a: \(=\dfrac{-4\cdot13\cdot9\cdot5}{3\cdot4\cdot5\cdot2\cdot13}=\dfrac{3}{2}\)

b: \(=\dfrac{1}{2}\cdot\dfrac{1}{3}\cdot5=\dfrac{5}{6}\)

\(\frac{2^3.3^3.5^3.7.8}{3.2^4.5^3.14}=\frac{2^3.3^3.5^3.7.2^3}{3.2^4.5^3.2.7}\)

\(=\frac{2^6.3^3.5^3.7}{2^5.3.5^3.7}=\frac{2.3^2.1.1}{1.1.1.1}=2.3^2\)

\(=2.9=18\)

\(\frac{2^3.3^3.5^3.7.8}{3.2^4.5^3.14}=\frac{1.3^2.1.1.8}{1.2.1.2}=\frac{3.3.4}{2.1}=\frac{3.3.2}{1}=18\)

\(\frac{2^3.3^3.5^3.7.8}{3.2^4.5^3.14}\)mk viết như vầy bn cố gắng hiểu nhé:.

2³.3³.5³.7.8/ 3.2⁴.5³.14

= 2³.3³.5³.7.8/3.2⁴.5³.7.2

=2³.3³.5³.8/3.2⁴.5³.2

=2³.3³.5³.2³/2⁴.2.5³

= 2³.2³.3³.5³/ 2⁴.2.5³

= 2⁶. 3³.5³/ 2⁵. 5³

= 2⁵.2.3³.5³/2⁵.5³

= 2.3³/1

^{= 2. 27/1}

^{= 54/1 = 54}

^{Thế nha bn, chúc bn học tốt!}

/ là phân số hả bạn

b) \(\dfrac{6\cdot9-2\cdot17}{63\cdot3-119}\)

\(=\dfrac{2\left(3\cdot9-17\right)}{7\cdot\left(3\cdot9-17\right)}\)

\(=\dfrac{2}{7}\)

\(\frac{2^6.3.5^3.3.2.7}{2^4.3.5^3.6.7}=\frac{2^2}{1}=4\)

B/  2³3×3²×5³×7×8

3×2³×2 ×5³×7×2

=3²×8

2×2

=18

a) \(\dfrac{2^4\cdot5^2\cdot7}{2^3\cdot5\cdot7^2\cdot11}=\dfrac{2^3\cdot5\cdot10\cdot7}{2^3\cdot5\cdot7\cdot77}=\dfrac{10}{77}\)

\(\dfrac{2^3\cdot3^3\cdot5^3\cdot7\cdot8}{3\cdot2^4\cdot5^3\cdot14}=\dfrac{2^3\cdot3\cdot5^3\cdot7\cdot3^2\cdot8}{3\cdot2^3\cdot2\cdot5^3\cdot14}=\dfrac{7\cdot3^2\cdot8}{2\cdot14}=\dfrac{63\cdot8}{2\cdot14}=18=\dfrac{1386}{77}\)

rút gọn rồi quy đồng nha

\(N=1+\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+...+\left(\dfrac{1}{2}\right)^{100}\)

\(\Rightarrow2N=2+1+\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+...+\left(\dfrac{1}{2}\right)^{99}\)

\(\Rightarrow N=2N-N=2+1+\dfrac{1}{2}+...+\left(\dfrac{1}{2}\right)^{99}-1-\dfrac{1}{2}-...-\left(\dfrac{1}{2}\right)^{100}=2-\left(\dfrac{1}{2}\right)^{100}\)

\(N=1+\left(\dfrac{1}{2}\right)+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+...+\left(\dfrac{1}{2}\right)^{100}\)

\(\dfrac{1}{2}N=\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+\left(\dfrac{1}{2}\right)^4+...+\left(\dfrac{1}{2}\right)^{101}\)

\(\dfrac{1}{2}N-N=\left(\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+\left(\dfrac{1}{2}\right)^4+...+\left(\dfrac{1}{2}\right)^{101}\right)\)

               \(-\left(1+\left(\dfrac{1}{2}\right)+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+...+\left(\dfrac{1}{2}\right)^{100}\right)\)

\(-\dfrac{1}{2}N=\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^{101}-1\)

\(N=\dfrac{-\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^{101}}{-\dfrac{1}{2}}\)