Câu 1 xem kỉ đề

\(B,\frac{49^6.5-7^{11}}{\left(-7\right)^{10}.5-2.49^5}=\frac{7^{12}.5-7^{11}}{7^{10}.5-2.7^{10}}=\frac{7^{11}.\left(7.5-1\right)}{7^{10}.\left(5-2\right)}=\frac{7.34}{3}=\frac{238}{3}\)

a) A=2¹².3⁵-\(\frac{2^{12}.3^6}{2^{12}}\)+9³+8⁴.3⁵

=2¹².3⁵-3⁶+3⁶+2¹².3⁵

=2¹³.3⁵

^b)B=49⁶.5-5.\(\frac{7^{11}}{\left(-7\right)^{10}}-2.49^5\)

=49⁶.5-7.5-2.49⁵

⁼7¹².5-7.5-2.7¹⁰

a)2.3.5.13/26.35

=2.3.5.13/2.13.5.7

=3/7

b)18.6-18/-36.(-35)

=18.6-18/-(36.25)=18(6-1)/-(36.35)

=18.5/-(36.35)=2.3.3.5/-(2.2.3.3.5.7)

=-1/14

c)6.4+6.7/6.5+12=6.4+6.7/6.5+6.2

=6(4+7)/6(5+2)

=11/10

d)25.9-25.17/-8.80-8.10=25(9-17)/8(-80-10)

=25.(-8)/8(-90)

=5/18

e) 3⁴.5-3⁶/3⁴.13-3⁴

⁼3⁴(5-3²)/3⁴(13-1)

=5-3²/13-1=-1/3

a. - \(\frac{-1}{2}\)

b. \(\frac{238}{3}\)

Kotori Minami bạn có thể giải chi tiết ra duoc ko

TL:

\(\frac{4.5+4.11}{4.3-8.7}=\frac{20+44}{12-56}=\frac{64}{-44}=-\frac{16}{1}=-16\)

\(\frac{4^6.3^4.9^5}{6^{12}.15}=\frac{2^{12}.3^{14}}{2^{12}.3^{13.5}}=\frac{3}{5}\)

Học tốt

-16/11và 3/5

                  Tham khảo đề bài và cách làm sau nha bạn ! 

Đề bài : Rút gọn phân số : \(\frac{4.5+4.11}{4.3-8.7}\)và \(\frac{10.7-15.8}{20.3+5.6}\)

                                                      Giải

Đặt A = \(\frac{4.5+4.11}{4.3-8.7}\)

        B = \(\frac{10.7-15.8}{20.3+5.6}\)

Ta có : A = \(\frac{4.5+4.11}{4.3-8.7}=\frac{4.\left(5+11\right)}{4.\left(3-2.7\right)}=\frac{16}{-11}\)

          : B = \(\frac{10.7-15.8}{20.3+5.6}=\frac{10.7-2.4.3.5}{2.5.2.3+5.2.3}=\frac{10.\left(7-4.3\right)}{30.\left(2+1\right)}=\frac{-5}{9}\)

Vậy A = \(\frac{16}{-11}\) ; B = \(\frac{-5}{9}\)

`4/7` giữ nguyên

`15/21=(15:3)/(21:3)=5/7`

`----`

`12/36 =(12:12)/(36:12)=1/3`

`5/12` giữ nguyên

`----`

`12/28=(12:4)/(28:4)=3/7`

`6/7` giữ nguyên

\(M=\frac{131.145+100}{45-132.145}\)

\(=\frac{131-\frac{100}{145}}{\frac{45}{145}-132}\)

\(=\frac{131-\frac{20}{29}}{\frac{9}{29}-132}\)

\(=\frac{131\frac{-20}{29}}{-132\frac{9}{29}}\)

\(\frac{-5}{12}+\frac{-1}{4}=\frac{-5}{12}+\frac{-3}{12}=\frac{-5+-3}{12}=\frac{-8}{12}=\frac{-2}{3}\)

\(\frac{5}{12}+\frac{-3}{28}=\frac{35}{84}+\frac{-9}{84}=\frac{35+\left(-9\right)}{84}=\frac{26}{84}=\frac{13}{42}\)

\(\frac{-7}{15}+\frac{3}{35}=\frac{-49}{105}+\frac{9}{105}=\frac{-49+9}{105}=\frac{-40}{105}=\frac{-8}{21}\)

\(\frac{-5}{7}+\frac{-3}{4}=\frac{-20}{28}+\frac{-21}{28}=\frac{-20+\left(-21\right)}{28}=\frac{-41}{28}\)

a) \(\dfrac{2^6\cdot5^5}{10^5}\)

\(=\dfrac{2^6\cdot5^5}{2^5\cdot5^5}\)

\(=2\)

b) \(\dfrac{3^6\cdot5^7}{15^6}\)

\(=\dfrac{3^6\cdot5^7}{3^6\cdot5^6}\)

\(=5\)

\(A=\frac{49^{24}.125^{10}.2^8-5^{30}.7^{49}.4^5}{5^{29}.16^2.7^{48}}\)

\(A=\frac{\left(7^2\right)^{24}.\left(5^3\right)^{10}.2^8-5^{30}.7^{49}.\left(2^2\right)^5}{5^{29}.\left(2^4\right)^2.7^{48}}\)

\(A=\frac{7^{49}.5^{30}.2^8-5^{30}.7^{49}.2^{10}}{5^{29}.2^8.7^{48}}\)

\(A=\frac{7^{48}.5^{30}.2^8\left(1-28\right)}{5^{29}.2^8.7^{48}}\)

\(A=5.\left(-27\right)\)

\(A=-135\)