\(a\le0,tính\sqrt{\dfrac{-2a}{3}}.\sqrt{\dfrac{-3a}{8}}\)

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27 tháng 9 2021

\(\sqrt{\dfrac{-2a}{3}}.\sqrt{\dfrac{-3a}{8}}=\sqrt{\dfrac{-2a}{3}.\dfrac{-3a}{8}}=\sqrt{\dfrac{a^2}{4}}=\dfrac{\left|a\right|}{2}=-\dfrac{a}{2}\left(do.a\le0\right)\)

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.

14 tháng 10 2022

\(A=\dfrac{2\sqrt{a}\left(a+1\right)-3\left(a+1\right)}{2\sqrt{a}-3}=a+1\)

\(B=\dfrac{2a\left(a-1\right)}{\sqrt{a}\left(a-1\right)}=2\sqrt{a}\)

\(A-B=a+1-2\sqrt{a}=\left(\sqrt{a}-1\right)^2>=0\)

=>A>=B

AH
Akai Haruma
Giáo viên
2 tháng 1 2020

Bạn tham khảo lời giải tại đây:

Câu hỏi của Phác Chí Mẫn - Toán lớp 9 | Học trực tuyến

19 tháng 8 2018

B=\(\dfrac{\sqrt{a.6}}{\sqrt{6.6}}+\dfrac{\sqrt{2a.3}}{\sqrt{3.3}}+\dfrac{\sqrt{3a.2}}{\sqrt{2.2}}\)

=\(\dfrac{\sqrt{6a}}{6}+\dfrac{\sqrt{6a}}{3}+\dfrac{\sqrt{6a}}{2}\)

=\(\dfrac{\sqrt{6a}+2\sqrt{6a}+3\sqrt{6a}}{6}\)

=\(\dfrac{6\sqrt{6a}}{6}=\sqrt{6a}\)

b: \(B=\dfrac{\sqrt{6}}{6}\cdot\sqrt{a}+\dfrac{\sqrt{6}}{3}\cdot\sqrt{a}+\dfrac{\sqrt{6}}{2}\cdot\sqrt{a}\)

\(=\sqrt{a}\cdot\sqrt{6}=\sqrt{6a}\)

e: \(=2-x-x=2-2x\)

i: \(=\left|x-\left(1-x\right)\right|-2x=\left|x-1+x\right|-2x\)

\(=\left|2x-1\right|-2x\)

=1-2x-2x=1-4x

26 tháng 5 2017

a)\(\dfrac{\sqrt{243a}}{\sqrt{3a}}=\dfrac{\sqrt{24}.\sqrt{3a}}{\sqrt{3a}}=2\sqrt{6}\)

b)\(\dfrac{3\sqrt{18a^2b^4}}{\sqrt{2a^2b^2}}=3\sqrt{9b^2}=\left[{}\begin{matrix}9b\\-9b\end{matrix}\right.\)

a: \(\dfrac{5}{3\sqrt{8}}=\dfrac{5\sqrt{2}}{3\cdot4}=\dfrac{5\sqrt{2}}{12}\)

\(\dfrac{2}{\sqrt{b}}=\dfrac{2\sqrt{b}}{b}\)

b: \(\dfrac{5}{5-2\sqrt{3}}=\dfrac{25+10\sqrt{3}}{13}\)

\(\dfrac{2a}{1-\sqrt{a}}=\dfrac{2a\left(1+\sqrt{a}\right)}{1-a}\)

c: \(\dfrac{4}{\sqrt{7}+\sqrt{5}}=\dfrac{4\left(\sqrt{7}-\sqrt{5}\right)}{2}=2\sqrt{7}-2\sqrt{5}\)

\(\dfrac{6a}{2\sqrt{a}-\sqrt{b}}=\dfrac{6a\left(2\sqrt{a}+\sqrt{b}\right)}{4a-b}\)

26 tháng 7 2018

a. \(\sqrt{\dfrac{3a}{2}}.\sqrt{\dfrac{2a}{75}}=\sqrt{\dfrac{3a.2a}{2.75}}=\sqrt{\dfrac{3a^2}{75}}=\sqrt{\dfrac{a^2}{25}}=\dfrac{\sqrt{a^2}}{\sqrt{25}}=\dfrac{a}{5}\)

b.\(\sqrt{5a}.\sqrt{\dfrac{2a}{a}}=\sqrt{5a}.\sqrt{2}=\sqrt{10a}\)

