b}\)
f) \(ab\sqrt{\dfrac{1}{a}+\dfrac{1}{b}}=\sqrt{a^2b^2\left(\dfrac{1}{a}+\dfrac{1}{b}\right)=\sqrt{ab^2+a^2b}}\)
a)\(-\dfrac{2}{3}\sqrt{ab}=-\sqrt{\dfrac{4ab}{9}}\left(a>0;b\ge0\right)\)
b)\(a\sqrt{\dfrac{3}{a}}=\sqrt{\dfrac{3a^2}{a}}=\sqrt{3a}\left(a>0;b\ge0\right)\)
c)\(a\sqrt{7}=\sqrt{7a^2}\left(a\ge0\right)\)
d)\(b\sqrt{3}=\sqrt{3b^2}\left(b< 0\right)\)
a, 2/3 căn (2/3)^2 nên
-2/3 căn ab =- căn (2/3)^2. căn ab = - căn 4ab/9
b, a căn a^2 do a> 0 nên a căn 3/a = căn a^2.căn 3/a= căn a^2.3/a= căn 3a
c, a căn 7 = căn 7a^2
d, b căn 3=-(-b) căn 3 = - căn 3b^2
e, ab căn a/b = căn a^2b^2 a/b = căn a^3b
f, ab căn 1/a+1/b = căn a^2b^2( 1/a + 1/b )=căn ab^2+a^2b
a)= -\(\sqrt{\dfrac{4ab}{9}}\)
b)= \(\sqrt{\dfrac{3a^2}{a}}\) = \(\sqrt{3a}\)
c) = \(\sqrt{7a^2}\)
d) = \(\sqrt{3b^2}\)
e) = \(\sqrt{\dfrac{a^3b^2}{b}}\) = \(\sqrt{a^2b}\)
f) = \(\sqrt{\dfrac{a^2b^2}{a}}\) + \(\sqrt{\dfrac{a^2b^2}{b}}\) = \(\sqrt{ab^2+a^2b}\)