a)\(\sqrt{45.80}=\sqrt{9.400}=\sqrt{9}.\sqrt{400}=3.20=60\)

b) \(\sqrt{75.48}=\sqrt{25.3.16.3}=\sqrt{5^2.3^2.4^2}=5.4.3=60\)

c)\(\sqrt{90.6,4}=\sqrt{10.9.4.1,6}=\sqrt{4^2.3^2.2^2}=4.3.2=24\)

d) \(\sqrt{2,5.14,4}=\sqrt{\dfrac{25}{10}.\dfrac{144}{10}}=\sqrt{\dfrac{25.144}{100}}=\sqrt{\left(\dfrac{5.12}{10}\right)^2}=\dfrac{5.12}{10}=6\)

a) \(\sqrt{45.80}=\sqrt{9.400}=\sqrt{9}.\sqrt{400}=3.20=60\)

b)\(\sqrt{75.48}=\sqrt{25.3.3.16}=5.3.4=60\)

c)\(\sqrt{90.6,4}=\sqrt{9.64}=3.8=24\)

d)\(\sqrt{2,5.14,4}=\sqrt{\dfrac{25}{10}.\dfrac{144}{10}}=\sqrt{\dfrac{25.144}{100}=\dfrac{5.12}{10}=\dfrac{60}{10}=6}\)

a)\(\sqrt{75\cdot48}=\sqrt{25\cdot3\cdot48}=\sqrt{25\cdot144}=\sqrt{25}\cdot\sqrt{144}=5\cdot12=60\)

b) \(\sqrt{2,5\cdot14,4}=\sqrt{25\cdot144\cdot\frac{1}{100}}=\sqrt{25}\cdot\sqrt{144}\cdot\sqrt{\frac{1}{100}}=5\cdot12\cdot\frac{1}{10}=6\)

Câu 1 nè

\(\sqrt{11}+6\sqrt{2}-3+\sqrt{2}=\sqrt{11}+7\sqrt{2}-3.\)

Câu 2 nè :

a) đề không rõ.

b) \(\sqrt{75.48}=\sqrt{25.16.3^2}=5.4.3=60\)

a)  45 . 80

= 5 . 9 . 5 . 16 = 5 2 . 3 2 . 4 2 = 5 . 3 . 4 = 60

Đây là đề chứng minh hả !

Phần a , b đúng r 

Nhưng phần b chỗ \(\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)=\left(\sqrt{a}\right)^2-\left(\sqrt{b}\right)^2\) = a - b 

Dùng hằng đẳng thức thức 3 như vậy sẽ hay hơn !

Chúc bạn học tốt!

nhưng bn làm đúng rùi mà

Đăt\(\sqrt{a}\)=x, \(\sqrt{b}\)=y (x,y>0)
=>xy+1=4y => 4y≥ \(2\sqrt{xy}\)=>\(2\sqrt{y}\)≥\(\sqrt{x}\)=> 4y≥x=> 4≥ \(\dfrac{x}{y}\)=> \(\dfrac{1}{4}\)≤\(\dfrac{y}{x}\)=>\(\dfrac{-1}{4}\)≥\(\dfrac{-y}{x}\)
Xét:A=(\(\dfrac{xy+y}{x+y}\)+\(\dfrac{xy+x}{y-x}\)+1):(\(\dfrac{xy+y}{x+y}\)+\(\dfrac{xy+x}{x-y}\)-1)
         = \(\dfrac{-2y^2\left(x+1\right)}{\left(x-y\right)\left(x+y\right)}\).\(\dfrac{\left(x-y\right)\left(x+y\right)}{2xy\left(x+1\right)}\)
=> A= \(\dfrac{-y}{x}\)≤\(\dfrac{-1}{4}\)
Dấu "=" xảy ra <=> xy=1 và x=4y <=> x=2, y=\(\dfrac{1}{2}\) <=> a =4, b=\(\dfrac{1}{4}\)

Vậy Max A =\(\dfrac{-1}{4}\) <=> a=4, b=\(\dfrac{1}{4}\)

B