Min_A = 29 \(\Leftrightarrow x=0\)

Min _B= 625 \(\Leftrightarrow x=\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)

làm rất ngu

ngu như t

Tu \(-\sqrt{30}\) den \(\sqrt{30}\) co 5 so nguyen chia het cho 2 la -4;-2;0;2;4

Tu \(\sqrt{5}\) den \(\sqrt{60}\) co 2 so nguyen chia het cho 3 la 3;6

Tu $-\sqrt{30}$ den $\sqrt{30}$

 co 5 so nguyen chia het cho 2 la -4;-2;0;2;4

Tu $\sqrt{5}$

 den $\sqrt{60}$ co 2 so nguyen chia het cho 3 la 3;6

\(A\left(\sqrt{2};\sqrt{2}\right)\Rightarrow x=\sqrt{2};y=\sqrt{2}\) Thay vào hàm số \(y=\left(\sqrt{a}-2\right)x\) ta được :

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

\(\Rightarrow\sqrt{a}-2=1\)

\(\Rightarrow\sqrt{a}=3\)

\(\Rightarrow a=9\)

Vậy \(a=9\)

a,>

b,vô lí

c,>

d,>

e<

a) 26 lớn hơn 5

b) -4 nhỏ hơn (-2)^2

c) a+b lớn hơn a+b

d)9.16 lớn hơn 9.16

e)12+20+3042 nhỏ hơn 20

Câu a)
\(A=\sqrt{20+1}+\sqrt{40+2}+\sqrt{60+3}\)
\(=\sqrt{1\left(20+1\right)}+\sqrt{2\left(20+1\right)}+\sqrt{3\left(20+1\right)}\)
\(=\sqrt{20+1}\left(\sqrt{1}+\sqrt{2}+\sqrt{3}\right)\)

\(B=\sqrt{1}+\sqrt{2}+\sqrt{3}+\sqrt{20}+\sqrt{40}+\sqrt{60}\)
\(=1\left(\sqrt{1}+\sqrt{2}+\sqrt{3}\right)+\left(\sqrt{1}\cdot\sqrt{20}+\sqrt{2}\cdot\sqrt{20}+\sqrt{3}\cdot\sqrt{20}\right)\)
\(=\sqrt{1}\left(\sqrt{1}+\sqrt{2}+\sqrt{3}\right)+\sqrt{20}\left(\sqrt{1}+\sqrt{2}+\sqrt{3}\right)\)
\(=\left(\sqrt{20}+\sqrt{1}\right)\left(\sqrt{1}+\sqrt{2}+\sqrt{3}\right)\)

Ta thấy: \(\hept{\begin{cases}\left(\sqrt{20+1}\right)^2=20+1\\\left(\sqrt{20}+\sqrt{1}\right)^2=20+1+2\sqrt{20}\end{cases}}\)
\(\Rightarrow\left(\sqrt{20+1}\right)^2< \left(\sqrt{20}+\sqrt{1}\right)^2\Rightarrow\sqrt{20+1}< \sqrt{20}+\sqrt{1}\)
Vậy A < B.

a) A<B

A= ( \(\sqrt{1}\)+\(\sqrt{2}\)+\(\sqrt{3}\) ) + (\(\sqrt{20}\) + \(\sqrt{40}\) + \(\sqrt{60}\))

= (1+1,4+1,7)+(4,4+6,3+7,7)

= 4,1+18,4

=22,5