Question

Giải giúp

Answer 1

Câu 2:

1: 

a: \(x+\dfrac{-3}{7}=\dfrac{5}{21}\)

=>\(x=\dfrac{5}{21}+\dfrac{3}{7}=\dfrac{5}{21}+\dfrac{9}{21}=\dfrac{14}{21}=\dfrac{2}{3}\)

b: \(\dfrac{1}{4}\cdot\dfrac{2}{6}\cdot\dfrac{3}{8}\cdot...\cdot\dfrac{14}{30}\cdot\dfrac{15}{32}=\dfrac{1}{2^{2x+1}}\)

=>\(1\cdot\dfrac{2}{4}\cdot\dfrac{3}{6}\cdot\dfrac{4}{8}\cdot...\cdot\dfrac{15}{30}\cdot\dfrac{1}{32}=\dfrac{1}{2^{2x+1}}\)

=>\(1\cdot\dfrac{1}{2}\cdot...\cdot\dfrac{1}{2}\cdot\dfrac{1}{32}=\dfrac{1}{2^{2x+1}}\)

=>\(\dfrac{1}{2^{14}\cdot2^5}=\dfrac{2}{2^{2x+1}}\)

=>2x+1=19

=>2x=18

=>x=9

Câu 1:

1:

a: \(\left(-\dfrac{3}{7}+\dfrac{4}{11}\right):\dfrac{5}{11}+\left(-\dfrac{4}{7}+\dfrac{-3}{11}\right):\dfrac{5}{11}\)

\(=\left(-\dfrac{3}{7}+\dfrac{4}{11}\right)\cdot\dfrac{11}{5}+\left(-\dfrac{4}{7}+\dfrac{-3}{11}\right)\cdot\dfrac{11}{5}\)

\(=\dfrac{11}{5}\left(-\dfrac{3}{7}+\dfrac{4}{11}+\dfrac{-4}{7}-\dfrac{3}{11}\right)\)

\(=\dfrac{11}{15}\left(-1+\dfrac{1}{11}\right)=\dfrac{11}{15}\cdot\dfrac{-10}{11}=-\dfrac{2}{3}\)

b: \(\left(\dfrac{10\cdot\sqrt{1,44}}{3}+\dfrac{24\cdot\sqrt{0,25}}{7}\right):\left(\dfrac{12}{7}+\dfrac{\sqrt{144}}{9}\right)\)

\(=\left(\dfrac{10\cdot1,2}{3}+\dfrac{24\cdot0,5}{7}\right):\left(\dfrac{12}{7}+\dfrac{12}{9}\right)\)

\(=\left(\dfrac{12}{3}+\dfrac{12}{7}\right):\left(\dfrac{12}{7}+\dfrac{12}{9}\right)\)

\(=\left(\dfrac{1}{3}+\dfrac{1}{7}\right):\left(\dfrac{1}{7}+\dfrac{1}{9}\right)=\dfrac{10}{21}:\dfrac{16}{63}=\dfrac{10}{21}\cdot\dfrac{63}{16}\)

\(=3\cdot\dfrac{5}{8}=\dfrac{15}{8}\)

2: \(A=\dfrac{1}{4\cdot9}+\dfrac{1}{9\cdot14}+...+\dfrac{1}{94\cdot99}\)

\(=\dfrac{1}{5}\left(\dfrac{5}{4\cdot9}+\dfrac{5}{9\cdot14}+...+\dfrac{5}{94\cdot99}\right)\)

\(=\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+...+\dfrac{1}{94}-\dfrac{1}{99}\right)\)

\(=\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{99}\right)=\dfrac{1}{5}\cdot\dfrac{95}{396}=\dfrac{19}{396}\)