Bài 1

a: 11/12=1-1/12

23/24=1-1/24

mà -1/12>-1/24

nên 11/12>23/24

b: -3/20=-9/60

-7/12=-35/60

mà -9>-35

nên -3/20>-7/12

a)\(2,4  =\frac{24}{10}=\frac{{12}}{5}\) và \(2\frac{3}{5} = \frac{{13}}{5}\)

Ta có: \(\frac{{12}}{5} < \frac{{13}}{5} \Rightarrow 2,4 < 2\frac{3}{5}\).             

b) \( - 0,12 = -\frac{12}{100}= - \frac{3}{{25}}\) và \( - \frac{2}{5} =  - \frac{{10}}{{25}}\)        

Ta có: -3 > -10 nên \( - \frac{3}{{25}} >  - \frac{{10}}{{25}}\) nên \( - 0,12 >  - \frac{2}{5}\).

c)\(\frac{{ - 2}}{7} = \frac{{ - 20}}{{70}}\) và \( - 0,3 = \frac{{ - 3}}{{10}} = \frac{{ - 21}}{{70}}\).

Do -20 > -21 nên \(\frac{{ - 20}}{{70}} > \frac{{ - 21}}{{70}}\) nên \(\frac{{ - 2}}{7} >  - 0,3.\)

Xin lỗ nhé thừa số 4 bé ở câu a

\(a,\sqrt{2}+\sqrt{11}< \sqrt{3}+\sqrt{16}=\sqrt{3}+4\)

a) Ta có:

\( - \frac{1}{3} = \frac{{ - 5}}{{15}};\frac{{ - 2}}{5} = \frac{{ - 6}}{{15}}\)

Vì -5 > -6 nên \(\frac{{ - 5}}{{15}} > \frac{{ - 6}}{{15}}\) hay \( - \frac{1}{3}\) > \(\frac{{ - 2}}{5}\)

b) 0,125 < 0,13 vì chữ số hàng phần trăm của 0,125 là 2 nhỏ hơn chữ số hàng phần trăm của 0,13 là 3

c) Ta có:

\(\begin{array}{l} - 0,6 = \frac{{ - 6}}{{10}} = \frac{{ - 3}}{5} = \frac{{ - 9}}{{15}};\\\frac{{ - 2}}{3} = \frac{{ - 10}}{{15}}\end{array}\)

Vì -9 > -10 nên \(\frac{{ - 9}}{{15}} > \frac{{ - 10}}{{15}}\) hay - 0,6 > \(\frac{{ - 2}}{3}\)

A<B

c.

(\sqrt{5}-\sqrt{3})-(\sqrt{10}-\sqrt{7})=(\sqrt{5}+\sqrt{7})-(\sqrt{3}+\sqrt{10})

Mà:

\((\sqrt{5}+\sqrt{7})^2=12+\sqrt{35}< 12+\sqrt{36}=18\)

\((\sqrt{3}+\sqrt{10})^2=13+\sqrt{30}>13+\sqrt{25}=18\)

\(\Rightarrow \sqrt{3}+\sqrt{10}> \sqrt{5}+\sqrt{7}\Rightarrow \sqrt{5}-\sqrt{3}< \sqrt{10}-\sqrt{7}\)

Lời giải:

a.

$5+\sqrt{2}>5+\sqrt{1}=6$

$4+\sqrt{3}< 4+\sqrt{4}=6$

$\Rightarrow 5+\sqrt{2}>4+\sqrt{3}$

b.

$\sqrt{8}-\sqrt{2}=2\sqrt{2}-\sqrt{2}=\sqrt{2}$

$\sqrt{5}-\sqrt{3}=\frac{5-3}{\sqrt{5}+\sqrt{3}}=\frac{2}{\sqrt{5}+\sqrt{3}}< \frac{2}{\sqrt{2}}=\sqrt{2}$

Vậy $\sqrt{8}-\sqrt{2}>\sqrt{5}-\sqrt{2}$

Bài 2:1-2+3-4+...+2011-2012

=1+2+3+4+...+2011+2012-2(2+4+6+...+2012)

=2025078-2(1012036)

=2025078-2024072

=1006

Học giỏi!

a) Ta có: \(\left(\dfrac{1}{243}\right)^6=\left(\dfrac{1}{3}\right)^{5\cdot6}=\left(\dfrac{1}{3}\right)^{30}\)

\(\Leftrightarrow\left(\dfrac{1}{3}\right)^{28}>\left(\dfrac{1}{243}\right)^6\)

\(\Leftrightarrow\left(\dfrac{1}{3^4}\right)^7>\left(\dfrac{1}{243}\right)^6\)

\(\Leftrightarrow\left(\dfrac{1}{81}\right)^7>\left(\dfrac{1}{243}\right)^6\)

mà \(\left(\dfrac{1}{80}\right)^7>\left(\dfrac{1}{81}\right)^7\)

nên \(\left(\dfrac{1}{80}\right)^7>\left(\dfrac{1}{243}\right)^6\)

\(\left(\dfrac{3}{8}\right)^5\&\left(\dfrac{5}{243}\right)^3\)
\(\left(\dfrac{3}{8}\right)^5=\left(\dfrac{90}{240}\right)^5=\dfrac{90^5}{240^5}\)

\(\left(\dfrac{5}{243}\right)^3=\dfrac{5^3}{243^3}\)

\(=>\dfrac{90^5}{240^5}>\dfrac{5^3}{243^3}\)

\(=>\left(\dfrac{3}{8}\right)^5>\left(\dfrac{5}{243}\right)^3\)

\(\left(\dfrac{1}{80}\right)^7\&\left(\dfrac{1}{243}\right)^6\)

\(\dfrac{1}{80}>\dfrac{1}{81}=\dfrac{1}{3^4}\)

\(=>\left(\dfrac{1}{80}\right)^7>\left(\dfrac{1}{3^4}\right)^7=\dfrac{1}{3^{7.4}}=\dfrac{1}{3^{28}}>\dfrac{1}{3^{30}}\)

\(=\dfrac{1}{\left(3^5\right)^6}=\left(\dfrac{1}{243}\right)^6\)

\(=>\left(\dfrac{1}{80}\right)^7>\left(\dfrac{1}{243}\right)^6\)

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