\(\frac{2^6\times3^{31}+2^{40}\times3^6}{2^{11}\times3^{31}+2^{41}\times3^6}\)
giúp tui
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a,\(A=1993^{1^{2\times3\times4\times...\times1994}}=1993^1=1993\)
b,\(B=1994^{\left(225-1^2\right)\times\left(225-2^2\right).....\left(225-50^2\right)}\)
\(=1994^{\left(225-1^2\right)\times\left(225-2^2\right)...\left(225-15^2\right)...\left(225-50^2\right)}\)
\(=1994^{\left(225-1^2\right)\times\left(225-2^2\right)...\left(225-225\right)...\left(225-50^2\right)}\)
\(=1994^{\left(225-1^2\right)\times\left(225-2^2\right)...\times0\times...\left(225-50^2\right)}\)
\(=1994^0=1\)
c, \(C=\frac{2^{10}\times3^{31}+2^{40}\times3^6}{2^{11}\times3^{31}+2^{41}\times3^6}\)
\(=\frac{2^{10}\times3^6\times\left(1\times3^{25}+2^{30}\times1\right)}{2^{11}\times3^6\times\left(1\times3^{25}+2^{30}\times1\right)}\)
\(=\frac{2^{10}}{2^{11}}=\frac{1}{2}\)
Ta có : D = (1 + 2 + 22 + 23 + ....... + 22004) - 22005
Đặt A = 1 + 2 + 22 + 23 + ....... + 22004
=> 2A = 2 + 22 + 23 + ....... + 22005
=> 2A - A = 22005 - 1
=> A = 22005 - 1
Thay vào ta có : D = (1 + 2 + 22 + 23 + ....... + 22004) - 22005
=> D = 22005 - 1 - 22005
=> D = -1
\(A=\frac{7x\left(2x2\right)^5x3^{11}+2^{13}x\left(3x3\right)^5}{\left(2x3\right)^{10}+2^{12}x3^{10}}\)
\(A=\frac{7x2^{10}x3^{11}+2^{13}x3^{10}}{2^{10}x3^{10}+2^{12}x3^{10}}\)
tự làm tiếp
a) \(\dfrac{9}{11}\times8=\dfrac{9\times8}{11}=\dfrac{72}{11}\)
b) \(\dfrac{4}{5}\times1=\dfrac{4\times1}{5}=\dfrac{4}{5}\)
c) \(\dfrac{15}{8}\times0=\dfrac{15\times0}{8}=\dfrac{0}{8}=0\)
a: 9/11*8=(9*8)/11=72/11
b: 4/5*1=(4*1)/5=4/5
c: 15/8*0=(15*0)/8=0/8=0
\(\frac{2^{12}.3^{12}+6^9}{4^6.3^{12}+6^{11}}\)
= \(\frac{2^{12}.3^{12}+6^9}{2^{12}.3^{12}+6^{11}}\)
= \(\frac{6^{12}+6^9}{6^{12}+6^{11}}\)
= \(\frac{6^9\left(6^3+1\right)}{6^9\left(6^3+6^2\right)}\)
= \(\frac{217}{252}\)
= \(\frac{31}{36}\)
Hk tốt
\(\frac{2^{12}\times3^{12}+6^9}{4^6\times3^{12}+6^{11}}=\frac{2^{12}\times3^{12}+6^9}{\left(2^2\right)^6\times3^{12}+6^{11}}\)
\(=\frac{2^{12}\times3^{12}+6^9}{2^{12}\times3^{12}+6^{11}}=\frac{6^{12}+6^9}{6^{12}+6^{11}}\)
\(=\frac{6^9\left(6^3+1\right)}{6^9\left(6^3+6^2\right)}=\frac{6^3+1}{6^3+6^2}\)
\(=1+\frac{1}{36}=\frac{37}{36}\)
\(\dfrac{11.3^{22}.3^7-9^{15}}{\left(2.3^{14}\right)^2}=6^x\)
\(\Rightarrow\dfrac{11.3^{29}-3^{30}}{2^2.3^{28}}=6^x\)
\(\Leftrightarrow\dfrac{3^{29}\left(11-3\right)}{2^2.3^{28}}=6^x\)
\(\Leftrightarrow\dfrac{3.8}{4}=6^x\)
\(\Leftrightarrow6^1=6^x\)
\(\Leftrightarrow x=1\)
Vậy..
xem lại cái đề đúng chưa bn