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a) Ta có: \(\sqrt{27\cdot48\left(1-a^2\right)}\)

\(=\sqrt{3^4\cdot4^2\cdot\left(1-a^2\right)}\)

\(=36\sqrt{1-a^2}\)

c) Ta có: \(\sqrt{5a}\cdot\sqrt{45a}-3a\)

\(=15a-3a=12a\)

b) Ta có: \(B=\dfrac{1}{a-b}\cdot\sqrt{a^4\cdot\left(a-b\right)^2}\)

\(=\dfrac{1}{a-b}\cdot a^2\cdot\left(a-b\right)\)

\(=a^2\)

d) Ta có: \(D=\left(3-a\right)^2-\sqrt{0.2}\cdot\sqrt{180a^2}\)

\(=a^2-6a+9-\sqrt{36a^2}\)

\(=a^2-6a+9-\left|6a\right|\)

\(=\left[{}\begin{matrix}a^2-6a+9-6a\left(a\ge0\right)\\a^2-6a+9+6a\left(a< 0\right)\end{matrix}\right.\)

\(=\left[{}\begin{matrix}a^2-12a+9\\a^2+9\end{matrix}\right.\)

NV
19 tháng 9 2019

\(A=\sqrt{9.3.3.16\left(1-a^2\right)}=3.3.4.\left|1-a\right|=36\left(a-1\right)\)

\(B=\frac{1}{a-b}a^2.\left|a-b\right|=\frac{a^2\left(a-b\right)}{a-b}=a^2\)

\(C=\sqrt{5.45.a^2}-3a=\sqrt{5^2.3^2.a^2}-3a=15\left|a\right|-3a=15a-3a=12a\)

\(D=\left(3-a\right)^2-\sqrt{\frac{2.180}{10}a^2}=\left(3-a\right)^2-6\left|a\right|\)

6 tháng 7 2018

2)

\(\sqrt{12,1.360}=\sqrt{12,1}.\sqrt{36}.\sqrt{10}\)

\(=\sqrt{12,1.36.10}\)

= \(\sqrt{121.36}\)

\(=\sqrt{4356}\)

\(=66\)

3)

\(\sqrt{5a}.\sqrt{45a}-3a\)

\(=\sqrt{5.45a^2}-3a\)

\(=\sqrt{225a^2}-3a\)

\(=\sqrt{\left(15a\right)^2}-3a\)

\(=-15a-3a\) ( vì \(a\le0\))

\(=-18a\)

5)

\(\sqrt{0,36a^2}\)

\(=\sqrt{\left(0,6a\right)^2}\)

\(=-0,6a\) ( vì \(a< 0\) )

Để tối mình rảnh lên coi có làm tiếp được nữa hông thì mình làm ha.

Chúc bạn học tốt!

6 tháng 7 2018

1)

\(\sqrt{3a^3}.\sqrt{12}\)

\(=\sqrt{3}.\sqrt{a^3}.\sqrt{12}\)

\(=\sqrt{3.12}.\sqrt{a^3}\)

\(=6\sqrt{a^3}\)

4)

\(\left(3-a\right)^2-\sqrt{0,2}.\sqrt{180a^2}\)

\(=9.6a.a^2-\sqrt{0,2}.\sqrt{18}.\sqrt{10}.\sqrt{a^2}\)

\(=54a^3-\sqrt{2}.\sqrt{18}.\sqrt{a^2}\)

\(=34a^3-\sqrt{2.18}.\sqrt{a^2}\)

\(=54a^3-6\sqrt{a^2}\)

\(=54a^3-6a^2\) ( vì a<0)

6)

\(\sqrt{a^4.\left(3-a^{ }\right)^2}\)

\(=\sqrt{\left(a^2\right)^2.\left(3-a\right)^2}\)

\(=\sqrt{\left(a^2\right)^2}.\sqrt{\left(3-a\right)^2}\)

\(=\left|a^2\right|\left|3-a\right|\) ( vì a>3 => a>3 nên 3-a<0)

\(\left|3-a\right|=-\left(-3-a\right)=-3+a=a-3\)

\(=a^2\left(a-3\right)\)

\(=a^3-3a^2\)

Còn lại bạn làm tương tự nha, trể quá rùi :)))))

31 tháng 3 2017

a) ĐS: ; b) ĐS: 26; c) ĐS: 12a

d) - = - 6a + 9 -

= - 6a + 9 - = - 6a + 9 - 6│a│.

Khi a ≥ 0 thì │a│= a.

Do đó - = - 6a + 9 -6a = - 12a + 9.

Khi a < 0 thì │a│= a.

Do đó - = - 6a + 9 + 6a = + 9.

