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\(M=\sqrt{\dfrac{2^{30}-2^{20}}{2^{22}-2^{12}}}=\sqrt{\dfrac{2^{20}\left(2^{10}-1\right)}{2^{12}\left(2^{10}-1\right)}}=\sqrt{2^8}=\sqrt{16^2}=16\)
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!!
\(M=\sqrt{\dfrac{8^{10}-4^{10}}{4^{11}-8^4}}\)
\(M=\sqrt{\dfrac{\left(2^3\right)^{10}-\left(2^2\right)^{10}}{\left(2^2\right)^{11}-\left(2^3\right)^4}}\)
\(M=\sqrt{\dfrac{2^{30}-2^{20}}{2^{22}-2^{12}}}\)
\(M=\sqrt{\dfrac{2^{20}\left(2^{10}-1\right)}{2^{12}\left(2^{10}-1\right)}}\)
\(M=\sqrt{2^8}=16\)
a, \(1< 2\Rightarrow\sqrt{1}< \sqrt{2}\Rightarrow1+1< \sqrt{2}+1\Rightarrow2< \sqrt{2}+1\)
c, \(4>3=>\sqrt{4}>\sqrt{3}=>\sqrt{4}-1>\sqrt{3}-1\Rightarrow1>\sqrt{3}-1\)
d, \(16>11=>\sqrt{16}>\sqrt{11}\Rightarrow4>\sqrt{11}=>4.\left(-3\right)< \sqrt{11}.\left(-3\right)\)
\(=>-12< -3.\sqrt{11}\)
Ta có: 11 < 16 ⇒ 11 < 16 ⇒ 11 < 4
Suy ra: -3. 11 > -3.4
Vậy -3 11 > -12
Không dùng máy tính bỏ túi, tính \(M=\sqrt{\dfrac{8^{10}-4^{10}}{4^{11}-8^4}}\)
Ta có:
\(8^{10}-4^{10}=4^{10}\left(2^{10}-1\right)=4^6.4^4\left(2^{10}-1\right)=2^{12}.4^4\left(2^{10}-1\right)\)
\(4^{11}-8^4=4^4\left(4^7-2^4\right)=4^4\left(2^{14}-2^4\right)=4^4.2^4\left(2^{10}-1\right)\)
Do đó: \(\dfrac{8^{10}-4^{10}}{4^{11}-8^4}=\dfrac{2^{12}.4^4\left(2^{10}-1\right)}{2^4.4^4\left(2^{10}-1\right)}=\dfrac{2^{12}}{2^4}=2^{12-4}=2^8\)
Vậy \(M=\sqrt{\dfrac{8^{10}-4^{10}}{4^{11}-8^4}}=\sqrt{2^8}=\sqrt{\left(2^4\right)^2}=2^4=16\)
cảm ơn bạn