`2 sqrt 3 = sqrt(3. 2^2) = sqrt 12.`

`3 sqrt 2 = sqrt(2 . 3^2) = sqrt 18`.

Vì `18 > 12 => sqrt 18 > sqrt 12`.

Nếu đề là `-2 sqrt 3` thì đơn giản lắm luôn, bạn so sánh 2 phân số với không thì dễ dàng thấy `-2 sqrt 3 < 0` và `3 sqrt 2 > 0` nên `-2 sqrt 3 < 3 sqrt 2`.

Theo bà ra ta có : 

\(f\left(2\sqrt{3}\right)=\left(m+1\right)x-2=\left(m+1\right)\left(2\sqrt{3}\right)-2\)

\(=\sqrt{12}\left(m+1\right)-2\)

\(f\left(3\sqrt{2}\right)=\left(m+1\right)x-2=\left(m+1\right)3\sqrt{2}-2\)

\(=\sqrt{18}\left(m+1\right)-2\)

vì 12 < 18 => \(\sqrt{12}< \sqrt{18}\)

hay \(f\left(2\sqrt{3}\right)< f\left(3\sqrt{2}\right)\)

Bài 1:

b=15cm nên AC=15cm

\(\widehat{B}=90^0-42^0=48^0\)

Xét ΔABC vuông tại A có 

\(\sin B=\dfrac{AC}{BC}\)

nên \(BC=15:\sin48^0\simeq20.18\left(cm\right)\)

\(AB=\sqrt{BC^2-AB^2}=13.50\left(cm\right)\)

a: ĐKXĐ: x-10>=0

=>x>=10

b: \(\sqrt{9a^2b}=\sqrt{\left(3a\right)^2\cdot b}=3a\cdot\sqrt{b}\)

c: \(\left(2\sqrt{3}+1\right)^2=13+4\sqrt{3}\)

\(\left(2\sqrt{2}+\sqrt{5}\right)^2=8+5+2\cdot2\sqrt{2}\cdot\sqrt{5}=13+4\sqrt{10}\)

mà \(4\sqrt{3}< 4\sqrt{10}\left(3< 10\right)\)

nên \(\left(2\sqrt{3}+1\right)^2< \left(2\sqrt{2}+\sqrt{5}\right)^2\)

=>\(2\sqrt{3}+1< 2\sqrt{2}+\sqrt{5}\)

tổng quát nhé \(\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}=\frac{1}{\sqrt{\left(k+1\right)k}}>\frac{1}{\left(k+1\right)\sqrt{k}}>\frac{1}{\left(k+1\right)k}=\frac{1}{k}-\frac{1}{k+1}\)
Đặt A= \(\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+...+\frac{1}{2011\sqrt{2010}}\)
\(\Rightarrow1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2010}}-\frac{1}{\sqrt{2011}}>A>1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2010}-\frac{1}{2011}\)\(\Leftrightarrow1-\frac{1}{\sqrt{2011}}>A>1-\frac{1}{2011}\Leftrightarrow\frac{88}{45}>\frac{2011-\sqrt{2011}}{2011}>A>\frac{2010}{2011}>\frac{87}{89}\)

ban chung minh tong quat ro hon ko

\(A=2\sqrt{6}\)

\(B=2\sqrt{4}=4\)

\(C=2\sqrt{7}\)

`3\sqrt2+4\sqrt{18}`
`=3\sqrt2+4.\sqrt{9.2}`
`=3\sqrt2+12\sqrt2`
`=15\sqrt2`

\(3\sqrt{2}+4\sqrt{18}\)

= \(3\sqrt{2}+12\sqrt{2}\)

= \(15\sqrt{2}\)

Bài 1 :

\(A=\sqrt{4-2\sqrt{3}}+\sqrt{27}\)

\(=\sqrt{3-2\sqrt{3}+1}+\sqrt{27}\)

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

\(=\left|\sqrt{3}-1\right|+3\sqrt{3}\)

\(=\sqrt{3}-1+3\sqrt{3}\)

\(=4\sqrt{3}-1\)

\(B=\sqrt{14-6\sqrt{5}}+\sqrt{125}\)

\(=\sqrt{9-6\sqrt{5}+5}+\sqrt{125}\)

\(=\sqrt{\left(3-\sqrt{5}\right)}^2+5\sqrt{5}\)

\(=\left|3-\sqrt{5}\right|+5\sqrt{5}\)

\(=3-\sqrt{5}+5\sqrt{5}\)

\(=3+4\sqrt{5}\)