a) \(\sqrt{3a^3}\cdot\sqrt{12a}=\sqrt{3a^3\cdot12a}=\sqrt{36a^4}=6a^2\)

b) \(\sqrt{2a\cdot32ab^2}=\sqrt{64a^2b^2}=8ab\)

\(\sqrt{20}\cdot\sqrt{72}\cdot\sqrt{4,9}=\sqrt{20\cdot72\cdot4,9}=\sqrt{2\cdot10\cdot72\cdot4,9}\\ =\sqrt{144\cdot49}=\sqrt{144}\cdot\sqrt{49}=12\cdot7=84\)

Bài 2:

a) \(\sqrt{3a^3}\cdot\sqrt{12a}=\sqrt{3a^3\cdot12a}=\sqrt{36a^4}=6a^2\)

b) \(\sqrt{2a\cdot32ab^2}=\sqrt{64a^2b^2}=8ab\)

\(P=\dfrac{9\sqrt{a}-\sqrt{25a}+\sqrt{4a^3}}{a^2+2a}=\dfrac{9\sqrt{a}-5\sqrt{a}+2a\sqrt{a}}{a\left(a+2\right)}=\dfrac{4\sqrt{a}+2a\sqrt{a}}{a\left(a+2\right)}=\dfrac{2\sqrt{a}\left(2+a\right)}{a\left(2+a\right)}=\dfrac{2\sqrt{a}}{a}=\dfrac{2.\sqrt{a}}{\sqrt{a}.\sqrt{a}}=\dfrac{2}{\sqrt{a}}\)

Lời giải:

\(A=\frac{2a^2+4}{(1-a)(1+a)}-\frac{1-\sqrt{a}+1+\sqrt{a}}{(1+\sqrt{a})(1-\sqrt{a})}=\frac{2a^2+4}{(1-a)(1+a)}-\frac{2}{1-a}\)

\(=\frac{2a^2+4}{(1-a)(1+a)}-\frac{2(1+a)}{(1-a)(1+a)}=\frac{2a^2-2a+2}{(1-a)(1+a)}=\frac{2(a^2-a+1)}{1-a^2}\)

\(A=\sqrt{1-4a+4a^2}-2a=\sqrt{\left(1-2a\right)^2}-2a=\left|1-2a\right|-2a\)

Nếu  \(a\le\frac{1}{2}\)thì:  \(A=1-2a-2a=1-4a\)

Nếu  \(a>\frac{1}{2}\)thì:  \(A=2a-1-2a=-1\)

ta có:\(\sqrt{\left(1-2a\right)^2}-2a=|1-2a|-2a\)

th1:neu 1-2a <0 <=>1<2a<=>1/2<a:

l1-2al=2a-1

=>2a-1-2a=-1

th2:neu 1-2a>=0=>1>=2a=>1/2>a ta co:

l1-2al=1-2a

=>1-2a-2a=1-4a

A=\(\frac{\sqrt{a}\left(a\sqrt{a}+1\right)}{a-\sqrt{a}+1}\) \(-\frac{\sqrt{a}\left(2\sqrt{a}+1\right)}{\sqrt{a}}+1\) (dk \(a\ge0\)

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

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

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

\(A=1+"\frac{2a+\sqrt{a}-1}{1-a}-\frac{2a\sqrt{a}-\sqrt{a}+a}{1-a\sqrt{a}}"\times\frac{a-\sqrt{a}}{2\sqrt{a}-1}=\)

\(A="\frac{1a+\sqrt{a}-1}{1-a}-\frac{1a\sqrt{a}-\sqrt{a}+a}{1-a\sqrt{a}}"\times\frac{a-\sqrt{a}}{1\sqrt{a}-1}\)

P/s: Ko chắc đâu nhé 

Đọc tiếp

.......

\(M=\frac{2\sqrt{a}\left(\sqrt{a}+\sqrt{2a}-\sqrt{3b}\right)+\sqrt{3b}\left(2\sqrt{a}-\sqrt{3b}\right)-2a\sqrt{a}}{a\sqrt{2}+\sqrt{3ab}}\left(đkxđ:a,b\ge0;mau\ne0\right)\)[tự tìm cái sau :)) ]

\(VP=\frac{2\sqrt{a}\left(\sqrt{a}+\sqrt{2}.\sqrt{a}-\sqrt{3}.\sqrt{b}\right)}{a\sqrt{2}+\sqrt{3ab}}+\frac{\sqrt{3b}\left(2\sqrt{a}-\sqrt{3b}\right)}{a\sqrt{2}+\sqrt{3ab}}-\frac{2a\sqrt{a}}{a\sqrt{2}+\sqrt{3ab}}\)

\(=\frac{2a+2a\sqrt{2}-2\sqrt{3ab}}{a\sqrt{2}+\sqrt{3ab}}+\frac{2\sqrt{3ab}-3b}{a\sqrt{2}+\sqrt{3ab}}-\frac{2a\sqrt{a}}{a\sqrt{2}+\sqrt{3ab}}\)

\(=\frac{2a+2a\sqrt{2}-3b+2a\sqrt{a}}{a\sqrt{2}+\sqrt{3ab}}\)

mình làm được đến đây , bạn làm được tiếp thì làm =))

M=\(M=6\sqrt{B\hept{\begin{cases}\\\end{cases}}3,6}\)