Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\frac{3^2.4^2.2^{32}}{11.2^{13}.4^{11}-16^9}=\frac{3^2.2^4.2^{32}}{11.2^{13}.2^{22}-2^{36}}=\frac{3^2.2^{36}}{11.2^{35}-2^{36}}=\frac{3^2.2^{36}}{2^{35}.\left(11-2\right)}=\frac{9.2}{9}=2\)
\(\frac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}=\frac{2^{19}.3^9+3.5.2^{18}.3^8}{6^9.2^{10}+6^{10}.2^{10}}=\frac{2^{19}.3^9+3^9.5.2^{18}}{6^9.2^{10}.\left(1+6\right)}=\frac{2^{18}.3^9.\left(2+5\right)}{2^9.3^9.2^{10}.7}=\frac{2^{18}.7}{2^{19}.7}=\frac{1}{2}\)
A = \(\frac{\left(2^4\right)^3.3^{10}+2^3.3.5.\left(2.3\right)^9}{\left(2^2\right)^6.3^{12}+\left(2.3\right)^{11}}\)= \(\frac{2^{12}.3^{10}+2^3.3.5.2^9.3^9}{2^{12}.3^{12}+2^{11}.3^{11}}\)
= \(\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^{12}+2^{11}.3^{11}}\)= \(\frac{2^{12}.3^{10}.\left(1+5\right)}{2^{11}.3^{11}.\left(2.3+1\right)}\)= \(\frac{2.6}{3.7}=\frac{4}{7}\)
c, theo đề bài ta có :
x2 = yz, y2 = xz , z2 = xy
\(\Rightarrow\frac{x}{y}=\frac{z}{x},\frac{y}{x}=\frac{z}{y},\frac{z}{x}=\frac{y}{z}\Rightarrow\frac{x}{y}=\frac{z}{x}=\frac{y}{z}\)
AD t/c DTSBN, ta có
\(\frac{x}{y}=\frac{z}{x}=\frac{y}{z}\Rightarrow\frac{X+z+y}{y+x+z}=1\)
x= 1y
z= 1x
y= 1z
=> x = y = x
sorry mn phần a em viết lộn
đây mới đúng đề bài nha
a, 1+6+8=2+4+9
ai giải hết em tk nhưng phải chi tiết
2) = 19683 . 3 : 59049 + 32 : 16 . 4 - 9 . 1
= 59049 : 59049 + 2 . 4 - 9
= 1 + 8 - 9
= 9 - 9
= 0
1) = {53^3 - 67 . [(169+144)].5 +7.3^4]} :2011
=[53^3 - 67 . (313 . 5 + 7 . 81 ] :2011
= [53^3 - 67. ( 1565 + 567 )] : 2011
= (53^3 - 67 . 2132) :2011
=(148877 - 142844 ) : 2011
= 6033 : 2011
= 3
Bài 1 :
a/ \(a^3.a^9=a^{3+9}=a^{12}\)
b/\(\left(a^5\right)^7=a^{5.7}=a^{35}\)
c/ \(\left(a^6\right).4.a^{12}=a^{24}.a^{12}.4=a^{24+12}.4=a^{36}.4\)
d/ \(\left(2^3\right)^5.\left(2^3\right)^3=2^{15}.2^9=2^{15+9}=2^{24}\)
e/ \(5^6:5^3+3^3.3^2\)
\(=5^3+3^5=125+243=368\)
i/ \(4.5^2-2.3^2\)
\(=2^2.5^2-2.3^2\)
\(=2^2.25-2^2.14\)
\(=2^2.\left(25-14\right)\)
\(=2^2.11\)
\(=4.11=44\)
\(10.4^6.9^5+6^9.120:8^4.3^{12}-6^{11}\)
\(=2.5.2^{12}.3^{10}+3^9.2^9.2^3.3.5:2^{12}.3^{12}-3^{11}.2^{11}\)
\(=5.2^{13}.3^{10}+2^{12}.5:2^{12}.3^{22}-3^{11}.2^{11}\)
\(=5.2^{13}.3^{10}+5.3^{22}-3^{11}.2^{11}\)
\(=5.3^{10}.\left(2^{13}+3^{12}\right)-3^{11}.2^{11}\)
Ta có: \(A=\dfrac{16^3\cdot3^{10}+120\cdot6^9}{4^6\cdot3^{12}+6^{11}}\)
\(=\dfrac{2^{12}\cdot3^{10}+2^3\cdot3\cdot5\cdot2^9\cdot3^9}{2^{12}\cdot3^{12}+6^{11}}\)
\(=\dfrac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{2^{12}\cdot3^{12}+2^{11}\cdot3^{11}}\)
\(=\dfrac{2^{12}\cdot3^{10}\left(1+5\right)}{2^{11}\cdot3^{11}\cdot\left(2\cdot3+1\right)}\)
\(=\dfrac{2\cdot6}{3\cdot7}=\dfrac{12}{21}=\dfrac{4}{7}\)
Ta có : \(A=\dfrac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{11}}=\dfrac{2^{12}.3^{10}+2^3.3.5.2^9.3^9}{2^{12}.3^{12}+2^{11}.3^{11}}\)
\(=\dfrac{2^{12}.3^{10}+5.2^{12}.3^{10}}{2^{12}.3^{12}+2^{11}.3^{11}}=\dfrac{2^{13}.3^{11}}{2^{11}.3^{11}.\left(1+6\right)}=\dfrac{2^2}{7}=\dfrac{4}{7}\)