Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(=\frac{7}{2}x\frac{1269}{46}x\frac{1}{17}x\frac{22}{9}=\frac{7x1269x22}{2x46x17x9}=\frac{10857}{782}\)
a) X = 15
b) X = 4
c ) X= 23
d) X= 11
( Chỉ là ý kiến riêng thôi nhé, nhận gạch đá )
a) \(\frac{6+x}{33}=\frac{7}{11}\)
=> (6 + x). 11 = 33.7
=> 66 + 11x = 231
=> 11x = 231 - 66
=> 11x = 165
=> x = 165 : 11
=> x = 15
b) 15/26 + x/13 = 46/52
=> x/13 = 23/26 - 15/26
=> x/13 = 4/13
=> x = 4
c) 121/27 x 54/11 < x < 100/21 : 25/126
=> 22 < x < 24
=> x = 23 (vì x là số tự nhiên)
d) 1 < 11/x < 12
=> 11/x \(\in\){2; 3; 4 ; ...; 11}
=> x \(\in\) {11/2; 11/3; ...; 1}
Vì x là số tự nhiên => x = 1
\(\left(X+\frac{1}{1.3}\right)+\left(X+\frac{1}{3.5}\right)+...+\left(X+\frac{1}{23.25}\right)=11.X+\)\(\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\right)\)
\(\Leftrightarrow12X+\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{23.25}\right)+11X\)\(+\frac{\left(1+\frac{1}{3}+...+\frac{1}{81}\right)-\left(\frac{1}{3}+\frac{1}{9}+...+\frac{1}{243}\right)}{2}\)
\(\Leftrightarrow X+\frac{1}{2}\times\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{23}+\frac{1}{23}-\frac{1}{25}\right)=\frac{242}{243}:2\)
\(\Leftrightarrow X+\frac{12}{25}=\frac{121}{243}\)
\(\Leftrightarrow X=\frac{109}{6075}\)
Vậy X=109/6075
Chắc Sai kết quả chứ công thức đúng nha!!!...
Fighting!!!...
Đặt:
\(A=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{23.25}\)
\(2A=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{23.25}=\frac{3-1}{1.3}+\frac{5-3}{3.5}+...+\frac{25-23}{23.25}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{23}-\frac{1}{25}=1-\frac{1}{25}=\frac{24}{25}\)
=> \(A=\frac{12}{25}\)
Đặt \(B=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\)
\(3B=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}\)
=> \(3B-B=\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\right)=1-\frac{1}{3^5}=\frac{242}{243}\)
=> \(2B=\frac{242}{243}\Rightarrow B=\frac{121}{243}\)
Giải phương trình:
\(\left(x+\frac{1}{1.3}\right)+\left(x+\frac{1}{3.5}\right)+...+\left(x+\frac{1}{23.25}\right)=11x+\left(\frac{1}{3}+\frac{1}{9}+...+\frac{1}{243}\right)\)
\(12x+\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{23.25}\right)=11x+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{242}\right)\)
\(12x+\frac{12}{25}=11x+\frac{121}{243}\)
\(12x-11x=\frac{121}{243}-\frac{12}{25}\)
\(x=\frac{109}{6075}\)
\(3\frac{3}{11}x27\frac{27}{46}x1\frac{6}{17}x2\frac{4}{9}=\frac{34x1269x23x22}{11x46x17x9}=\frac{2x17x141x9x23x2x11}{11x23x17x9}=282\)
\(\frac{3\cdot11+1}{11}\cdot\frac{27\cdot46+27}{46}=\frac{34}{11}\cdot\frac{1269}{46}\)\(=\frac{21573}{253}\)
K mk nha thanks
21573
253