tính giá trị của biểu thức:
(3.4.2^16)^2/11.2^13.4^11-16^9
bạn nào biết giải giúp mình bài này với
(3.4.2^16)^2 . (11.2^13.4^11-16^9)=?
giúp mik cách giải nha
Tính bằng cách hợp lí
(3.4.2^16)/11.2^13.4^11-16^9 các bạn ơn giúp minh với minh đang gấp nhớ ghi cách giải cho minh với nhé
Tính:
A= ( 3.4.2^16)^2 / 11.2^13.4^11-16^9
bạn sai rồi
Oh
\(\frac{\left(3.4.2^{16}\right)^2}{11.2^{13}.4^{11}-16^9}\)
\(=\frac{3^2.2^4.2^{32}}{11.2^{13}.2^{22}-2^{36}}\)
\(=\frac{3^2.2^{36}}{11.2^{35}-2^{36}}\)
\(=\frac{3^2.3^{36}}{2^{35}.\left(11-2\right)}\)
\(=\frac{9.2}{9}=2\)
\(A=\frac{\left(3.4.2^{16}\right)^2}{11.2^{13}.4^{11}-16^9}\)
\(A=\frac{3^2.4^2.2^{32}}{11.2^{13}.4^{11}-16^9}\)
\(A=\frac{3^2.2^4.2^{32}}{11.2^{13}.\left(2^2\right)^{11}-\left(2^4\right)^9}\)
\(A=\frac{3^2.2^{36}}{11.2^{13}.2^{22}-2^{36}}\)
\(A=\frac{3^2.2^{36}}{11.2^{35}-2.2^{35}}\)
\(A=\frac{3^2.2^{36}}{\left(11-2\right).2^{35}}\)
\(A=\frac{3^2.2}{9}=\frac{18}{9}=2\)
Tính :
A=(3.4.2^16)^2 / 11.2^13.4^11-16^9
\(A=\frac{\left(3.4.2^{16}\right)^2}{11.2^{13}.4^{11}-16^9}\)
\(=\frac{3^2.4^2.2^{32}}{11.2^{13}.4^{11}-16^9}\)
\(=\frac{3^2.\left(2^2\right)^2.2^{32}}{11.2^{13}.\left(2^2\right)^{11}-\left(2^4\right)^9}\)
\(=\frac{3^2.2^4.2^{32}}{11.2^{13}.2^{22}-2^{36}}\)
\(=\frac{3^2.2^{36}}{11.2^{35}-2^{36}}\)
\(=\frac{3^2.2^{36}}{2^{35}.\left(11-2\right)}\)
\(=\frac{9.2}{2}=2\)
Tính giá trị của biểu thức:
\(\frac{\left(3.4.2^{16}\right)^{^2}}{11.2^{13}.4^{11}-16^9}\)
\(\frac{\left(3.4.2^{16}\right)^2}{11.2^{13}.4^{11}-16^9}\)
\(=\frac{3^2.4^2.2^{32}}{11.2^{13}.\left(2^2\right)^{11}-\left(2^4\right)^9}\)
\(=\frac{3^2.\left(2^2\right)^2.2^{32}}{11.2^{13}.2^{22}-2^{36}}\)
\(=\frac{3^2.2^4.2^{32}}{11.2^{35}-2^{36}}\)
\(=\frac{3^2.2^{36}}{2^{35}.\left(11-2\right)}\)
\(=\frac{3^2.2^{36}}{2^{35}.9}=2\)
(3.4.2^16)^2/11.2^13.4^11-16^9
\(\dfrac{\left(3.4.2^{16}\right)^2}{11.2^{13}.4^{11}-16^9}\)
\(=\dfrac{3^2.\left(2^2\right)^2.\left(2^{16}\right)^2}{11.2^{13}.\left(2^2\right)^{11}-\left(2^4\right)^9}\)
\(=\dfrac{3^2.2^4.2^{32}}{11.2^{13}.2^{22}-2^{36}}\)
\(=\dfrac{3^2.2^{36}}{11.2^{35}-2.2^{35}}\)
\(=\dfrac{3^2.2^{36}}{2^{35}.\left(11-2\right)}\)
\(=\dfrac{3^2.2^{36}}{2^{35}.9}\)
\(=\dfrac{3^2.2^{36}}{2^{35}.3^2}\)
\(=2\)
các bạn giúp mik với : (3.4.2^16)^2/11.2^13.4^11-16^9
thank mấy bạn nhak
\(\frac{\left(3.4.2^{16}\right)^2}{11.2^{13}.4^{11}-16^9}\)\(=\frac{3^2.4^2.2^{32}}{11.2^{12}.2^{22}-2^{36}}\)\(=\frac{3^2.2^4.2^{32}}{11.2^{34}-2^{34}.2^2}\)\(=\frac{3^2.2^{34}.2^2}{2^{34}.\left(11-2^2\right)}\)\(=\frac{2^{34}.36}{2^{34}.7}=\frac{36}{7}\)
Tính giá trị biểu thức theo cách hợp lí:
1. B=20182018.2017-20172017.2018
2. C=1^2 +2^2 +3^2 +...+ 100^2
3. D= (3.4.2^16)^2 : 11.2^13 . 4^11-16^9
Ai giải được câu nào thì hộ mình nha,mình cảm ơn nhiều!
1.
\(B=20182018.2017-20172017.2018\)
\(B=2018.10001.2017-2017.10001.2018\)
\(B=0\)
1 B=20182018.2017-20172017.2018
B=2018.10001.2017-2017.10001.2018
B=0
2 C=12+22+32+...+1002
C=1(1+0)+2(1+1)+3(1+2)+...+100(1+99)
C=1+2+1.2+3+2.3+...+100+99.100
C=(1+2+3+...+100)+(1.2+2.3+...+99.100)
C=[(1+100).100:2]+[(99.100.101):3]
C=5050+333300
C=338350
1) \(B=20182018.2017-20172017.2018\)
\(B=2018.10001.2017-2017.10001.2018\)
\(B=0-0\)
\(B=0\)
3) \(D=\left(3.4.2^{16}\right)^2:11.2^{13}.4^{11}-16^9\)
\(D=3^2.2^4.2^{32}:11.2^{13}.2^{22}-2^{36}\)
\(D=3^2.2^{71}:11-2^{36}\)
\(D=2^{36}.\left(9.11.2^{35}-1\right)\)
(3.4.2^16)^2/11.2^13.4^11-16^9
(3.4.2^16)^2/11.2^13.4^11-16^9
=3^2.4^2.(2^16)^2/11.2^13.(2^2)^11-(2^4)^9
=3^2.(2^2)^2.2^32/11.2^13.2^12-2^36
=3^2.2^4.2^32/11.2^13.2^12-2^36
=3^2.2^32+4/11.2^13+12-2^36
=3^2.2^36/11.2^35-2^36
=3^2.2^35.2/11.2^35-2^35.2
=3^2.2^35.2/2^35.(11-2)
=3^2.2^35.2/2^35.9
=3^2.2/9
=9.2/9
=2