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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, \(\left(a-b\right).\sqrt{\dfrac{ab}{\left(a-b\right)^2}}=\left(a-b\right).\dfrac{\sqrt{ab}}{\sqrt{\left(a-b\right)^2}}=\left(a-b\right).\dfrac{\sqrt{ab}}{\left(a-b\right)}=\sqrt{ab}\)
a)\(\sqrt{\dfrac{9+12a+4a^2}{b^2}}=\sqrt{\dfrac{\left(2a+3\right)^2}{b^2}}=\dfrac{\left|2a+3\right|}{\left|b\right|}=\dfrac{-\left(2a+3\right)}{b}\)
b) \(\left(a-b\right).\sqrt{\dfrac{ab}{\left(a-b\right)^2}}\)
\(\Leftrightarrow\left(a-b\right).\dfrac{\left|ab\right|}{\left|a-b\right|}=-ab\)
c,\(\left(\frac{\sqrt{1+a}}{\sqrt{1+a}-\sqrt{1-a}}+\frac{1-a}{\sqrt{1-a^2}-1+a}\right)\left(\sqrt{\frac{1}{a^2}-1}-\frac{1}{a}\right)\)
\(=\left(\frac{\sqrt{1+a}}{\sqrt{1+a}-\sqrt{1-a}}+\frac{\sqrt{1-a}.\sqrt{1-a}}{\sqrt{1-a}\left(\sqrt{1+a}-\sqrt{1-a}\right)}\right)\left(\frac{\sqrt{1-a^2}-1}{a}\right)\)
\(=\frac{\left(\sqrt{1+a}+\sqrt{1-a}\right)^2}{\left(1+a\right)-\left(1-a\right)}.\frac{\left(\sqrt{1-a^2}-1\right)}{a}=-1\)
M chỉ làm tiếp thôi nha, ko chép lại đề với đk đâu
a,
\(=\frac{a+2\sqrt{ab}+b-4\sqrt{ab}}{\sqrt{a}-\sqrt{b}}-\)\(\frac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\frac{a-2\sqrt{ab}+b}{\sqrt{a}-\sqrt{b}}-\left(\sqrt{a}-\sqrt{b}\right)\)
\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}-\sqrt{a}+\sqrt{b}\)
\(=\sqrt{a}-\sqrt{b}-\sqrt{a}+\sqrt{b}\)
\(=0\)
b,
\(=\left(a-b\right)\left(\sqrt{\frac{a+b}{a-b}}-1\right)\left(a-b\right)\left(\sqrt{\frac{a+b}{a-b}}+1\right)\)
\(=\left(a-b\right)^2\left(\frac{a+b}{a-b}-1\right)\)
\(=\left(a-b\right)^2\cdot\frac{a+b-a+b}{a-b}\)
\(=\left(a-b\right)2b=2ab-2b^2\)
\(\frac{\sqrt{9+12a+4a^2}}{\sqrt{b^2}}\)
\(=\frac{\sqrt{\left(2a+3\right)^2}}{\sqrt{b^2}}\)
\(=\frac{2a+3}{-b}\)( theo điều kiện )
b: \(=\left|b\cdot\left(b-1\right)\right|=b\cdot\left|b-1\right|\)
c: \(=\left|a\right|\cdot\left|a+1\right|=a\left(a+1\right)=a^2+a\)
d: \(=1-2a-4a=-6a+1\)
Lời giải:
a)
\(\sqrt{36(b-2)^2}=\sqrt{6^2(b-2)^2}=6\sqrt{(b-2)^2}=6|b-2|=6(2-b)\) do \(b<2\)
b)
\(\sqrt{b^2(b-1)^2}=\sqrt{b^2}\sqrt{(b-1)^2}=|b||b-1|\)
Do \(b< 0\Rightarrow b,b-1< 0\)
\(\Rightarrow \sqrt{b^2(b-1)^2}=|b||b-1|=-b(1-b)=b(b-1)\)
c) \(\sqrt{a^2(a+1)^2}=\sqrt{a^2}\sqrt{(a+1)^2}=|a||a+1|\)
\(=a(a+1)\) do \(a>0\)
d) \(\sqrt{(2a-1)^2}-4a=|2a-1|-4a\)
Vì \(a< \frac{1}{2}\Rightarrow 2a-1< 0\)
\(\Rightarrow \sqrt{(2a-1)^2}-4a=|2a-1|-4a=(1-2a)-4a=1-6a\)
a/ \(\frac{2}{a}.\frac{4\left|a\right|}{3}=\frac{-8a}{3a}=-\frac{8}{3}\)
b/ \(\frac{3}{a-1}\sqrt{\frac{4\left(a-1\right)^2}{25}}=\frac{3}{\left(a-1\right)}.\frac{2\left|a-1\right|}{5}=\frac{6\left(a-1\right)}{5\left(a-1\right)}=\frac{6}{5}\)
c/ \(\frac{3\sqrt{9a^2b^4}}{\sqrt{a^2b^2}}=\frac{9.\left|a\right|.b^2}{\left|a\right|\left|b\right|}=9\left|b\right|\)
d/ \(\left(1+\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\left(1-\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)=1-a\)
a/ \(=\frac{2}{a}.\frac{4\left|a\right|}{3}=\frac{2}{a}.\frac{-4a}{3}=\frac{-8}{3}\)
b/ \(=\frac{3}{a-1}.\frac{\left|2a-2\right|}{5}=\frac{3}{a-1}.\frac{2\left(a-1\right)}{5}=\frac{6}{5}\)
c/ \(=\sqrt{\frac{162a^2b^4}{2a^2b^2}}=\sqrt{81b^2}=9\left|b\right|\)
d/ \(=\left(1+\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\left(1-\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)\)
\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)=1-a\)
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 $a-b< 0\rightarrow |a-b|=b-a$
\(\Rightarrow (a-b)\sqrt{\frac{ab}{(a-b)^2}}=(a-b).\frac{\sqrt{ab}}{|a-b|}=(a-b).\frac{\sqrt{ab}}{b-a}=-\sqrt{ab}\)