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a: \(=\left(-\dfrac{25}{140}+\dfrac{245}{140}+\dfrac{32}{140}\right)\cdot\dfrac{-69}{20}\)
\(=\dfrac{252}{140}\cdot\dfrac{-69}{20}\)
\(=\dfrac{9}{5}\cdot\dfrac{-69}{20}=\dfrac{-621}{100}\)
b: \(=\left(6-2-\dfrac{4}{5}\right)\cdot\dfrac{25}{8}-\dfrac{8}{5}\cdot4\)
\(=\dfrac{16}{5}\cdot\dfrac{25}{8}-\dfrac{32}{5}=\dfrac{18}{5}\)
c: \(=\left(\dfrac{2}{24}+\dfrac{18}{24}+\dfrac{14}{24}\right):\dfrac{-17}{8}\)
\(=\dfrac{34}{24}\cdot\dfrac{-8}{17}=\dfrac{-1}{3}\cdot2=-\dfrac{2}{3}\)
\(a.\)
\(A=\)\(\frac{10^{15}+1}{10^{16}+1}\)
\(10A=\) \(\frac{10\left(10^{15}+1\right)}{10^{16}+1}\)
\(10A=\) \(\frac{10^{16}+10}{10^{16}+1}\)
\(10A=\)\(\frac{10^{16}+1+9}{10^{16}+1}\)
\(10A=\frac{10^{16}+1}{10^{16}+1}+\frac{9}{10^{16}+1}\)
\(10A=1+\frac{9}{10^{16}+1}\)
\(B=\frac{10^{16}+1}{10^{17}+1}\)
\(10B=\frac{10\left(10^{16}+1\right)}{10^{17}+1}\)
\(10B=\frac{10^{17}+10}{10^{17}+1}\)
\(10B=\frac{10^{17}+1+9}{10^{17}+1}\)
\(10B=\frac{10^{17}+1}{10^{17}+1}+\frac{9}{10^{17}+1}\)
\(10B=1+\frac{9}{10^{17}+1}\)
\(\Rightarrow10B< 10A\Rightarrow B< A\)\(\text{( vì tự làm ) }\)
xin lỗi hôm qua mk đang làm thì phải đy học zoom học xong quên h mới nhơ ra làm típ :)
b
\(A=\frac{3}{8^3}+\frac{7}{8^4}=\frac{3}{8^3}+\frac{3}{8^4}+\frac{4}{8^4}\)
\(B=\frac{3}{8^4}+\frac{7}{8^3}=\frac{3}{8^4}+\frac{3}{8^3}+\frac{4}{8^3}\)
Vì \(\frac{4}{8^4}< \frac{4}{8^3}\)=.> A < B
A. \(\frac{3}{4}\) x \(\frac{8}{9}\)x \(\frac{15}{16}\)x .... x \(\frac{899}{900}\)
= \(\frac{1.3}{2^2}\) x \(\frac{2.4}{3^3}\)x \(\frac{3.5}{4^2}\)x ... x \(\frac{29.31}{30^2}\)
= \(\left(\frac{1.2.3...29}{2.3.4...30}\right).\left(\frac{3.4.5...31}{2.3.4...30}\right)\)
= \(\frac{1}{30}.\frac{31}{2}\)= \(\frac{31}{60}\)
B.
\(\frac{1}{3}+\frac{3}{8}-\frac{7}{12}=\frac{8}{24}+\frac{9}{24}-\frac{14}{24}=\frac{8+9-14}{24}=\frac{3}{24}=\frac{1}{8}\)
\(A=-1-4=-5\)
\(B=\frac{4}{3}.\frac{15}{7}-16\)
\(B=\frac{20}{7}-16\)
\(B=\frac{-92}{7}\)
\(C=\frac{28}{15}.0,25.3+\left(\frac{8}{15}-\frac{1}{4}\right)\div1\frac{23}{24}\)
\(C=1,4+\frac{17}{60}\div\frac{47}{24}\)
\(C=1,4+\frac{34}{235}\)
\(C=\frac{363}{235}\)
\(A=\frac{-15}{8}+\frac{7}{8}-4\)
\(=-1-4=-5\)
\(B=\left(4-2\frac{2}{3}\right).2\frac{1}{7}-1\frac{3}{5}:\frac{1}{10}\)
\(=\frac{4}{3}.\frac{15}{7}-\frac{8}{5}:\frac{1}{10}\)
\(=\frac{20}{7}-16=\frac{-92}{7}\)
\(C=1\frac{13}{15}.\left(0,5\right)^2.3+\left(\frac{8}{15}-25\%\right):1\frac{23}{24}\)
\(=\frac{28}{15}.\frac{1}{4}.3+\frac{17}{60}:\frac{47}{24}\)
\(=\frac{7}{15}.3+\frac{17}{60}:\frac{47}{24}\)
\(=\frac{7}{5}+\frac{34}{235}=\frac{363}{235}\)
\(A=\frac{15^8-1+3}{15^8-1}=1+\frac{3}{15^8-1}\)
\(B=\frac{15^8-3+3}{15^8-3}=1+\frac{3}{15^8-3}\)
Ta có: \(15^8-3< 15^8-1\)
=> B>A