\(A=\sqrt[]{50}+\sqrt[]{65}\Rightarrow A^2=50+65+2\sqrt[]{50.65}=115+2\sqrt[]{5.10.5.13=}115+10\sqrt[]{130}\left(1\right)\)

\(B=\sqrt[]{15}+\sqrt[]{115}\Rightarrow B^2=15+115+2\sqrt[]{15.115}=15+115+2\sqrt[]{3.5.5.23}=15+115+10\sqrt[]{69}\left(2\right)\)Ta có  \(10\sqrt[]{130}< 10\sqrt[]{69.2}=10\sqrt[]{2}\sqrt[]{69}< 15+10\sqrt[]{69}\left(3\right)\)

\(\left(1\right),\left(2\right),\left(3\right)\Rightarrow A^2< B^2\Rightarrow A< B\)

\(\Rightarrow\sqrt[]{50}+\sqrt[]{65}< \sqrt[]{15}+\sqrt[]{115}\)

So sánh gì thế em, em nhập đủ đề vào hi

\(\sqrt{8}+\sqrt{15}< \sqrt{9}+\sqrt{16}=3+4=7\)

\(\sqrt{65}-1>\sqrt{64}-1=8-1=7\)

Vậy \(\sqrt{8}+\sqrt{15}< \sqrt{65}-1\)

a) \(\sqrt{8}+\sqrt{15}< \sqrt{9}+\sqrt{16}=3+4=7\)

\(\sqrt{65}-1>\sqrt{64}-1=8-1=7\)

\(\Rightarrow\sqrt{8}+\sqrt{15}< \sqrt{65}-1\)

b) \(\frac{13-2\sqrt{3}}{6}>\frac{13-2\sqrt{4}}{6}=1,5\)

mà 1,5² = 2,25 ; \(\sqrt{2}^2=2\)

\(\Rightarrow1,5>\sqrt{2}\)hay \(\frac{13-2\sqrt{3}}{6}>\sqrt{2}\)

Ta có: \(\sqrt{8}+\sqrt{15}< \sqrt{9}+\sqrt{16}=3+4=7\) (1)

\(\sqrt{65}-1>\sqrt{64}-1=8-1=7\) (2)

Từ (1) và (2) suy ra \(\sqrt{8}+\sqrt{15}< \sqrt{65}-1\)

\(\sqrt{7}+\sqrt{15}<\sqrt{9}+\sqrt{25}=3+5=8=\sqrt{64}=\sqrt{65-1}\)

\(\sqrt{65-1}=\sqrt{64}=8\)

\(\sqrt{7}<\sqrt{9};\sqrt{15}<\sqrt{16}\rightarrow\sqrt{7}+\sqrt{15}<\sqrt{9}+\sqrt{16}=3+4=7<8\)

Do đó phải điền dấu < 

Lời giải:

a.

$\sqrt{8}+\sqrt{15}+1<\sqrt{9}+\sqrt{16}+1=3+4+1=8=\sqrt{64}< \sqrt{65}$

$\Rightarrow \sqrt{8}+\sqrt{15}< \sqrt{65}-1$
b.

$(2\sqrt{3}+6\sqrt{2})^2=84+24\sqrt{6}< 84+24\sqrt{9}< 169$

$\Rightarrow 2\sqrt{3}+6\sqrt{2}< 13$

$\Rightarrow \frac{13-2\sqrt{3}}{6}> \sqrt{2}$

\(VT^2=23+4\sqrt{30}\)

\(VP^2=66-2\sqrt{65}\)

Ta phải so sánh:

\(4\sqrt{30}\) và \(43-2\sqrt{65}\)

\(480\) và \(2109-172\sqrt{65}\)

\(0\) và \(1629-172\sqrt{65}\)

\(0< 1629-172\sqrt{65}\)

Vậy \(\sqrt{8}+\sqrt{15}< \sqrt{65}-1\)

Ta có: \(\sqrt{8}+\sqrt{15}< \sqrt{9}+\sqrt{16}=3+4=7\)

    và  \(\sqrt{65}-1=\sqrt{64}-1=8-1=7\)

      Vậy \(\sqrt{8}+\sqrt{15}< \sqrt{65}-1\)

Ta có:

Vì \(\sqrt{8}< \sqrt{9}\)và \(\sqrt{15}< \sqrt{16}\)

\(\Rightarrow\sqrt{8}+\sqrt{15}< \sqrt{9}+\sqrt{16}\)

mà \(\sqrt{9}+\sqrt{16}=3+4=7\) 

\(\Rightarrow\sqrt{8}+\sqrt{15}< 7\)(1)

Vì \(\sqrt{65}>\sqrt{64}\)

\(\Rightarrow\sqrt{65}-1>\sqrt{64}-1\)

mà \(\sqrt{64}-1=8-1=7\) 

\(\Rightarrow\sqrt{64}-1>7\)(2)

Từ (1) và (2) =>\(\sqrt{8}+\sqrt{15}< 7< \sqrt{65}-1\)

\(\Rightarrow\sqrt{8}+\sqrt{15}< \sqrt{65}-1\)

a) 

Ta có:

\(\left(\sqrt{26}+\sqrt{5}\right)^2=26+2\sqrt{26}\sqrt{5}+5\)

\(=31+2\sqrt{130}\)(1)

Mặt khác: \(\left(\sqrt{7}\right)^2=7\) (2)

Từ (1) và (2) =>\(\sqrt{26}+\sqrt{5}>\sqrt{7}\)

a) \(\sqrt{26}+\sqrt{5}< \sqrt{25}+\sqrt{4}=5+2=7\)

b) \(\sqrt{8}+\sqrt{24}< \sqrt{9}+\sqrt{25}=3+5=8\)

\(\sqrt{65}>\sqrt{64}=8\)

\(\Rightarrow\sqrt{8}+\sqrt{24}< \sqrt{65}\)

1)  \(A^2=2+2.\frac{\sqrt{\left(8+\sqrt{15}\right)\left(8-\sqrt{15}\right)}}{2}\)

              \(2+\sqrt{64-15}=2+\sqrt{49}=2+7=9\) mà A>0

=> A=3

2) \(A=\sqrt{4-\sqrt{15}}\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right).\)

 \(A=\sqrt{\left(4-\sqrt{15}\right)\left(4+\sqrt{15}\right)}\sqrt{4+\sqrt{15}}\left(\sqrt{10}-\sqrt{6}\right).\)

​​\(A=\sqrt{4+\sqrt{15}}\left(\sqrt{10}-\sqrt{6}\right).\)

\(A^2=\left(4+\sqrt{15}\right)\left(16-4\sqrt{15}\right)\)

       \(=4\left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right)=4\)

Mà A >0 

=> A=2

Mà 4>3

=> \(\sqrt{4}=2>\sqrt{3}\)

=> \(A>\sqrt{3}\)