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19 tháng 7 2019
\(\left(\frac{a\sqrt{a}-b\sqrt{b}}{\sqrt{a}-\sqrt{b}}+\sqrt{ab}\right)\left(\frac{\sqrt{a}-\sqrt{b}}{a-b}\right)^2\) \(ĐKXĐ:\hept{\begin{cases}a\ge0\\b\ge0\\a\ne b\end{cases}}\)
\(=\left(\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}{\sqrt{a}-\sqrt{b}}+\sqrt{ab}\right)\left(\frac{\sqrt{a}-\sqrt{b}}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\right)^2\)
\(=\left(\left(a+\sqrt{ab}+b\right)+\sqrt{ab}\right)\left(\frac{1}{\left(\sqrt{a}+\sqrt{b}\right)}\right)^2\)
\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\left(\sqrt{a}+\sqrt{b}\right)^2}\)
\(=1\)
A
19 tháng 7 2019
\(\left(\frac{a\sqrt{a}-b\sqrt{b}}{\sqrt{a}-\sqrt{b}}+\sqrt{ab}\right)\left(\frac{\sqrt{a}-\sqrt{b}}{a-b}\right)^2.\)
\(=\left(\frac{a\sqrt{a}-b\sqrt{b}}{\sqrt{a}-\sqrt{b}}+\frac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\right)\left(\frac{\sqrt{a}-\sqrt{b}}{a-b}\right)^2.\)
\(=\left(\frac{a\sqrt{a}-b\sqrt{b}}{\sqrt{a}-\sqrt{b}}+\frac{a\sqrt{b}-b\sqrt{a}}{\sqrt{a}-\sqrt{b}}\right)\left(\frac{\sqrt{a}-\sqrt{b}}{a-b}\right)^2.\)
\(=\left(\frac{a\sqrt{a}-b\sqrt{b}+a\sqrt{b}-b\sqrt{a}}{\sqrt{a}-\sqrt{b}}\right)\left(\frac{\sqrt{a}-\sqrt{b}}{a-b}\right)^2.\)
\(=\left(\frac{a\left(\sqrt{a}+\sqrt{b}\right)-b\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\right)\left(\frac{\sqrt{a}-\sqrt{b}}{a-b}\right)^2.\)
\(=\left(\frac{\left(a-b\right)\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\right)\left(\frac{\sqrt{a}-\sqrt{b}}{a-b}\right)^2.\)
\(=\left(\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}\right)\left(\frac{\sqrt{a}-\sqrt{b}}{a-b}\right)^2.\)
\(=\left(\sqrt{a}+\sqrt{b}\right)^2\left(\frac{\sqrt{a}-\sqrt{b}}{a-b}\right)^2.\)
\(=\left(\sqrt{a}+\sqrt{b}\right)^2\left(\frac{\sqrt{a}-\sqrt{b}}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\right)^2.\)
\(=\left(\sqrt{a}+\sqrt{b}\right)^2\cdot\frac{1}{\left(\sqrt{a}+\sqrt{b}\right)^2}.\)\(=1\)
AH
Akai Haruma
Giáo viên
30 tháng 6 2019
Lời giải:
\(\sqrt{\frac{9+12a+4a^2}{b^2}}=\sqrt{\frac{(2a)^2+2.2a.3+3^2}{b^2}}=\sqrt{\frac{(2a+3)^2}{b^2}}\)
\(=|\frac{2a+3}{b}|\)
