Ta có x=\(\frac{30-2\sqrt{45}}{4}< \frac{30-2\sqrt{49}}{4}\)

  \(\Leftrightarrow x=\frac{30-2\sqrt{45}}{4}< \frac{30-14}{4}< 4\)

Ta có x<4       (1)

lại có y=\(\sqrt{17}>\sqrt{16}\Rightarrow\sqrt{17}>4\)

=> y>4          (2)

từ (1) và (2) =>x<y

Ta có : x = \(\frac{30-2\sqrt{45}}{4}\)= \(\frac{15-\sqrt{45}}{2}\)> 0

           y = \(\sqrt{17}>0\)

\(\Rightarrow\)\(x^2\)= \(\frac{\left(15-\sqrt{45}\right)^2}{4}\)= \(\frac{225-30\sqrt{45}+45}{4}\)= \(\frac{270-30\sqrt{45}}{4}\)

        \(y^2\)= 17

Xét hiệu : \(x^2-y^2\)= \(\frac{270-30\sqrt{45}}{4}\)\(-\)17 = \(\frac{202-30\sqrt{45}}{4}\)= \(\frac{\sqrt{40804}-\sqrt{40500}}{4}>0\)

              ( vì 40804\(>\)40500 \(\ge\)0 )

\(\Rightarrow\)\(x^2>y^2\)\(\Rightarrow\)\(x>y\) ( vì \(x,y>0\))

Giả sử

\(\frac{30-2\sqrt{45}}{4}>\sqrt{17}\)

\(\Leftrightarrow15>2\sqrt{17}+\sqrt{45}\)

\(\Leftrightarrow225>113+4\sqrt{765}\)

\(\Leftrightarrow28>\sqrt{765}\)

\(\Leftrightarrow784>765\) (đúng)

Vậy \(\frac{30-2\sqrt{45}}{4}>\sqrt{17}\)

Giả sử:

\(\frac{30-2\sqrt{45}}{4}>\sqrt{17}\)

\(\Leftrightarrow15>2\sqrt{17}+\sqrt{45}\)

\(\Leftrightarrow225>113+4\sqrt{765}\)

\(\Leftrightarrow28>\sqrt{765}\)

\(\Leftrightarrow784>765\)(đúng)

Vậy \(\frac{30-2\sqrt{45}}{4}>\sqrt{17}\)

a] < b] < c] >

a/ \(\sqrt{17}+\sqrt{5}+1>\sqrt{16}+\sqrt{4}+1=4+2+1=7\)

\(\sqrt{45}< \sqrt{49}=7\)

\(\Rightarrow\sqrt{17}+\sqrt{5}+1>\sqrt{45}\)

b/ Ta có:

\(\sqrt{n}< \sqrt{n+1}\)

\(\Rightarrow2\sqrt{n}< \sqrt{n+1}+\sqrt{n}\)

\(\Rightarrow\dfrac{1}{\sqrt{n}}>\dfrac{2}{\sqrt{n+1}+\sqrt{n}}=2\left(\sqrt{n+1}-\sqrt{n}\right)\)

Áp dụng vào bài toán được

\(1+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{36}}>2\left(\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{37}-\sqrt{36}\right)\)

\(=2\left(\sqrt{37}-1\right)>6\)

mình chỉ giải được phần này thôi

b.A = \(\sqrt{17}\)+\(\sqrt{26}\)+ 1 > \(\sqrt{16}\)+\(\sqrt{25}\)+ 1 = 4 + 5 +1 = 10

B = \(\sqrt{99}\)<\(\sqrt{100}\)= 10

=> A > B

Ta có :

1) 45^10 . 5^30= (5.9)^10 . 5^30 = 5^10 . 5^30 . 9^10 = 5^40 . 3^20 = 25^20 . 3^20=75^20

2)\(\sqrt{40+2}=\sqrt{42}<\sqrt{49}=7=6+1=\sqrt{36}+\sqrt{1}<\sqrt{40}+\sqrt{2}\)

Vậy \(\sqrt{40+2}<\sqrt{40}+\sqrt{2}\)

3)\(Cho\frac{x}{3}=\frac{y}{4}=k\Rightarrow x=3k;y=4k\)

Ta lại có:

\(xy=12\Rightarrow3k.4k=12\)

\(12.k^2=12\Rightarrow k^2=1\Rightarrow k=1:-1\)

\(Vơik=1\Rightarrow x=1.3=3;y=1.4=4\)

\(k=-1\Rightarrow x=-1.3=-3;y=-1.4=-4\)