VÌ A = 1/2010 > 1/2011 > 1/2012  (1)

     B = 1/2009 <1/1007 (2) 

TỪ (1) VÀ (2)  => 1/2010 < 1/1007

VẬY A < B

Sao bạn biết

ta có: A = 1990 x 2010 + 1 = 1990 x 2000 + 1990 x 10 + 1 = 1 990 x 2000 + 19 900 + 1

B = 2000 x 2000 + 1 = 2000 x 1990 + 2000 x 10 +1 = 2000 x 1990 + 20 000 + 1 > 1 990 + 2000 + 19 000 + 1

=> A < B

Giả sử A > B

<=> 19 >\(5\sqrt{3}+6\sqrt{2}\)

<=> (6 + 3 - \(2\sqrt{3}\sqrt{6}\)

) + (10 - 5\(\sqrt{3}\))>0

<=> (\(\sqrt{6}-\sqrt{3}\))² + (10 - \(5\sqrt{3}\))>0

Mà 10 - 5\(\sqrt{3}\)> 10 - 5\(\sqrt{4}\) = 0

Vậy A > B

a: \(6\sqrt{3}=\sqrt{108}>\sqrt{54}=3\sqrt{6}\)

\(\Rightarrow5^{6\sqrt{3}}>5^{3\sqrt{6}}\)

b: \(\sqrt{2}\cdot2^{\dfrac{2}{3}}=2^{\dfrac{1}{2}}\cdot2^{\dfrac{2}{3}}=2^{\dfrac{1}{2}+\dfrac{2}{3}}=2^{\dfrac{7}{6}}\)

\(\left(\dfrac{1}{2}\right)^{-\dfrac{4}{3}}=2^{\left(-1\right)\cdot\left(-\dfrac{4}{3}\right)}=2^{\dfrac{4}{3}}\)

mà \(\dfrac{7}{6}< \dfrac{8}{6}=\dfrac{4}{3}\).

nên \(\sqrt{2}\cdot2^{\dfrac{2}{3}}< \left(\dfrac{1}{2}\right)^{-\dfrac{4}{3}}\).

tại sao ko sử dụng máy tính

Vì \(\sqrt{2,5+6,5}\ge0\Rightarrow\sqrt{2,5+6,5}< \sqrt{2,5+6,5}+1\)

a) Ta có: \(\frac{1}{5}\sqrt{150}=\frac{1}{5}\cdot5\sqrt{6}=\sqrt{6}=\frac{1}{3}\cdot\sqrt{6\cdot9}=\frac{1}{3}\sqrt{54}>\frac{1}{3}\sqrt{51}\)

b) Ta có: \(\frac{1}{2}\sqrt{6}=\sqrt{\frac{6}{4}}< \sqrt{\frac{36}{2}}=6\sqrt{\frac{1}{2}}\)

a) Vì  \(5,\left(6\right)< 6\)\(\Rightarrow\)\(\frac{51}{9}< \frac{150}{25}\)

                                    \(\Rightarrow\)\(\sqrt{\frac{51}{9}}< \sqrt{\frac{150}{25}}\)

                                    \(\Rightarrow\)\(\frac{1}{3}\sqrt{51}< \frac{1}{5}\sqrt{150}\)

b) Vì  \(1,5< 18\)\(\Rightarrow\)\(\frac{6}{4}< \frac{36}{2}\)

                                 \(\Rightarrow\)\(\sqrt{\frac{6}{4}}< \sqrt{\frac{36}{2}}\)

                                 \(\Rightarrow\)\(\frac{1}{2}\sqrt{6}< 6\sqrt{\frac{1}{2}}\)

\(A=2000\times2000=1990\times2000+10\times2000\)

\(B=1990\times2010=1990\times2000+10\times1990\)

Do đó A > B và lớn hơn là :

\(1990\times2000+10\times2000\)\(-\)\(1990\times2000+10\times1990\) \(=10\times2000-10\times1990=10\times\left(2000-1990\right)=10\times10=100\)

A=2000.2000                           B=1990.2010

=2000(1990+10)                           =1990(2000+10)

=2000.1990+2000.10                     =1990.2000+1990.10

Vì 2000.1990=1990.2000;2000.10>1990.10

=>A>B

\(a,\sqrt{42}=\sqrt{3\cdot14}>\sqrt{3\cdot12}=6\\ \sqrt[3]{51}=\sqrt[3]{17}< \sqrt[3]{3\cdot72}=6\\ \Rightarrow\sqrt{42}>\sqrt[3]{51}\\ b,16^{\sqrt{3}}=4^{2\sqrt{3}}\\ 18>12\Rightarrow3\sqrt{2}>2\sqrt{3}\Rightarrow4^{3\sqrt{2}}>4^{2\sqrt{3}}\\ \Rightarrow4^{3\sqrt{2}}>16^{\sqrt{3}}\)

\(c,\left(\sqrt{16}\right)^6=16^3=4^6=4^2\cdot4^4=4^2\cdot16^2\\ \left(\sqrt[3]{60}\right)^6=60^2=4^2\cdot15^2\\ 4^2\cdot16^2>4^2\cdot15^2\Rightarrow\sqrt{16}>\sqrt[3]{60}\Rightarrow0,2^{\sqrt{16}}< 0,2^{\sqrt[3]{60}}\)

Ta có:
\(A=2000.2000\)

\(\Rightarrow A=\left(1997+3\right).2000\)

\(\Rightarrow A=1997.2000+2000.3\)

\(\Rightarrow A=1997.2000+6000\)

\(B=1997.2003\)

\(\Rightarrow B=\left(2000+3\right).1997\)

\(\Rightarrow B=1997.2000+1997.3\)

\(\Rightarrow B=1997.2000+5991\)

Vì \(1997.2000+6000>1997.2000+5991\) nên A > B

Vậy A > B

Ta có :

\(B=1997.2003=\left(2000-3\right)\left(200+3\right)\)

\(\Rightarrow B=\left(2000-3\right)\left(2000+3\right)\)

\(\Rightarrow B=2000.\left(2000+3\right)-3\left(2000+3\right)\)

\(\Rightarrow B=2000^2+3.200-3.200-9\)

\(\Rightarrow B=2000^2-9< 2000.2000=A\)

=> B<A

\(B=1997.2003=\left(2000-3\right)\left(2000+3\right)=2000^2-9\)

\(A=2000.2000=2000^2\)

=> A > B