xét 2  hiệu sau

(\(3\sqrt{2}-1\))²-(2\(\sqrt{3}\))²=(19-6\(\sqrt{2}\))-12=7-6\(\sqrt{2}\)=18-(11+6\(\sqrt{2}\)) = (3\(\sqrt{2}\))²-(3+\(\sqrt{2}\))² <0

(vì \(3\sqrt{2}\)<3+\(\sqrt{2}\) <=>2\(\sqrt{2}\)<3 <=>8<9 đúng)

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

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

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

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

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

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

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

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

= \(3\sqrt{2}+12\sqrt{2}\)

= \(15\sqrt{2}\)

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

`Answer:`

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

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

\(=2\sqrt{3}+2-\sqrt{3}\) (Do `2>\sqrt{3}`)

\(=\sqrt{3}+2\)

\(\sqrt{4}+2\sqrt{3}-\sqrt{13}+4\sqrt{3}\approx8.8\)