\(M=\sqrt{\dfrac{2^{30}-2^{20}}{2^{22}-2^{12}}}=\sqrt{\dfrac{2^{20}\left(2^{10}-1\right)}{2^{12}\left(2^{10}-1\right)}}=\sqrt{2^8}=\sqrt{16^2}=16\)

a) \(9=6+3=6+\sqrt{9}\)

\(6+2\sqrt{2}=6+\sqrt{8}\)

\(\sqrt{8}< \sqrt{9}\) nên \(6+\sqrt{8}=6+2\sqrt{2}< 6+\sqrt{9}=9\)

b) \(\left(\sqrt{2}+\sqrt{3}\right)^2=5+2\sqrt{6}=5+\sqrt{24}\)

\(3^2=9=5+4=5+\sqrt{16}\)

\(\sqrt{16}< \sqrt{24}\Rightarrow3^2< \left(\sqrt{2}+\sqrt{3}\right)^2\Rightarrow3< \sqrt{2}+\sqrt{3}\)

c) \(9+4\sqrt{5}=\left(2+\sqrt{5}\right)^2\)

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

\(\sqrt{4}< \sqrt{5}\Rightarrow2+\sqrt{4}< 2+\sqrt{5}\Rightarrow\left(2+\sqrt{4}\right)^2=16< \left(2+\sqrt{5}\right)^2=9+4\sqrt{5}\)

d) \(\left(\sqrt{11}-\sqrt{3}\right)^2=14-2\sqrt{33}=14-\sqrt{132}\)

\(2^2=14-10=14-\sqrt{100}\)

\(\sqrt{100}< \sqrt{132}\Leftrightarrow-\sqrt{100}>-\sqrt{132}\Leftrightarrow14-\sqrt{100}>14-\sqrt{132}\)

\(\Rightarrow2>\sqrt{11}-\sqrt{3}\)

2 + 3  và  10

Ta có: 2 + 3 2 = 2 + 2 6 + 3 = 5 + 2 6

10 2  = 10 = 5 + 5

So sánh 26 và 5:

Ta có: 2 6 2 = 2 2 . 6 2  = 4.6 = 24

5 2  = 25

Vì 2 6 2 < 5 2  nên 2 6 < 5

Vậy 5 + 2 6  < 5 + 5 ⇒ 2 + 3 2 < 10 2  ⇒  2 + 3 <  10

Ta có: 31 > 25 ⇒  31  >  25  ⇒  31  > 5

Suy ra: 2. 31  > 2.5

Vậy 2. 31  > 10

9 + 4 5  và 16

So sánh 4 5  và 5

Ta có: 16 > 5 ⇒  16  >  5  ⇒ 4 >  5

Vì  5  > 0 nên 4.  5  >  5 . 5  ⇒ 4 5  > 5 ⇒ 9 + 4 5  > 5 + 9

Vậy 9 + 4 5  > 16

\(M=\sqrt{\frac{8^{10}-4^{10}}{4^{11}-8^4}}\)

\(=\sqrt{\frac{2^{30}-2^{20}}{2^{22}-2^{12}}}\)

\(=\sqrt{\frac{2^{20}\left(2^{10}-1\right)}{2^{12}\left(2^{10}-1\right)}}\)

\(=\sqrt{\frac{2^{20}}{2^{12}}}\)

\(=\sqrt{2^8}\)

\(=2^4\)

\(=16\)

=.= hok tốt!!

\(M=\sqrt{\dfrac{8^{10}-4^{10}}{4^{11}-8^4}}\)

\(M=\sqrt{\dfrac{\left(2^3\right)^{10}-\left(2^2\right)^{10}}{\left(2^2\right)^{11}-\left(2^3\right)^4}}\)

\(M=\sqrt{\dfrac{2^{30}-2^{20}}{2^{22}-2^{12}}}\)

\(M=\sqrt{\dfrac{2^{20}\left(2^{10}-1\right)}{2^{12}\left(2^{10}-1\right)}}\)

\(M=\sqrt{2^8}=16\)

\(X+\sqrt{25}=10\)

\(X+5=10\)

\(X=10-5\)

\(X=5\)

\(\dfrac{10}{11}+\dfrac{4}{11}:4-\dfrac{1}{8}=\dfrac{10}{11}+\dfrac{1}{11}-\dfrac{1}{8}=1-\dfrac{1}{8}=\dfrac{8}{8}-\dfrac{1}{8}=\dfrac{7}{8}\)

=\(=\dfrac{10}{11}+\dfrac{1}{11}-\dfrac{1}{8}=\dfrac{11}{11}-\dfrac{1}{8}=1-\dfrac{1}{8}=\dfrac{8}{8}-\dfrac{1}{8}=\dfrac{7}{8}\)