Tính:
a,2^3.6-(15+5^8x :5^6)
b,3^2.64+36.3^2-450
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Ta có x =7
=>x+1=8
\(\Rightarrow\)\(A=x^{15}-8x^{14}+8x^{13}-8x^{12}+.......8x^2+8x-5\)
\(\Rightarrow x^{15}-\left(x+1\right)x^{14}+\left(x+1\right)x^{13}-\left(x+1\right)x^{12}+...\left(x+1\right)x^2\)
\(+\left(x+1\right)x^5\)
\(\Rightarrow x^{15}-x^{15}-x^{14}+x^{14}+x^{13}-x^{13}-x^{12}+...-x^3-x^2+x-5\)
\(\Rightarrow x-5\Leftrightarrow A=7-5=2\Rightarrow A=2\)
Vậy A=2 khi x=7
a: \(A=\dfrac{3^3\cdot2^3+3^3\cdot2^2+3^3\cdot1}{-13}=\dfrac{27\left(2^3+2^2+1\right)}{-13}=-27\)
b: \(B=\dfrac{2\cdot2^{12}\cdot3^6+2^{11}\cdot3^9}{2^3\cdot2^7\cdot3^7+2^7\cdot2^3\cdot5\cdot3^8}\)
\(=\dfrac{2^{13}\cdot3^6+2^{11}\cdot3^9}{2^{10}\cdot3^7+2^{10}\cdot5\cdot3^8}\)
\(=\dfrac{2^{11}\cdot3^6\left(2^2+3^3\right)}{2^{10}\cdot3^7\left(1+5\cdot3\right)}=\dfrac{2}{3}\cdot\dfrac{4+27}{1+15}=\dfrac{2}{3}\cdot\dfrac{31}{16}=\dfrac{31}{24}\)
c: \(C=\dfrac{5\cdot2^{30}\cdot3^{18}-2^{29}\cdot3^{20}}{5\cdot2^{35}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}}\)
\(=\dfrac{2^{29}\cdot3^{18}\left(5\cdot2-3^2\right)}{2^{29}\cdot3^{18}\left(5\cdot2^6-7\right)}=\dfrac{10-9}{5\cdot64-7}=\dfrac{1}{313}\)
a, 32 . 5 - 2x . 6 + 183
= 9 . 5 - 2x . 6 + 1
= 45 - 2x . 6 + 1
= 45 - 2x + 1 . 3 + 1
= 3( 15 - 2x + 1) + 1
b, 100 : { 280 : [ 450 - ( 480 - 4 . 52 ) ] }
= 100 : { 280 : [ 450 - 380 ] }
= 100 : { 280 : 70 }
= 100 : 4 = 25
c, 8 . { 24 - [ 3 . ( 5 + 25 ) + 15 ] : 15 }
= 8 . { 24 - [ 3 . 30 + 15 ] : 15 }
= 8 . { 24 - 105 : 15 }
= 8 . { 24 - 7 }
= 8 . 17 = 136
a: \(A=\dfrac{16^5\cdot15^5}{2^{10}\cdot3^5\cdot5^4}=\dfrac{2^{20}\cdot3^5\cdot5^5}{2^{10}\cdot3^5\cdot5^4}=2^{10}\cdot5=5120\)
b: \(B=\dfrac{2^{15}\cdot3+2^{19}\cdot10}{2^{12}\cdot26}=\dfrac{2^{15}\left(3+2^4\cdot10\right)}{2^{13}\cdot13}=2^2\cdot\dfrac{163}{13}=\dfrac{652}{13}\)
\(A=\dfrac{6^3+3\cdot6^2+3^3}{13}\)
\(=\dfrac{3^3\cdot8+3^3\cdot4+3^3}{13}\)
=27
a; 23.6 - (15 + 58 : 56)
= 8.6 - (15 + 52)
= 48 - (15+ 25)
= 48 - 40
= 8
b; 32.64 + 36.32 - 450
= 9.64 + 36.9 - 9.50
= 9.(64 + 36 - 50)
= 9.(100 - 50)
= 9.50
= 450