Ta có: \(\frac{A}{5}=\frac{15^{23}-3}{15^{23}-690}=\frac{15^{23}-690+687}{15^{23}-690}=1+\frac{687}{15^{23}-690}\)

\(\frac{B}{5}=\frac{15^{22}+4}{15^{22}-25}=\frac{15^{22}-25+29}{15^{22}-25}=1+\frac{29}{15^{22}-25}\)

Ta có: \(\frac{687}{15^{23}-690}=\frac{687\cdot29}{29\left(15^{23}-690\right)}=\frac{19923}{29\left(15^{23}-690\right)}\)

\(\frac{29}{15^{22}-25}=\frac{29\cdot687}{687\left(15^{22}-25\right)}=\frac{19923}{687\left(15^{22}-25\right)}\)

Ta có: \(29\left(15^{23}-690\right)-687\left(15^{22}-25\right)\)

\(=29\cdot15^{23}-29\cdot690-687\cdot15^{22}+687\cdot25\)

\(=15^{22}\left(29\cdot15-687\right)+687\cdot25-29\cdot690=15^{22}\cdot\left(-252\right)-1915<0\)

=>\(29\left(15^{23}-690\right)<687\left(15^{22}-25\right)\)

=>\(\frac{19923}{29\left(15^{23}-690\right)}>\frac{19923}{687\left(15^{22}-25\right)}\)

=>\(\frac{687}{15^{23}-690}>\frac{29}{15^{22}-25}\)

=>\(\frac{687}{15^{23}-690}+1>\frac{29}{15^{22}-25}+1\)

=>\(\frac{A}{5}>\frac{B}{5}\)

=>A>B

1/4

11

1/4

11

a) − 6 11 + 13 22 + − 3 17 + 5 − 11 + 9 22   = − 6 11 + − 5 11 + 13 22 + 9 22 + − 3 17 = − 1 + 1 + − 3 17 = − 3 17

b)  16 3 13 − 3 5 7 + 10 3 13   = 16 3 13 − 10 3 13 − 3 5 7 = 6 − 3 5 7 = 5 7 7 − 3 5 7 = 2 2 7

\(1234 + 678 + 766 + 1322=(1234+766)+(678+1322)=2000+2000=4000\)

\(427\times42+427\times35+427\times23=427\times\left(42+35+23\right)=427\times100=42700\)

\(248\times125-248\times24-248=248\times\left(125-24-1\right)=248\times100=24800\)

Giải:

478 + 3 277 + 123 + 1 522

= (478+1522) +(3277 + 123)

= 2000 + 3400

= 5400

Đáp án C

a) x -2143=1341

x              = 1341 + 2143

x               = 3484

b) x-2146=4146

x             = 4146+2146

x             =6292

c) x+528 = 1322 + 249

x     + 528=1571

x              =1571-528

x             =1043

d) x -3145=3915

x             =3915+3145

x                =7060