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a)\(\)https://www.cymath.com/answer?q=2sqrt(27)-6sqrt(4%2F3)%2B3%2F5sqrt(75)
\(M=2\sqrt{27}-6\sqrt{\frac{4}{3}}+\frac{3}{5}\sqrt{75}=2\sqrt{3^2.3}-6\sqrt{\frac{2^2.3}{3^2}}+\frac{3}{5}\sqrt{5^2.3}=.\)
\(=6\sqrt{3}-4\sqrt{3}+3\sqrt{3}=5\sqrt{3}\)
\(P=\frac{2}{x-1}\sqrt{\frac{x^2-2x+1}{4x^2}}.Với...0< x< 1\Leftrightarrow\) \(P=\frac{2}{x-1}\sqrt{\frac{\left(x-1\right)^2}{\left(2x\right)^2}}=\frac{2}{(x-1)}.\frac{\left(1-x\right)}{2x}=\frac{-1}{x}.\)
a) \(\sqrt{4\left(a-3\right)^2}=\sqrt{2^2\left(a-3\right)^2}=2\sqrt{\left(a-3\right)^2}=2.\left|a-3\right|=2\left(a-3\right)=2a-6\) (Vì \(a\ge3\) )
b) \(\sqrt{9\left(b-2\right)^2}=\sqrt{3^2\left(b-2\right)^2}=3\sqrt{\left(b-2\right)^2}=3\left|b-2\right|=3\left(2-b\right)\)
\(=6-3b\) (vì b < 2 )
b) \(\sqrt{27.48\left(1-a\right)^2}=\sqrt{27.3.16.\left(1-a\right)^2}=\sqrt{81.16.\left(1-a\right)^2}\)
\(=\sqrt{9^2.4^2.\left(1-a\right)^2}=9.4\sqrt{\left(1-a\right)^2}=36.\left|1-a\right|=36\left(1-a\right)=36-36a\) (vì a > 1)
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) \(A=\sqrt{11+6\sqrt{2}}-3+\sqrt{2}=\sqrt{9+2.3\sqrt{2}+2}-3+\sqrt{2}\)
\(=\sqrt{\left(3+\sqrt{2}\right)^2}-3+\sqrt{2}=3+\sqrt{2}-3+\sqrt{2}=2\sqrt{2}\)
b) x<0
\(B=\sqrt{9x^2}-2x=\left|3x\right|-2x=-3x-2x=-5x\)
c) x>4
\(C=x-4+\sqrt{16-8x+x^2}=x-4+\sqrt{\left(4-x\right)^2}\)
\(=x-4+\left|4-x\right|=x-4+x-4=2x-8\)
a) \(\frac{b-16}{4-\sqrt{b}}\left(b\ge0,b\ne16\right)\)
\(=\frac{\left(\sqrt{b}-4\right)\left(\sqrt{b}+4\right)}{4-\sqrt{b}}\)
\(=-\sqrt{b}-4\)
b) \(\frac{a-4\sqrt{a}+4}{a-4}\left(a\ge0;a\ne4\right)\)
\(=\frac{a-2.\sqrt{a}.2+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\)
\(=\frac{\left(\sqrt{a}-2\right)^2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}=\frac{\sqrt{a}-2}{\sqrt{a}+2}\)
c) \(2x+\sqrt{1+4x^2-4x}\) với \(x\le\frac{1}{2}\)
\(=2x+\sqrt{\left(1-2x\right)^2}\)
\(=2x+\left|1-2x\right|=2x+1-2x=1\)
d) \(\frac{4a-4b}{\sqrt{a}-\sqrt{b}}\left(a,b\ge0;a\ne b\right)\)
\(=\frac{4\left(a-b\right)}{\sqrt{a}-\sqrt{b}}=\frac{4\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\)
\(=4\left(\sqrt{a}+\sqrt{b}\right)\)
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) \(\sqrt{4a^2}=2\left|a\right|=-2a\) ( do a<0)
b) \(\sqrt{4x^2-12x+9}=\sqrt{\left(2x-3\right)^2}=\left|2x-3\right|=3-2x\)(do \(x< \dfrac{3}{2}\Leftrightarrow2x-3< 0\))