1 tháng 8 2018

a)\(\sqrt{4\left(a-3\right)^2}=\sqrt{2^2\left(a-3\right)^2}=\sqrt{\left(2a-6\right)^2}=2a-6\)

b) \(\sqrt{9\left(b-2\right)^2}=\sqrt{3^2\left(b-2\right)^2}=\sqrt{\left[3\left(b-2\right)\right]^2}=3b-6\)

c) bạn xem lại đề

d)
\(\sqrt{5a}.\sqrt{45a}-3a=\sqrt{225a^2}-3a=\sqrt{\left(15a\right)^2}-3a=15a-3a=12a\)

1 tháng 8 2018

e) \(\dfrac{\sqrt{48x^3}}{\sqrt{3x^5}}=\sqrt{\dfrac{48x^3}{3x^5}}=\sqrt{\dfrac{16}{x^2}}=\dfrac{\sqrt{16}}{\sqrt{x^2}}=\dfrac{4}{x}\)

NV
30 tháng 7 2021

\(A=\left|a-3\right|-3a=3-a-3a=3-4a\)

\(B=4a+3-\left|2a-1\right|=4a+3-2a+1=2a+4\)

\(C=\dfrac{4}{a^2-4}\left|a-2\right|=\dfrac{-4\left(a-2\right)}{\left(a-2\right)\left(a+2\right)}=\dfrac{-4}{a+2}\)

\(D=\dfrac{a^2-9}{12}:\sqrt{\dfrac{\left(a+3\right)^2}{16}}=\dfrac{a^2-9}{12}:\dfrac{\left|a+3\right|}{4}=\dfrac{\left(a-3\right)\left(a+3\right).4}{-12\left(a+3\right)}=\dfrac{3-a}{3}\)

\(A=\sqrt{\left(a-3\right)^2}-3a\)

=3-a-3a

=3-4a

 

a) Ta có: \(A=\dfrac{a^2-1}{3}\cdot\sqrt{\dfrac{9}{\left(1-a\right)^2}}\)

\(=\dfrac{\left(a+1\right)\cdot\left(a-1\right)}{3}\cdot\dfrac{3}{\left|1-a\right|}\)

\(=\dfrac{\left(a+1\right)\left(a-1\right)}{1-a}\)

=-a-1

b) Ta có: \(B=\sqrt{\left(3a-5\right)^2}-2a+4\)

\(=\left|3a-5\right|-2a+4\)

\(=5-3a-2a+4\)

=9-5a

c) Ta có: \(C=4a-3-\sqrt{\left(2a-1\right)^2}\)

\(=4a-3-\left|2a-1\right|\)

\(=4a-3-2a+1\)

\(=2a-2\)

d) Ta có: \(D=\dfrac{a-2}{4}\cdot\sqrt{\dfrac{16a^4}{\left(a-2\right)^2}}\)

\(=\dfrac{a-2}{4}\cdot\dfrac{4a^2}{\left|a-2\right|}\)

\(=\dfrac{a^2\left(a-2\right)}{-\left(a-2\right)}\)

\(=-a^2\)

29 tháng 7 2018

a) \(\sqrt{4\left(a-3\right)^2}=\sqrt{2^2\left(a-3\right)^2}=2\sqrt{\left(a-3\right)^2}=2.\left|a-3\right|=2\left(a-3\right)=2a-6\) (Vì \(a\ge3\) )

29 tháng 7 2018

b) \(\sqrt{9\left(b-2\right)^2}=\sqrt{3^2\left(b-2\right)^2}=3\sqrt{\left(b-2\right)^2}=3\left|b-2\right|=3\left(2-b\right)\)

                                                         \(=6-3b\) (vì b < 2 )

b) \(\sqrt{27.48\left(1-a\right)^2}=\sqrt{27.3.16.\left(1-a\right)^2}=\sqrt{81.16.\left(1-a\right)^2}\) 

                                         \(=\sqrt{9^2.4^2.\left(1-a\right)^2}=9.4\sqrt{\left(1-a\right)^2}=36.\left|1-a\right|=36\left(1-a\right)=36-36a\) (vì a > 1)

a) \(\sqrt{27\cdot48\cdot\left(1-a\right)^2}\)

\(=3\sqrt{3}\cdot4\sqrt{3}\cdot\left|1-a\right|\)

\(=36\cdot\left(a-1\right)=36a-36\)

b) \(\dfrac{1}{a-b}\cdot\sqrt{a^4\left(a-b\right)^2}\)

\(=\dfrac{1}{a-b}\cdot\left(a-b\right)\cdot a^2\)

\(=a^2\)