26 tháng 7 2018

a.\(\sqrt{\dfrac{3a}{2}}.\sqrt{\dfrac{2a}{75}}=\dfrac{\sqrt{3a}}{\sqrt{2}}.\dfrac{\sqrt{2a}}{\sqrt{25}.\sqrt{3}}=\dfrac{a}{5}\) b. \(\sqrt{5a}.\sqrt{\dfrac{2a}{a}}=\dfrac{\sqrt{5}.\sqrt{a}.\sqrt{2a}}{\sqrt{a}}=\sqrt{10a}\)

7 tháng 8 2017

\(\dfrac{\sqrt{14-6\sqrt{5}}}{\sqrt{5}-3}\)

\(=\dfrac{\sqrt{\left(3-\sqrt{5}\right)^2}}{\sqrt{5}-3}\)

\(=\dfrac{3-\sqrt{5}}{\sqrt{5}-3}\)

= - 1

\(\dfrac{\sqrt{3+\sqrt{5}}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{6+2\sqrt{5}}}{2}\)

\(=\dfrac{\sqrt{\left(\sqrt{5}+1\right)^2}}{2}\)

\(=\dfrac{\sqrt{5}+1}{2}\)

\(\dfrac{2+\sqrt{2}}{\sqrt{1,5+\sqrt{2}}}\)

\(=\dfrac{2\sqrt{2}+2}{\sqrt{3+2\sqrt{2}}}\)

\(=\dfrac{2\left(\sqrt{2}+1\right)}{\sqrt{\left(\sqrt{2}+1\right)^2}}\)

\(=\dfrac{2\left(\sqrt{2}+1\right)}{\sqrt{2}+1}\)

= 2

\(\dfrac{\sqrt{20}}{\sqrt{5}}+\dfrac{\sqrt{117}}{\sqrt{13}}+\dfrac{\sqrt{272}}{\sqrt{17}}+\dfrac{\sqrt{105}}{\sqrt{2\dfrac{1}{7}}}\)

\(=4+9+16+49\)

= 78

7 tháng 8 2017

\(\dfrac{x\sqrt{x}-y\sqrt{y}}{x+\sqrt{xy}+y}\)

\(=\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{x+\sqrt{xy}+y}\)

\(=\sqrt{x}-\sqrt{y}\)

\(\dfrac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\dfrac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)

\(=\dfrac{\left(2+\sqrt{3}\right)\left(\sqrt{2}-\sqrt{2-\sqrt{3}}\right)+\left(2-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{2+\sqrt{3}}\right)}{\left(\sqrt{2}+\sqrt{2+\sqrt{3}}\right)\left(\sqrt{2}-\sqrt{2-\sqrt{3}}\right)}\)

\(\left[-\text{tử}-\right]=\sqrt{2}\left(2+\sqrt{3}\right)-\sqrt{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)^2}+\sqrt{2}\left(2-\sqrt{3}\right)+\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)^2}\)

\(=4\sqrt{2}-\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\)

\(\left[-\text{mẫu}-\right]=2-\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}-\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}\)

\(=2-\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{4-3}\)

\(=2-\left(\sqrt{3}-1\right)+\left(\sqrt{3}+1\right)-1\)

= 3

Ta có:

\(\dfrac{4\sqrt{2}-\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}}{3}\)

\(=\dfrac{8-\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}}{3\sqrt{2}}\)

\(=\dfrac{8-\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}}{3\sqrt{2}}\)

\(=\dfrac{8-\left(\sqrt{3}+1\right)+\left(\sqrt{3}-1\right)}{3\sqrt{2}}=\dfrac{6}{3\sqrt{2}}=\sqrt{2}\)

\(\sqrt{\dfrac{2+a-2\sqrt{2a}}{a+3-2\sqrt{3a}}}\)

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

\(=\dfrac{\left|\sqrt{a}-\sqrt{2}\right|}{\left|\sqrt{a}-\sqrt{3}\right|}\)