\(\frac{2^{13}+2^5}{2^{10}+2^2}=\)
\(\frac{2^{10}\cdot13+2^{10}\cdot65}{2^8\cdot104}=\)
CẢ CÁCH LÀM NỮA NHÉ<-_->
\(\frac{2^{13}+2^5}{2^{10}+2^2}\)
\(\frac{2^{10}\cdot13+2^{10}\cdot65}{2^8\cdot104}\)
\(\frac{2^{13}+2^5}{2^{10}+2^2}=\frac{2^5.\left(2^8+1\right)}{2^2.\left(2^8+1\right)}=2^3=8\)
\(\frac{2^{10}.13+2^{10}.65}{2^8.104}=\frac{2^{10}.\left(13+65\right)}{2^8.104}=\frac{2^2.78}{104}=3\)
b\(^{\frac{2^{10\cdot13+2^{10}\cdot65}}{2^8\cdot104}}\)
c) \(\frac{5\cdot4^{15}-99-4\cdot3^{20}\cdot89}{5\cdot29\cdot6^{19}\cdot7\cdot2^{29}\cdot27^6}\)
b\(^{\frac{2^{10\cdot13+2^{10}\cdot65}}{2^8\cdot104}}\)
c) \(\frac{5\cdot4^{15}-99-4\cdot3^{20}\cdot89}{5\cdot29\cdot6^{19}\cdot7\cdot2^{29}\cdot27^6}\)
Rút gọn biểu thức:
\(\frac{2^{12}\cdot13+2^{12}\cdot65}{2^{10}\cdot104}+\frac{3^{10}\cdot11+3^{10}\cdot5}{3^9\cdot2^4}\)
\(\frac{2^{12}.13+2^{12}.65}{2^{10}.104}+\frac{3^{10}.11+3^{10}.5}{3^9.2^4}\)
\(\Rightarrow\frac{2^{12}.\left(13+65\right)}{2^{10}.104}+\frac{3^{10}.\left(11+5\right)}{3^9.2^4}\)
\(\Rightarrow\frac{2^{12}.78}{2^{10}.104}+\frac{3^{10}.2^4}{3^9.2^4}\)
\(=\frac{2^2.3}{4}+3\)
\(=3+3=6\)
Hãy so sánh B và C: \(B=\frac{3^{10}\cdot11+3^{10}\cdot5}{3^9\cdot2^4}\) và \(C=\frac{2^{10}\cdot13+2^{10}\cdot65}{2^8\cdot104}\)
B=3^10.11+3^10.5/3^9.2^4
= 3^10( 11+5)/3^9.16
= 3^10.16/3^9.16
= 3^10/3^9
= 3
Vậy B = 3 (1)
C = 2^10.13+2^10.65/2^8.104
= 2^10(13+65)/2^8.2^2.26
= 2^10.78/2^10.26
= 78/26
= 3
Vậy C = 3 (2)
Từ (1) v (2) suy ra B=C
Tính gíá trị biểu thức :
A= \(\frac{3^{10}\cdot11+\left(3^2\right)^5:5}{24\cdot27}\)
B= \(\frac{2^{10}\cdot13+2^{10}\cdot65}{28\cdot104}\)
AI NHANH MK K CHO
cách làm :
câu B
ta thấy có 2 lần 210 xuất hiện trên 1 phần tử
ta gộp lại như sau :
210 x ( 13 + 65 ) cho dễ
còn câu A
ta không thể tóm gọn nên phải tính như bình thường
A = 649539 + 45 / 648
A = 1002 . 44444444444444444
hay 649584 / 648
B = 0 + 66560 / 2912
B = 22 . 8571428571
hay 66560 / 2912
Đ/s : .........
nhé !
cho mình sửa lại câu B
B = 1024 x ( 13 + 65 ) / 2912
B = 79872 / 2912
a) \(\frac{2^{10}\cdot13+2^{10}\cdot65}{2^8\cdot104}\)
b) (1+2+3+4+....+100)x(\(1^2\) +\(2^2\) \(3^2\) +....+\(10^2\) )x(65 x 111 -13x15x37)
a.
\(\frac{2^{10}\times13+2^{10}\times65}{2^8\times104}=\frac{2^{10}\times\left(13+65\right)}{2^8\times104}=\frac{2^2\times78}{104}=\frac{4\times78}{104}=\frac{312}{104}=3\)
b.
\(\left(1+2+...+100\right)\times\left(1^2+2^2+...+10^2\right)\times\left(65\times111-15\times37\times13\right)\)
\(=\left(1+2+...+100\right)\times\left(1^2+2^2+...+10^2\right)\times\left(7215-7215\right)\)
\(=\left(1+2+...+100\right)\times\left(1^2+2^2+...+10^2\right)\times0\)
= 0
a,\(A=\frac{3^{10}\cdot11+\left(3^2\right)^5:5}{27\cdot24}\)
b, \(B=\frac{2^{10}+13+2^{10}\cdot65}{28\cdot104}\)
Tính:
\(\frac{2^{10}\cdot13+2^{10}\cdot65}{2^8\cdot104}\)
Nhớ ghi cả cách làm
\(\frac{2^{10}\cdot13+2^{10}\cdot65}{2^8\cdot104}\)
\(=\frac{2^{10}\cdot\left(13+65\right)}{2^8\cdot13\cdot8}\)
\(=\frac{2^{10}\cdot78}{2^8\cdot13\cdot8}\)
\(=\frac{2^{10}\cdot13\cdot2\cdot3}{2^8\cdot13\cdot2\cdot4}\)
\(=\frac{2^2\cdot3}{4}\)
\(=3\)
\(=\frac{2^{10}x\left(13+65\right)}{2^8x104}\)
\(=\frac{2^8x2^2x78}{2^8x104}\)
\(=\frac{4x78}{104}\)
\(=\frac{312}{104}=3\)
\(\frac{2^{10}.13+2^{10}.65}{2^8.104}\)
\(=\frac{2^{10}\left(13+65\right)}{2^8.13.2.4}\)
\(=\frac{2^{10}.78}{2^8.13.2.4}\)
\(=\frac{2^{10}.13.2.3}{2^8.13.2.4}\)
\(=\frac{2^2.3}{4}\)
\(=3\)