\(\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}+\frac{2}{192}\)
\(\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}+\frac{2}{96}+\frac{2}{192}\)
Ta thấy tất cả các phân số đều có mẫu chung là 192
=> \(\frac{128+64+32+16+8+4+2}{192}\)
= \(\frac{254}{192}\)= \(\frac{127}{96}\)
2/3+2/6+2/12+2/24+2/48+2/96+2/192
=2/3+1/3+1/6+1/12+1/24+1/48+1/96
=127/96
\(\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}+\frac{29}{96}+\frac{2}{192}\)
\(=\frac{2}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}\)
\(=\frac{254}{192}=\frac{127}{96}\)
\(C=\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}+\frac{2}{96}+\frac{2}{192}\)
\(C=\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}+\frac{2}{96}+\frac{2}{192}\)
\(2C=\frac{4}{3}+\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}+\frac{2}{96}\)
\(2C-C=\frac{4}{3}-\frac{2}{192}\)
\(C=\frac{127}{96}\)
\(C=\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}+\frac{2}{96}+\frac{2}{192}\)
\(C=\frac{2}{3}+\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}\)
\(C=\frac{254}{192}=\frac{127}{96}\)
\(\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}+\frac{2}{96}+\frac{2}{192}\)
=\(\frac{2}{3}+\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\right)\)
Ta xét phần trong ngoặc :
\(\frac{1}{2}+\frac{1}{4}=1-\frac{1}{4}\)
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}=1-\frac{1}{8}\)
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}=1-\frac{1}{64}\)
=\(\frac{63}{64}\)
=> \(1+\frac{63}{64}=\frac{127}{96}\)
Vậy tổng trên có kết quả là \(\frac{127}{96}\)
k nha
\(C=\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}+\frac{2}{96}+\frac{2}{192}\)
\(C=\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}+\frac{2}{96}+\frac{2}{192}\)
\(C=\frac{2}{3}+\frac{2}{3.2}+\frac{2}{3.4}+\frac{2}{3.8}+\frac{2}{3.16}+\frac{2}{3.32}+\frac{2}{3.64}\)
\(C=1-\frac{2}{3}+\frac{2}{3}-\frac{2}{4}+\frac{2}{4}-\frac{2}{8}+...+\frac{2}{64}\)
\(C=1-\frac{2}{64}\)
\(C=\frac{31}{32}\)
Làm mò, không biết đúng không nữa?
Tính nhanh \(\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}+\frac{2}{96}+\frac{2}{192}\)
Đặt bt bằng A
Ta có 2A= 2(2/3 + 2/6 + 2/12 +2/24 + 2/48 + 2/96 + 2/192)
2A= 4/3 +2/3 + 2/6 + 2/12 + 2/24 + 2/48 + 2/96
A= 2A-A= (4/3 +2/3 + 2/6 + 2/12 + 2/24 + 2/48 + 2/96) - (2/3 + 2/6 + 2/12 +2/24 + 2/48 + 2/96 2/192)
A=4/3 +2/3 + 2/6 + 2/12 + 2/24 + 2/48 + 2/96 - 2/3 - 2/6 - 2/12 - 2/24 - 2/48 - 2/96 - 2/192
A=(2/3 - 2/3) + (2/6 - 2/6) + ( 2/12 - 2/12) + (2/24 - 2/24) + (2/48 - 2/48) + ( 2/96 - 2/96) + (4/3 - 2/192)
A=0+0+0+0+0+0+ (256/192 - 2/192)
A=254/192
A=127/96(rút gọn phân số)
\(\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+...+\frac{2}{192}\)
\(=2\times\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{192}\right)\)
\(=2\times\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{12}+...+\frac{1}{96}-\frac{1}{192}\right)\)
\(=2\times\left(1-\frac{1}{192}\right)\)
\(=2\times\frac{191}{192}\)
\(=\frac{382}{192}=\frac{191}{96}\)
\(\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}+\frac{2}{96}+\frac{2}{192}\)
\(=2.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{6}+...+\frac{1}{96}-\frac{1}{192}\right)\)
\(=2.\left(1-\frac{1}{192}\right)\)
\(=2.\frac{191}{192}\)
\(=\frac{382}{192}=\frac{191}{96}\)
\(\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}+\frac{2}{96}+\frac{2}{192}\) bài này là bài tính nhanh nhé các bạn giải giùm minh để mình còn thi
\(\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+...+\frac{2}{192}.\)
\(=2\times\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+...+\frac{1}{192}\right)\)
\(=2\times\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{12}+...+\frac{1}{96}-\frac{1}{192}\right)\)
\(=2\times\left(1-\frac{1}{192}\right)\)
\(=2\times\frac{191}{192}=\frac{191}{68}\)
\(\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+...+\frac{2}{192}\)
\(=\frac{1}{3.1}+\frac{1}{3.2}+\frac{1}{3.2^2}+...+\frac{1}{3.2^6}\)
\(=\frac{1}{3}.\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^6}\right)\)
\(=\frac{1}{3}.A\)với \(A=\frac{1}{1}+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^6}\)
\(\Rightarrow2A=2.\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^6}\right)\)
\(\Rightarrow2A=2+\frac{1}{1}+\frac{1}{2}+...+\frac{1}{2^5}\)
\(\Rightarrow2A-A=\left(2+\frac{1}{1}+\frac{1}{2}+...+\frac{1}{2^5}\right)-\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^6}\right)\)
\(\Rightarrow A=2-\frac{1}{2^6}=2-\frac{1}{64}=\frac{127}{64}\)
\(\Rightarrow\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+...+\frac{2}{192}=\frac{1}{3}.\frac{127}{64}=\frac{127}{192}\)
Câu 1: Tính nhanh hợp lí \(\frac{119x83-183}{120x83x-266}\)
Câu 2: Tính nhanh hợp lí: \(\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}+\frac{2}{96}+\frac{2}{192}\)
Câu 1
Ta có \(\frac{119x83-183}{120x83-266}=\frac{119x83-183}{119x83+83-266}=\frac{119x83-183}{119x83-183}=1\)
Tính nhanh:
a) 2+5+8+11+...+104+107.
b) \(\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}+\frac{1}{192}\)
cái a bằng 1962
cái b bằng 127/192
à quên mình chưa rút gọn phân số đấy đâu bạn ạ
ban rút gọn phân số đấy hộ mình nha
Tính nhanh :A=\(\frac{2}{3}\)+\(\frac{2}{6}\)+\(\frac{2}{12}\)+\(\frac{2}{24}\)+\(\frac{2}{48}\)+\(\frac{2}{96}\)
\(\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}\)+\(\frac{2}{96}\)
=\(2\)x (\(\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}\)\(+\frac{1}{96}\))
=\(2\)x (\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{6}+...\)\(+\frac{1}{48}-\frac{1}{96}\))
=\(2\)x (\(1-\frac{1}{96}\))
=\(2\)x \(\frac{95}{96}\)
=\(\frac{190}{96}=\frac{95}{48}\)
TÍNH : E=\(\frac{2^2}{3}\times\frac{3^2}{8}\times\frac{4^2}{15}\times\frac{5^2}{24}\times\frac{6^2}{35}\times\frac{7^2}{48}\times\frac{8^2}{63}\times\frac{9^2}{80}\)
\(E=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.\frac{5^2}{4.6}...\frac{9^2}{8.10}=\frac{\left(2.3.4...9\right)^2}{1.2.\left(3.4...8\right)^2.9.10}=\frac{2^2.9^2}{1.2.9.10}=\frac{18}{10}=\frac{9}{5}\)