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

B =

K
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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

15 tháng 7 2017

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

\(=-10\sqrt{2}+5.2-\left(18-30\sqrt{2}+25\right)\)

\(=-10\sqrt{2}+10-18+30\sqrt{2}-25\)

\(=20\sqrt{2}-33\)

b) câu b đề sai

16 tháng 7 2017

câu a, \(\left(2-\sqrt{2}\right)\left(-5\sqrt{2}\right)-\left(3\sqrt{2}-5\right)^2=-10\sqrt{2}+5.2-\left(8-30\sqrt{2}+25\right)\)

= \(-33+20\sqrt{2}\)

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.

4 tháng 9 2017

a) \(\sqrt{\left(\sqrt{7-2}\right)^2}=\sqrt{5}\)

b)\(\sqrt{\left(\sqrt{2}-1\right)^2}-\sqrt{\left(2-3\sqrt{2}\right)^2}\)

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

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

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

d) hình như bn ghi sai

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

=\(\left(\dfrac{\sqrt{2+\sqrt{3}}}{\sqrt{4-2\sqrt{3}}}+\dfrac{\sqrt{2-\sqrt{3}}}{\sqrt{4+2\sqrt{3}}}\right):\sqrt{2}\)

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

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

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

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

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

f) \(\sqrt{9a^2}+3a-7=-3a+3a-7=-7\)

g)\(\dfrac{\sqrt{4x^2-4x+1}}{4x-2}+3x+2\)

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

=\(\dfrac{1}{2}+3x+2=\dfrac{5}{2}+3x\)

h)\(\sqrt{\left(5a-1\right)^2}+2a-3\)

nếu a<0 :\(-5a+1+2a-3=-3a-2\)

nếu a>0 : \(5a-1+2a-3=7a-4\)

i)\(\sqrt{\dfrac{2a}{5}}.\sqrt{\dfrac{5a}{18}}+2\left(a-1\right)\)

=\(\sqrt{\dfrac{10a^2}{90}}+2a-2=\sqrt{\dfrac{a^2}{9}}+2a-2\)

=\(\dfrac{a}{3}+2a-2=\dfrac{7a}{3}-2\)

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

15 tháng 10 2022

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

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

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

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

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

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

c: \(A-\dfrac{2}{3}=\dfrac{a+1}{a+\sqrt{a}+1}-\dfrac{2}{3}\)

\(=\dfrac{3a+3-2a-2\sqrt{a}-2}{3\left(a+\sqrt{a}+1\right)}=\dfrac{a-2\sqrt{a}+1}{3\left(a+\sqrt{a}+1\right)}>0\)

=>A>2/3

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:

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

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

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

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

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

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

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

11 tháng 9 2018

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

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

\(=1+\dfrac{2\sqrt{a}-1+0}{1+\sqrt{a}+a}.\dfrac{\sqrt{a}\left(-1\right)}{2\sqrt{a}-1}\\ =1+\dfrac{1}{1+\sqrt{a}+a}.\sqrt{a}.\left(-1\right)\)

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