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Ta có: \(\frac{9x}{4}=\frac{16}{x}\Rightarrow9.x^2=64\)
ủa 64 không chia hết cho 9 bạn ạ!
\(\frac{9x}{4}=\frac{16}{x}\)
=>\(x^2=\frac{16\cdot4}{9}=\frac{64}{9}\)
Mà x âm nên \(x=-\frac{8}{3}\)
Ta có : \(\frac{9x}{4}=\frac{16}{x}\)
=> 9x2 = 16 x 4
=> 9x2 = 64
=> x2 = 64 : 9
=> x2 = \(\frac{64}{9}\)
=> x = \(\frac{8}{3}\)
ta có: \(\frac{9x}{4}\)=\(\frac{16}{x}\)
suy ra 9xx=16.4
9x2=64
x2=64:9
x2=\(\frac{3}{8}\)
x=công trừ 3/8
mà theo đề bài thì x là số âm
vậy x=\(\frac{-3}{8}\)thì mới thỏa mãn 9x/4=16/x
ta có\(\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}\)
\(=\frac{1}{2}-\left(\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)\)
tách
\(B=\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\)
\(2B=\frac{1}{2}+\frac{1}{4}+...+\frac{1}{512}\)
\(2B-B=\frac{1}{2}-\frac{1}{1024}\)
thay vào B ta có
\(\frac{1}{2}-\left(\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)\)
\(=\frac{1}{2}-\frac{1}{2}+\frac{1}{1024}=\frac{1}{1024}\)
\(A=\frac{1}{2}-\frac{1}{4}-\cdot\cdot\cdot-\frac{1}{1024}\)
\(\Rightarrow A=\frac{1}{2}-\frac{1}{2^2}-\cdot\cdot\cdot-\frac{1}{2^{10}}\)
\(\Rightarrow2A=1-\frac{1}{2}-\cdot\cdot\cdot-\frac{1}{2^9}\)
\(\Rightarrow2A-A=\left(1-\frac{1}{2}-\cdot\cdot\cdot-\frac{1}{2^9}\right)-\left(\frac{1}{2}-\frac{1}{2^2}-\cdot\cdot\cdot-\frac{1}{2^{10}}\right)\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2^{10}}\)
\(\Rightarrow A=\frac{1}{2}+\frac{1}{2^{10}}\)
\(\Rightarrow A=\frac{2^9+1}{2^{10}}\)
\(\Rightarrow A=\frac{513}{1024}\)
3,(34) : 2,(03) =\(\frac{331}{99}\): \(\frac{67}{33}\)=\(\frac{331}{201}\)
Ta có : \(\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-.....-\frac{1}{1024}\)
\(=\frac{1}{2}-\left(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+.....+\frac{1}{1024}\right)\)
Đặt \(A=\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+.....+\frac{1}{1024}\)
=> \(2A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+.....+\frac{1}{512}\)
=> \(2A-A=\frac{1}{2}-\frac{1}{1024}\)
Thay A vào ta có : \(\frac{1}{2}-\left(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+.....+\frac{1}{1024}\right)\)
\(=\frac{1}{2}-\frac{1}{2}+\frac{1}{1024}=\frac{1}{1024}\)
Jenny123 tham khảo nhé
Đặt tổng trên là A, ta có:
\(A.2=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}+\frac{1}{512}\)
\(A.2-A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{512}-"\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\)
\(\frac{1}{64}+\frac{1}{128}+\frac{1}{256}+\frac{1}{512}+\frac{1}{1024}"\)
\(A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}+\frac{1}{512}\)
\(-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-\frac{1}{16}-\frac{1}{32}-\frac{1}{64}-\frac{1}{128}-\frac{1}{256}-\frac{1}{512}-\frac{1}{1024}\)
\(A=1-\frac{1}{1024}=\frac{1023}{1024}\)
P/s: Bn xem lại đề nha