a: \(0,75< 1\)

=>Hàm số \(y=0,75^x\) nghịch biến trên R

mà -2,3>-2,4

nên \(0,75^{-2,3}< 0,75^{-2,4}\)

b: \(\dfrac{1}{4}< 1\)

=>Hàm số \(y=\left(\dfrac{1}{4}\right)^x\) nghịch biến trên R

mà 2023<2024

nên \(\left(\dfrac{1}{4}\right)^{2023}>\left(\dfrac{1}{4}\right)^{2024}\)

c: Vì 3,5>1

nên hàm số \(y=3,5^x\) đồng biến trên R

mà 2023<2024

nên \(3,5^{2023}< 3,5^{2024}\)

\(a,\left(\dfrac{1}{256}\right)^{-0,75}+\left(\dfrac{1}{27}\right)^{-\dfrac{4}{3}}\\ =256^{\dfrac{3}{4}}+27^{\dfrac{4}{3}}\\ =\sqrt[4]{256^3}+\sqrt[3]{27^4}\\ =145\\ b,\left(\dfrac{1}{49}\right)^{-1,5}-\left(\dfrac{1}{256}\right)^{-\dfrac{2}{3}}\\ =49^{\dfrac{3}{2}}-256^{\dfrac{2}{3}}\\ \simeq343-40,3\\ \simeq302,7\)

   0,75 + \(\dfrac{9}{5}\) ( 1,5 - \(\dfrac{2}{3}\) )²

= 0,75 + \(\dfrac{9}{5}\) ( \(\dfrac{3}{2}\) - \(\dfrac{2}{3}\))²

= 0,75 + \(\dfrac{9}{5}\) (\(\dfrac{5}{6}\))²

= 0,75 + \(\dfrac{5}{4}\)

= 0,75 + 1,25

= 2 

\(\dfrac{-22}{25}\) + ( \(\dfrac{22}{7}\) - 0,12)

= \(\dfrac{-22}{25}\) + ( \(\dfrac{22}{7}\) - \(\dfrac{3}{25}\))

= \(\dfrac{-22}{25}\) + \(\dfrac{22}{7}\) - \(\dfrac{3}{25}\)

= - ( \(\dfrac{22}{25}\) + \(\dfrac{3}{25}\)) + \(\dfrac{22}{7}\)

= -1 + \(\dfrac{22}{7}\)

= \(\dfrac{-7}{7}\) + \(\dfrac{22}{7}\)

= \(\dfrac{15}{7}\)

a; 0,75 +9/5(1,5 - 2,3) = 

cvAjcfajshcfaSsjhcfzsjhcvZXVxhjcvzhjc

Vì \(\left|x\right|=1,5\)

\(\Rightarrow x=1,5\) hoặc \(x=-1,5\)

Thay \(x=1,5;y=-0,75\) vào P:

\(P=1,5-4.1,5.\left(-0,75\right)+\left(-0,75\right)\)

\(=5,25\)

Thay \(x=-1,5;y=-0,75\) vào P ta đc:

\(P=-1,5-4.\left(-1,5\right).\left(-0,75\right)+\left(-0,75\right)\)

\(=-6,75\)

Vậy P \(\left[\begin{matrix}=5,25\\=-6,75\end{matrix}\right.\)

A= 4/7.

Biết có cái

\(=\left(-\dfrac{2}{3}\right).\dfrac{3}{4}+\dfrac{5}{3}.\left(-\dfrac{9}{4}\right)+\dfrac{1}{4}=-\dfrac{1}{2}-\dfrac{15}{4}+\dfrac{1}{4}=-\dfrac{7}{2}-\dfrac{1}{2}=-4\)

11/45

=(7/8-1/4):(5/6-3/4)^2

=(7/8-2/8):(10/12-9/12)^2

=5/8-(1/12)^2

=5/8-1/144

=90/144-1/144

=89/144

\(a)\frac{2}{{15}} + \left( {\frac{{ - 5}}{{24}}} \right) = \frac{{16}}{{120}} + \left( {\frac{{ - 25}}{{120}}} \right) = \frac{{ - 9}}{{120}} = \frac{{ - 3}}{{40}}\)          

 b) \(\left( {\frac{{ - 5}}{9}} \right) - \left( { - \frac{7}{{27}}} \right) = \left( {\frac{{ - 15}}{{27}}} \right) + \frac{7}{{27}} = \frac{{ - 8}}{{27}}\)          

 c)\(\left( { - \frac{7}{{12}}} \right) + 0,75 = \left( { - \frac{7}{{12}}} \right) + \frac{75}{100} \\= \left( { - \frac{7}{{12}}} \right) + \frac{3}{4} \\= \left( { - \frac{7}{{12}}} \right) + \frac{9}{{12}} = \frac{2}{{12}} = \frac{1}{6}\)

d)\(\left( {\frac{{ - 5}}{9}} \right) - 1,25 =\left( {\frac{{ - 5}}{9}} \right) - \frac{125}{100} = \left( {\frac{{ - 5}}{9}} \right) - \frac{5}{4}\\ = \left( {\frac{{ - 20}}{{36}}} \right) - \frac{{45}}{{36}} = \frac{{ - 65}}{{36}}\)       

 e)\(0,34.\frac{{ - 5}}{{17}} =\frac{{34}}{{100}}.\frac{{ - 5}}{{17}} = \frac{{17}}{{50}}.\frac{{ - 5}}{{17}} = \frac{{ - 1}}{{10}}\)                       

g) \(\frac{4}{9}:\left( { - \frac{8}{{15}}} \right) = \frac{4}{9}.\left( { - \frac{{15}}{8}} \right) = \frac{{ - 5}}{6}\)

h)\(\left( {1\frac{2}{3}} \right):\left( {2\frac{1}{2}} \right) = \frac{5}{3}:\frac{5}{2} = \frac{5}{3}.\frac{2}{5} = \frac{2}{3}\)       

i) \(\frac{2}{5}.\left( { - 1,25} \right) = \frac{2}{5}.\frac{{ - 125}}{100} = \frac{2}{5}.\frac{{ - 5}}{4} = \frac{{ - 1}}{2}\)                     

k) \(\left( {\frac{{ - 3}}{5}} \right).\left( {\frac{{15}}{{ - 7}}} \right).3\frac{1}{9} = \left( {\frac{{ - 3}}{5}} \right).\left( {\frac{{15}}{{ - 7}}} \right).\frac{{28}}{9}\\ = \frac{{ - 3.3.5.7.4}}{{5.\left( { - 7} \right).3.3}} = 4\)