(1+32).32=?
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Gọi số thóc ở 3 kho lần lượt là a, b, c
Ta có: a + b + c = 278
Ta có: \(1-\frac{1}{3}a=1-\frac{3}{5}b=1-\frac{4}{7}c\)
\(\Rightarrow\frac{2}{3}a=\frac{2}{5}b=\frac{3}{7}c\)
\(\Rightarrow\frac{a}{\frac{3}{2}}=\frac{b}{\frac{5}{2}}=\frac{c}{\frac{7}{3}}\)
Áp dụng tính chất của dãy tỉ số bằng nhau:
\(\frac{a}{\frac{3}{2}}=\frac{b}{\frac{5}{2}}=\frac{c}{\frac{7}{3}}=\frac{a+b+c}{\frac{3}{2}+\frac{5}{2}+\frac{7}{3}}=\frac{228}{\frac{19}{3}}=36\)
\(\Rightarrow\hept{\begin{cases}a=36.\frac{3}{2}\\b=36.\frac{5}{2}\\c=36.\frac{7}{3}\end{cases}\Rightarrow\hept{\begin{cases}a=54\\b=90\\c=84\end{cases}}}\)
Vậy...
Chúc em học tốt!!!
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Dịch ra là: Ta có: 3A = 3. (1 + 3 + 32 + 33 + ... + 399 + 3100) (1 + 3 + 32 + 33 + ... + 399 + 3100) 3A = 3 + 32 + 33 + ... + 3100 + 31013 + 32 + 33 + ... + 3100 + 3101 Suy ra: 3A - A = (3 + 32 + 33 + ... + 3100 + 3101) - (1 + 3 + 32 + 33 + ... + 399 + 3100) (3 + 32 + 33 + ... + 3100 + 3101) - (1 + 3 + 32 + 33 + ... + 399 + 3100) ⇒⇒ A = 3101−123101−12 Vậy A = 3101−12
Mà đoạn 2A sai nhé bạn, sửa lại:
2A = 3101−13101−1 2A=-10001
A=-10001/2
A=-5000,5
Vậy A=-5000,5
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\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}\)
\(\Rightarrow\dfrac{A}{3}=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\)
\(\Rightarrow A-\dfrac{A}{3}=\dfrac{2A}{3}=\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\dfrac{2A}{3}=\left(\dfrac{1}{3^2}-\dfrac{1}{3^2}\right)+\left(\dfrac{1}{3^3}-\dfrac{1}{3^3}\right)+...+\left(\dfrac{1}{3^{99}}-\dfrac{1}{3^{99}}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)=\dfrac{1}{3}-\dfrac{1}{3^{100}}\)
\(\Rightarrow2A=3\cdot\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\text{A}=\dfrac{1-\dfrac{1}{3^{99}}}{2}\)
\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}< \dfrac{1}{2}\)
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\(32\left(\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.17}+...+\frac{1}{197.200}\right)-x=\frac{1}{2}\)
\(\frac{32}{3}\left(\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+....+\frac{3}{197.200}\right)-x=\frac{1}{2}\)
\(\frac{32}{3}\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+...+\frac{1}{197}-\frac{1}{200}\right)-x=\frac{1}{2}\)
\(\frac{32}{3}\left(\frac{1}{8}-\frac{1}{200}\right)-x=\frac{1}{2}\)
x=0.78