Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(\frac{9}{33-3}=\frac{1}{3}\)
b) \(\frac{7}{100+6\times100}=\frac{1}{100}\)
c) \(\frac{11\times22+33\times36+55\times60}{22\times24+66\times72+110\times120}=\frac{1}{4}\)
d) \(\frac{9^4\times27^5\times3^6\times4^4}{3^8\times81^4\times243\times8^2}=4\)
e) \(\frac{199919991999}{200020002000}=\frac{1999}{2000}\)
-( -x + 13 - 142 ) + 18 = 55
-( -x - 129) + 18 = 55
x + 129 + 18 = 55
x + 147 = 55
x = 55 - 147
x = -92
b, 25 - 3.(6 -x ) = 22
3(6-x) = 25 - 22
3(6-x) = 3
6 - x = 1
x =5
c, [ ( 2x - 11 ) :3 +1] .5 = 20
( 2x - 11) : 3 + 1 = 20 : 5
(2x - 11) : 3 + 1 = 4
( 2x - 11) : 3 = 4 - 1
(2x - 11 ) : 3 = 3
2x - 11 = 3 x 3
2x - 11 = 9
2x = 9 + 11
2x = 20
x = 10
d, 3(x+5) - x - 11 = 24
3(x+5) - x = 24 +11
3x + 15 - x = 35
2x = 35 - 15
2x = 20
x = 10
a) \(\dfrac{22}{55}=\dfrac{2}{5}\)
b) \(\dfrac{-63}{81}=\dfrac{-7}{9}\)
c) \(\dfrac{2.14}{7.8}=\dfrac{2.7.2}{7.2.2.2}=\dfrac{1}{2}\)
d) \(\dfrac{49+7.49}{49}=\dfrac{49.\left(7+1\right)}{49}=\dfrac{49.8}{49}=8\)
\(A=\dfrac{3^{32}\cdot2^{33}}{2^{33}\cdot3^{33}}=\dfrac{1}{3}\)
\(B=\dfrac{3^6\cdot5^{10}\cdot2^5}{2\cdot5^{14}\cdot3^{14}\cdot3^2\cdot2}=\dfrac{2^3}{3^8\cdot5^4}\)
\(\dfrac{11}{6}+\dfrac{1}{4}=\dfrac{22}{12}+\dfrac{3}{12}=\dfrac{25}{12}\)
\(\dfrac{2}{5}-\dfrac{3}{8}=\dfrac{16}{40}-\dfrac{15}{40}=\dfrac{1}{40}\)
\(\dfrac{3}{10}-\dfrac{4}{15}=\dfrac{9}{30}-\dfrac{8}{30}=\dfrac{1}{30}\)
\(3+\dfrac{2}{5}=\dfrac{15}{5}+\dfrac{2}{5}=\dfrac{17}{5}\)
\(\dfrac{333}{777}+\dfrac{22}{55}=\dfrac{3}{7}+\dfrac{2}{5}=\dfrac{15}{35}+\dfrac{14}{35}=\dfrac{29}{35}\)
a,= 6-55-55-33+(-9)+(-6).11+22
= 6-(-33)+(-9)+(-66)+22
= 39+(-9)+(-66)+22
= 30+(-44)
= -14
b,= a+(-b)+(-c)+b+(-a)+c+(-c)+b
=[a+(-a)]+[(-b)+b+b]+[-c+c]
= 0+b+0
=b
THE END. :) :3 :D