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a: \(\dfrac{1}{7}\cdot\dfrac{3}{8}+\dfrac{1}{7}\cdot\dfrac{5}{8}+\dfrac{\left(-1\right)^{2023}}{7}\)
\(=\dfrac{1}{7}\left(\dfrac{3}{8}+\dfrac{5}{8}\right)-\dfrac{1}{7}\)
\(=\dfrac{1}{7}-\dfrac{1}{7}=0\)
b: \(-3-\dfrac{16}{23}-\sqrt{\dfrac{4}{49}}-\dfrac{7}{23}+\dfrac{\left(-3\right)^2}{7}\)
\(=-3-\left(\dfrac{16}{23}+\dfrac{7}{23}\right)-\dfrac{2}{7}+\dfrac{9}{7}\)
\(=-3-\dfrac{23}{23}+\dfrac{7}{7}\)
=-3-1+1
=-3
c: \(\dfrac{4^2\cdot0,2^3}{2^6}\)
\(=\dfrac{2^4\cdot0,008}{2^6}=\dfrac{0.008}{4}=0.002\)
Bài 2:
\(A=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101}\)
\(\Leftrightarrow A=\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\)
\(\Leftrightarrow A=\dfrac{1}{1}-\dfrac{1}{101}\)
\(\Leftrightarrow A=\dfrac{100}{101}\)
Vậy ...
\(B=\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+\dfrac{1}{10.13}+\dfrac{1}{13.16}\)
\(\Leftrightarrow B=\dfrac{1}{3}\left(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{16}\right)\)
\(\Leftrightarrow B=\dfrac{1}{3}\left(\dfrac{1}{1}-\dfrac{1}{16}\right)\)
\(\Leftrightarrow B=\dfrac{1}{3}.\dfrac{15}{16}\)
\(\Leftrightarrow B=\dfrac{5}{16}\)
Vậy ...
Bài 1:
B=\(\dfrac{\left(1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}\right)}{\left(1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{16}\right)}\)
\(=\dfrac{\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}\right)}{\left(1-\dfrac{1}{2}+\dfrac{1}{2^2}-\dfrac{1}{2^3}+\dfrac{1}{2^4}\right)}\)
\(=\dfrac{\left(\dfrac{2^4+2^3+2^2+2+1}{2^4}\right)}{\left(2^4-2^3+2^2-2+1\right)}\)
\(=\dfrac{\left(2^3+2\right)\left(2+1\right)+1}{2^4}.\dfrac{2^4}{\left(2^3+2\right)\left(2-1\right)}\)
\(=\dfrac{2\left(2^2+1\right)\left(2+1\right)+1}{2\left(2^2+1\right)\left(2-1\right)+1}\)
\(=\dfrac{2.5.3+1}{2.5.1+1}\)
\(=\dfrac{31}{11}\)
\(=2,\left(81\right)\)
`@` `\text {Ans}`
`\downarrow`
`1,`
`3/16 - (x - 5/4) - (3/4 + (-7)/8 - 1) = 2 1/2`
`=> 3/16 - x + 5/4 - (-1/8 - 1) = 2 1/2`
`=> 3/16 - x + 5/4 - (-9/8) = 2 1/2`
`=> 3/16 - x + 19/8 = 2 1/2`
`=> 3/16 - x = 2 1/2 - 19/8`
`=> 3/16 - x =1/8`
`=> x = 3/16 - 1/8`
`=> x = 1/16`
Vậy, `x = 1/16`
`2,`
`1/2* (1/6 - 9/10) = 1/5 - x + (1/15 - (-1)/5)`
`=> 1/2 * (-11/15) = 1/5 - x + 4/15`
`=> -11/30 = x + 1/5 - 4/15`
`=> x + (-1/15) = -11/30`
`=> x = -11/30 + 1/15`
`=> x = -3/10`
Vậy, `x = -3/10.`
\(A=\frac{1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}}{1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\frac{1}{16}}\)
Đặt tử số là B, mẫu số là C
\(B=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\)
\(2B=2+1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}\)
\(2B-B=\left(2+1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}\right)-\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)\)
\(B=2-\frac{1}{16}\)
\(B=\frac{32}{16}-\frac{1}{16}=\frac{31}{16}\)
\(C=1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\frac{1}{16}\)
\(2C=2-1+\frac{1}{2}-\frac{1}{4}+\frac{1}{8}\)
\(2C+C=\left(2-1+\frac{1}{2}-\frac{1}{4}+\frac{1}{8}\right)+\left(1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\frac{1}{16}\right)\)
\(3C=2+\frac{1}{16}\)
\(3C=\frac{32}{16}+\frac{1}{16}\)
\(3C=\frac{33}{16}\)
\(C=\frac{33}{16}:3=\frac{11}{16}\)
=> \(A=\frac{B}{C}=\frac{31}{16}:\frac{11}{16}=\frac{31}{16}.\frac{16}{11}=\frac{31}{11}\)
tại ko có dấu phần và hỗn số sorry nha giải giúm mik ik mà