Vì $a>-1,5; b< 0$ nên \(\frac{2a+3}{b}< 0\Rightarrow \sqrt{\frac{9+12a+4a^2}{b^2}}= |\frac{2a+3}{b}|=\frac{-2a-3}{b}\)
\((a-b)\sqrt{\frac{ab}{(a-b)^2}}=(a-b)\sqrt{ab}.\frac{1}{|a-b|}\)
Do $a< b< 0$ nên $a-b< 0\rightarrow |a-b|=b-a$
\(\Rightarrow (a-b)\sqrt{\frac{ab}{(a-b)^2}}=(a-b).\frac{\sqrt{ab}}{|a-b|}=(a-b).\frac{\sqrt{ab}}{b-a}=-\sqrt{ab}\)
31 tháng 3 2017
a,\(ab^2\sqrt{\dfrac{3}{a^2b^4}}=ab^2.\dfrac{\sqrt{3}}{\sqrt{a^2b^4}}=ab^2.\dfrac{\sqrt{3}}{ab^2}=\sqrt{3}\)
b,\(\sqrt{\dfrac{27\left(a-3\right)^2}{48}}=\dfrac{3\sqrt{3}\left(a-3\right)}{4\sqrt{3}}=\dfrac{3}{4}\left(a-3\right)\)
c,\(\sqrt{\dfrac{9+12a+4a^2}{b^2}}=\dfrac{\sqrt{\left(3+2a\right)^2}}{\sqrt{b^2}}=\dfrac{3+2a}{b}\)
d, \(\left(a-b\right).\sqrt{\dfrac{ab}{\left(a-b\right)^2}}=\left(a-b\right).\dfrac{\sqrt{ab}}{\sqrt{\left(a-b\right)^2}}=\left(a-b\right).\dfrac{\sqrt{ab}}{\left(a-b\right)}=\sqrt{ab}\)
13 tháng 5 2021
a) ab2.√3a2b4=ab2.√3√a2b4ab2.3a2b4=ab2.3a2b4
=ab2.√3√a2.√b4=ab2.√3|a|.|b2|=ab2.3a2.b4=ab2.3|a|.|b2|
=ab2.√3
13 tháng 5 2021
a) ab2.√3a2b4=ab2.√3√a2b4ab2.3a2b4=ab2.3a2b4
=ab2.√3√a2.√b4=ab2.√3|a|.|b2|=ab2.3a2.b4=ab2.3|a|.|b2|
=ab2.√3
8 tháng 4 2021
a, \(A=\left(\sqrt{12}-2\sqrt{5}\right)\sqrt{3}+\sqrt{60}\)
\(=\left(2\sqrt{3}-2\sqrt{5}\right)\sqrt{3}+2\sqrt{15}\)
\(=2\sqrt{9}-2\sqrt{15}+2\sqrt{15}=2\sqrt{9}\)
b, \(B=\frac{\sqrt{4x}}{x-3}\sqrt{\frac{x^2-6x+9}{x}}=\frac{2\sqrt{x}}{x-3}.\sqrt{\frac{\left(x-3\right)^2}{x}}\)
\(=\frac{2\sqrt{x}}{x-3}.\frac{x-3}{\sqrt{x}}=2\)
AH
Akai Haruma
Giáo viên
3 tháng 3 2020
Lời giải:
\(x=\frac{1}{a}\sqrt{\frac{2a-b}{b}}\Rightarrow ax=\sqrt{\frac{2a-b}{b}}\)
\(\Rightarrow 1+ax=\frac{\sqrt{2a-b}+\sqrt{b}}{\sqrt{b}}; 1-ax=\frac{\sqrt{b}-\sqrt{2a-b}}{\sqrt{b}}\)
\(\Rightarrow \frac{1-ax}{1+ax}=\frac{\sqrt{b}-\sqrt{2a-b}}{\sqrt{b}+\sqrt{2a-b}}=\frac{(\sqrt{b}-\sqrt{2a-b})^2}{2(b-a)}\)
Lại có:
\(\frac{1+bx}{1-bx}=\frac{a+\sqrt{2ab-b^2}}{a-\sqrt{2ab-b^2}}=\frac{a^2-(2ab-b^2)}{(a-\sqrt{2ab-b^2})^2}=\frac{(a-b)^2}{(a-\sqrt{2ab-b^2})^2}\)
\(\Rightarrow \sqrt{\frac{1+bx}{1-bx}}=\frac{b-a}{a-\sqrt{2ab-b^2}}\)
Do đó:
$A=\frac{(\sqrt{b}-\sqrt{2a-b})^2}{2a-2\sqrt{2ab-b^2}}=\frac{2a-2\sqrt{2ab-b^2}}{2a-2\sqrt{2ab-b^2}}=1$
(vì a < b < 0 và b < 0 nên |a - b| = -(a - b), ab > 0)