So sánh
A=2008x2010
B=2009^2
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a: \(\left(4+\sqrt{33}\right)^2=49+8\sqrt{33}=49+2\cdot\sqrt{528}\)
\(\left(\sqrt{29}+\sqrt{14}\right)^2=43+2\cdot\sqrt{29\cdot14}=43+2\cdot\sqrt{406}\)
mà 49>43 và 528>406
nên \(\left(4+\sqrt{33}\right)^2>\left(\sqrt{29}+\sqrt{14}\right)^2\)
=>\(4+\sqrt{33}>\sqrt{29}+\sqrt{14}\)
a, Chia hết cho 3 thì nhóm 2 số thành 1 cặp ; chia hết cho 7 thì nhóm 3 số thành 1 cặp
b, Đề phải là A = 2009.2011
Có :A = 2009.(2010+1) = 2009.2010+2009
= 2009.2010+2010-1 = 2010.(2009+1)-1 = 2010^2-1
Vì 2010^2-1 < 2010^2 = B => A < B
c, A = (3^3)^150 = 27^150
B = (5^2)^150 = 25^150
Vì 27^150 > 25^150 => A > B
k mk nha
\(a,2\sqrt{2}=\sqrt{8}< \sqrt{9}=3\\ \Leftrightarrow6+2\sqrt{2}< 3+6=9\\ b,\left(\sqrt{11}-\sqrt{3}\right)^2=14-2\sqrt{33}\\ 2^2=4=14-10\\ \left(2\sqrt{33}\right)^2=132>100=10^2\Leftrightarrow-2\sqrt{33}< -10\\ \Leftrightarrow\sqrt{11}-\sqrt{3}< 2\)
a) \(2^6\) và \(8^2\)
\(2^6=\left(2^2\right)^3\)
\(8^2=\left(2^3\right)^2\)\(=2^6\)
\(\Rightarrow\) \(2^6=8^2\)
a) Ta có :\(20< 25\Rightarrow\sqrt{20}< \sqrt{25}\Leftrightarrow2\sqrt{5}< 5\)
b) Ta có : \(\dfrac{16}{9}< 12\Rightarrow\sqrt{\dfrac{16}{9}}< \sqrt{12}\Leftrightarrow\dfrac{1}{3}\cdot\sqrt{16}< \sqrt{12}\)
a: \(2\sqrt{5}=\sqrt{20}\)
\(5=\sqrt{25}\)
mà 20<25
nên \(2\sqrt{5}< 5\)
b: \(\dfrac{1}{3}\cdot\sqrt{16}=\sqrt{\dfrac{1}{9}\cdot16}=\sqrt{\dfrac{16}{9}}\)
\(\sqrt{12}=\sqrt{\dfrac{108}{9}}\)
mà 16<9
nên \(\dfrac{1}{3}\sqrt{16}< \sqrt{12}\)
A=1+2+2^2+2^3+....+2^9
2A=2+2^2+2^3+....+2^10
2A-A=2^10-1
A=2^10-1/2
B=5.2^8=(2^2+1).2^8=2^10+2^8
=>B>A
2A = 2(1 + 2 + 22 + .... + 29 )
= 2 + 22 + 23 + ..... + 210
2A - A = (2 + 22 + 23 + ..... + 210) - (1 + 2 + 22 + .... + 29 )
A = 210 - 1
B = 5.28 = (22 + 1).28 = 210 + 28
210 - 1 < 210 + 28
=> A < B
a) \(2=\sqrt{4}>\sqrt{3}\)
b) \(6=\sqrt{36}< \sqrt{41}\)
c) \(7=\sqrt{49}>\sqrt{47}\)
a, Ta có: \(\left(\dfrac{1}{2}\right)^{300}=\left[\left(\dfrac{1}{2}\right)^3\right]^{100}=\left(\dfrac{1}{8}\right)^{100}\)
\(\left(\dfrac{1}{3}\right)^{200}=\left[\left(\dfrac{1}{3}\right)^2\right]^{100}=\left(\dfrac{1}{9}\right)^{100}\)
=> \(\left(\dfrac{1}{8}\right)^{100}>\left(\dfrac{1}{9}\right)^{100}\)=> \(\left(\dfrac{1}{2}\right)^{300}>\left(\dfrac{1}{3}\right)^{200}\)
b, Ta có: \(\left(\dfrac{1}{3}\right)^{75}=\left[\left(\dfrac{1}{3}\right)^3\right]^{25}=\left(\dfrac{1}{27}\right)^{25}\)
\(\left(\dfrac{1}{5}\right)^{50}=\left[\left(\dfrac{1}{5}\right)^2\right]^{25}\)\(=\left(\dfrac{1}{25}\right)^{25}\)
Do \(\left(\dfrac{1}{27}\right)^{25}< \left(\dfrac{1}{25}\right)^{25}=>\left(\dfrac{1}{3}\right)^{75}< \left(\dfrac{1}{5}\right)^{50}\)
Kiểm tra lại bài nhé, học tốt!!
A<B
nhìn là bít rùi
\(A=2008.2010=2008.\left(2009+1\right)=2008.2009+2008\)
\(B=2009.2009=2009\left(2008+1\right)=2008.2009+2009\)
\(2008< 2009\)
\(\Rightarrow2008.2009+2008< 2008.2009+2009\)
\(\Rightarrow A< B\)