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a/ \(\sqrt{4a^4-12a^2+9}-\sqrt{a^4-8a^2+16}\)
= \(\sqrt{\left(2a^2-3\right)^2}-\sqrt{\left(a^2-4\right)^2}\)
= \(|2a^2-3|-|a^2-4|\)
= \(2a^2-3+a^2-4\)
= \(3a^2-7\)
Thay a=\(\sqrt{3}\).Ta có:
\(3.\left(\sqrt{3}\right)^2-7\)
= 3.3-7=2
b/ \(\sqrt{10a^2-12a\sqrt{10}+36}\)
= \(\sqrt{\left(a\sqrt{10}\right)^2-2.a\sqrt{10}.6+6^2}\)
= \(\sqrt{\left(a\sqrt{10}-6\right)^2}\)
= \(|a\sqrt{10}-6|\)
= \(-a\sqrt{10}+6\)
Thay a= \(\sqrt{\frac{5}{2}}-\sqrt{\frac{2}{5}}\)=\(\frac{3}{\sqrt{10}}\),Ta có:
\(-\frac{3}{\sqrt{10}}.\sqrt{10}+6\)
= -3+6 =3
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/ \(=\left(1+\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\left(1-\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)\)
\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)=1-a\)
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) \(\left(a-b\right).\sqrt{\dfrac{ab}{\left(a-b\right)^2}}\)
\(\Leftrightarrow\left(a-b\right).\dfrac{\left|ab\right|}{\left|a-b\right|}=-ab\)
\(\frac{\sqrt{9+12a+4a^2}}{\sqrt{b^2}}\)
\(=\frac{\sqrt{\left(2a+3\right)^2}}{\sqrt{b^2}}\)
\(=\frac{2a+3}{-b}\)( theo điều kiện )
tách 11 ra thành \(\sqrt{3}\) mũ 2 + căn 8 mũ 2
áp dụng hẳng đẳng thức đáng nhớ A^2+2AB +B^2=(A+B)^2
vào \(\sqrt{11+4\sqrt{6}}\)
.Bản thử đi nhé kết quả của mình là \(\sqrt{3}\)+\(\sqrt{8}\)
Vì ko gõ đc căn nên mình ko giải hẳn hoi ra đc .Bạn thông cảm ha.
Chúc bn hok tốt!
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) \(3\sqrt{a^2-4a+4}=3\sqrt{\left(a-2\right)^2}=3\left|a-2\right|=3\left(a-2\right)\) (vì \(a\ge2\))
b) \(2\sqrt{9a^2+12a+4}=2\sqrt{\left(3a+2\right)^2}=2\left|3a+2\right|=2\left(-3a-2\right)=-2\left(3a+2\right)\) (vì \(a< -\frac{2}{3}\))