a ) ( 165 + 99 + 165 x 1 ) - ( 163 x 101 - 163 x 1 ) 

= ( 165 x ( 99 + 1 ) ) - ( 163 x ( 101 - 1 ) )

= 165 x 100 - 163 x 100 

= 16500 - 16300

= 200

b ) 24 x 62 + 24 x 2 x 19

= 24 x 62 + 24 x 38

= 24 x ( 62 + 38 )

= 24 x 100

= 2400

= 

a) ( 165 x 99 + 165 x 1 ) - ( 163 x 101 - 163 x 1 )

= 165 x ( 99 + 1 )  -  163 x ( 101 - 1 )

= 165 x  100 - 163 x 100

= 100 x ( 165 - 163 )

= 100 x 2

= 200

b) 24 x 62 + 24 x 2 x 19

= 24 x 62 + 24  x 38

= 24 x ( 62 + 38 )

= 24 x 100

= 2400

( 165 x 99 + 165 x 1 ) - ( 163 x 101 - 163 x 1)

=[165 x ( 99 + 1 )] - [ 163 x ( 101 - 1)]

=165 x 100 - 163 x 100

= 100 x ( 165 - 163)

=100 x 2 = 200

24 x 62 + 24 x 2 x 19

= 24 x 62 + 24 x 38

= 24 x ( 62 + 38 )

=24 x 100 =2400

a) \(\left(165.99+165\right)-\left(163.101-163\right)\)

\(=165\left(99+1\right)-163\left(101-1\right)\)

\(=165.100-163.100\)

\(100\left(165-163\right)\)

\(=100.2\)

\(=200\)

b) \(24.62+48.19\)

\(=24.62+24.2.19\)

\(=24.62+24.38\)

\(=24\left(62+38\right)\)

\(=24.100=2400\)

\((165.99+165)-(163.101-163)\)

\(=(165.99+165.1)-(163.101-163.1)\)

\(=165(99+1)-163(100-1)\)

\(=165.100-163.100\)

\(=(165-163).100=2.100=200\)

\(24.62+48.19\)

\(=24.62+24.2.19 \)

\(=24.62+24.38\)

\(=24(62+38)\)

\(=24.100=2400\)

a: =274+184=458

b: =29+132+868+237+163

=1000+400+29

=1429

f: =A*135*137(1001-1001)=0

e: =24*25+24*37+24*38

=24*100=2400

c: =31(98+2)=31*100=3100

lay ong di qua lay ba di lai cho xin may tick

lay ong di qua lay ba di lai cho xin may tick

lay ong di qua lay ba di lai cho xin may tick

a) Ta có: \(\left(x-1\right)^{x+2}-\left(x-1\right)^{x+4}=0\)

\(\Leftrightarrow\left(x-1\right)^x\cdot\left(x-1\right)^2-\left(x-1\right)^x\cdot\left(x-1\right)^4=0\)

\(\Leftrightarrow\left(x-1\right)^{x+2}\cdot\left[1-\left(x-1\right)^2\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-1=1\\x-1=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=0\end{matrix}\right.\)

b) Ta có: \(\dfrac{1}{4}\cdot\dfrac{2}{6}\cdot\dfrac{3}{8}\cdot\dfrac{4}{10}\cdot\dfrac{5}{15}\cdot...\cdot\dfrac{30}{62}\cdot\dfrac{31}{64}=2x\)

\(\Leftrightarrow2x=\dfrac{1}{64}\)

hay \(x=\dfrac{1}{128}\)

Ta có :\(\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+...+\frac{1}{90}=\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{60}\right)+\left(\frac{1}{61}+\frac{1}{62}+...+\frac{1}{90}\right)\)

                   60 số hạng                                                              30 số hạng                                     30 số hạng

\(>\left(\frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}\right)+\left(\frac{1}{90}+\frac{1}{90}+...+\frac{1}{90}\right)=30.\frac{1}{60}+30.\frac{1}{90}=\frac{1}{2}+\frac{1}{3}=\frac{5}{6}\)

Vậy \(\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+...+\frac{1}{90}>\frac{5}{6}\)

Ta có: \(\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+...+\frac{1}{60}>\frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}=30.\frac{1}{60}=\frac{1}{2}\)

Lại có: \(\frac{1}{61}+\frac{1}{62}+\frac{1}{63}+...+\frac{1}{90}>\frac{1}{90}+\frac{1}{90}+...+\frac{1}{90}=30.\frac{1}{90}=\frac{1}{3}\)

\(\Rightarrow\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+...+\frac{1}{90}>\frac{1}{2}+\frac{1}{3}=\frac{5}{6}\)

\(\Rightarrow\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+...+\frac{1}{90}>\frac{5}{6}\) (đpcm)