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Lời giải:
$\frac{-3}{8}.\frac{2}{15}+\frac{-3}{8}.\frac{13}{15}+1\frac{3}{8}$
$=\frac{-3}{8}(\frac{2}{15}+\frac{13}{15})+1+\frac{3}{8}$
$=\frac{-3}{8}.\frac{15}{15}+1+\frac{3}{8}=\frac{-3}{8}+1+\frac{3}{8}=1$
rõ ràng ta chỉ cần so sánh giữa \(15^{30}+16^{12}+17^{50}-16^8\) và \(17^{30}+16^8+15^{50}-16^{12}\)
Áp dụng tính chất nếu a>b thì a-b>0 ta được:
\(15^{30}+16^{12}+17^{50}-16^8\)- \(\left(17^{30}+16^8+15^{50}-16^{12}\right)\)
= \(\left(17^{50}-17^{30}\right)+\left(16^{12}+16^{12}\right)+\left(15^{30}-15^{50}\right)-\left(16^8+16^8\right)\)
= \(\left(17^{50}-17^{30}\right)+\left(15^{30}-15^{50}\right)+2\left(16^{12}-16^8\right)\)
Vì 17^50 - 17^30 > l 15^30 - 15^50 l
nên \(\left(17^{50}-17^{30}\right)+\left(15^{30}-15^{50}\right)>0\)
=>\(15^{30}+16^{12}+17^{50}-16^8\)> \(17^{30}+16^8+15^{50}-16^{12}\)
=> Phân số thứ nhất > 1 và p/s thứ hai < 1
Lúc này bạn tự so sánh nha
\(\frac{-7}{15}.\frac{5}{8}-\frac{2}{8}.\frac{7}{15}+\frac{1}{8}.\frac{-23}{15}=\frac{-7}{24}-\frac{7}{60}-\frac{23}{120}=-\frac{3}{5}\)
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: \(=\left(\dfrac{2}{24}+\dfrac{18}{24}+\dfrac{14}{24}\right):\dfrac{-17}{8}\)
\(=\dfrac{34}{24}\cdot\dfrac{-8}{17}=\dfrac{-1}{3}\cdot2=-\dfrac{2}{3}\)
\(A=-1-4=-5\)
\(B=\frac{4}{3}.\frac{15}{7}-16\)
\(B=\frac{20}{7}-16\)
\(B=\frac{-92}{7}\)
\(C=\frac{28}{15}.0,25.3+\left(\frac{8}{15}-\frac{1}{4}\right)\div1\frac{23}{24}\)
\(C=1,4+\frac{17}{60}\div\frac{47}{24}\)
\(C=1,4+\frac{34}{235}\)
\(C=\frac{363}{235}\)
\(A=\frac{-15}{8}+\frac{7}{8}-4\)
\(=-1-4=-5\)
\(B=\left(4-2\frac{2}{3}\right).2\frac{1}{7}-1\frac{3}{5}:\frac{1}{10}\)
\(=\frac{4}{3}.\frac{15}{7}-\frac{8}{5}:\frac{1}{10}\)
\(=\frac{20}{7}-16=\frac{-92}{7}\)
\(C=1\frac{13}{15}.\left(0,5\right)^2.3+\left(\frac{8}{15}-25\%\right):1\frac{23}{24}\)
\(=\frac{28}{15}.\frac{1}{4}.3+\frac{17}{60}:\frac{47}{24}\)
\(=\frac{7}{15}.3+\frac{17}{60}:\frac{47}{24}\)
\(=\frac{7}{5}+\frac{34}{235}=\frac{363}{235}\)
\(A=\frac{15^8-1+3}{15^8-1}=1+\frac{3}{15^8-1}\)
\(B=\frac{15^8-3+3}{15^8-3}=1+\frac{3}{15^8-3}\)
Ta có: \(15^8-3< 15^8-1\)
=> B>A