Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
mik ko ghi lại đề nhé!
\(A=\left(\frac{18}{15}.\frac{1}{4}.3\right)+\left(-\frac{47}{60}\right).\frac{24}{47}\)
\(A=\frac{8}{5}+\left(-\frac{2}{5}\right)\)
\(A=\frac{6}{5}\)
\(B=\frac{3}{4}.\frac{28}{15}-\left(\frac{8}{15}+\frac{1}{4}\right).\frac{24}{47}-\frac{17}{13}\)
\(B=\frac{7}{5}-\frac{47}{60}.\frac{24}{47}-\frac{17}{13}\)
\(B=\frac{7}{5}-\frac{2}{5}-\frac{17}{13}\)
\(B=1-\frac{17}{13}\)
\(B=-\frac{4}{13}\)
THANKS
\(\dfrac{-5}{9}+1\dfrac{5}{9}.\left(\dfrac{3}{4}-\dfrac{2}{5}\right):7^2\\ =\dfrac{-5}{9}+\dfrac{14}{9}.\left(\dfrac{3}{4}-\dfrac{2}{5}\right):49\\ =\dfrac{-5}{9}+\dfrac{14}{9}.\left(\dfrac{15}{20}-\dfrac{8}{20}\right):49\\ =\dfrac{-5}{9}+\dfrac{14}{9}.\dfrac{7}{20}:49\\ =\dfrac{-5}{9}+\dfrac{49}{90}:49\\ =\dfrac{-5}{9}+\dfrac{1}{90}\\ =\dfrac{-50}{90}+\dfrac{1}{90}\\ =\dfrac{-49}{90}\)
\(1\dfrac{13}{15}.0,75-\left(\dfrac{104}{195}+25\%\right).\dfrac{24}{47}-3\dfrac{12}{13}:3\\ =\dfrac{28}{15}.\dfrac{3}{4}-\left(\dfrac{8}{15}+\dfrac{1}{4}\right).\dfrac{24}{47}-\dfrac{51}{13}:3\\ =\dfrac{7}{5}-\left(\dfrac{32}{60}+\dfrac{15}{60}\right).\dfrac{24}{47}-\dfrac{51}{13}.\dfrac{1}{3}\\ =\dfrac{7}{5}-\dfrac{47}{60}.\dfrac{24}{47}-\dfrac{17}{13}\\ =\dfrac{7}{5}-\dfrac{2}{5}-\dfrac{17}{13}\\ =1-\dfrac{17}{13}\\ =\dfrac{13}{13}-\dfrac{17}{13}\\ =\dfrac{-4}{13}\)
a,\(5^3\left[\left(-7\right)+\left(-2\right)^3\right]+4\)
\(=125.\left[\left(-7\right)+\left(-8\right)\right]+4\)
\(=125.\left(-15\right)+4\)
\(=-1875+4=-1871.\)
b,\(47\left(45-15\right)-47\left(45+15\right)\)
\(=47\left(45-15-45-15\right)\)
\(=47.\left(-30\right)=-1410.\)
c,\(71.64-32.\left(-7\right)-32.\left(-11\right)\)
\(=32.\left(142+7+11\right)\)
\(=32.160=5120.\)
1/Ta co :
\(\dfrac{15}{7}.\left(\dfrac{1}{5}-\dfrac{46}{45}\right)+\dfrac{46}{45}.\left(\dfrac{15}{7}-\dfrac{45}{46}\right)\)
=\(\dfrac{15}{7}.\dfrac{1}{5}-\dfrac{15}{7}.\dfrac{46}{45}+\dfrac{46}{45}.\dfrac{15}{7}-\dfrac{46}{45}.\dfrac{45}{46}\)
=\(\dfrac{15}{7}.\left(\dfrac{1}{5}-\dfrac{46}{45}+\dfrac{46}{45}\right)-1\)
=\(\dfrac{15}{7}.\dfrac{1}{5}-1\)
=\(\dfrac{3}{7}-1=\dfrac{-4}{7}\)
2/Ta co
\(\dfrac{43}{47}.\left(\dfrac{18}{37}+\dfrac{47}{43}\right)-\dfrac{18}{3}.\left(\dfrac{43}{47}+\dfrac{37}{36}\right)\)
=\(\dfrac{43}{47}.\dfrac{18}{37}+\dfrac{43}{47}.\dfrac{47}{43}-\dfrac{18}{37}.\dfrac{43}{47}+\dfrac{18}{37}.\dfrac{37}{36}\)
=\(\dfrac{18}{37}.\left(\dfrac{43}{37}-\dfrac{43}{37}+\dfrac{37}{36}\right)+1\)
=\(\dfrac{18}{37}.\dfrac{37}{36}+1\)
=\(\dfrac{1}{2}+1=\dfrac{3}{2}\)
tick cho mk nha
1: \(=\dfrac{15}{37}\cdot\dfrac{38}{41}-\dfrac{15}{37}\cdot\dfrac{74}{45}-\dfrac{38}{41}\cdot\dfrac{15}{37}-\dfrac{38}{41}\cdot\dfrac{82}{76}\)
\(=\dfrac{-2}{3}-1=-\dfrac{5}{3}\)
2: \(=\dfrac{47}{53}\cdot\dfrac{17}{3}-\dfrac{47}{53}\cdot\dfrac{53}{47}+\dfrac{17}{3}\cdot\dfrac{6}{17}-\dfrac{17}{3}\cdot\dfrac{47}{53}\)
\(=-1+2=1\)
\(1\dfrac{13}{15}.0,75-\left(\dfrac{8}{15}+25\%\right).\dfrac{24}{47}-3\dfrac{12}{13}:3\)
\(=\dfrac{28}{15}.\dfrac{3}{4}-\left(\dfrac{8}{15}+\dfrac{1}{4}\right).\dfrac{24}{47}-\dfrac{51}{13}:3\)
\(=\dfrac{7}{5}-\dfrac{47}{60}.\dfrac{24}{47}-\dfrac{17}{13}\)
\(=\dfrac{7}{5}-\dfrac{2}{5}-\dfrac{17}{13}\)
\(=1-\dfrac{17}{13}=-\dfrac{4}{13}\)
\(1\frac{13}{15}.0,75-\left(\frac{8}{15}+25\%\right).\frac{24}{47}-3\frac{12}{13}:3\)
\(=\frac{28}{15}.\frac{3}{4}-\left(\frac{8}{15}+\frac{1}{4}\right).\frac{24}{47}-\frac{51}{13}:3\)
\(=\frac{7}{5}-\frac{47}{60}.\frac{24}{47}-\frac{17}{13}\)
\(=\frac{7}{5}-\frac{2}{5}-\frac{17}{13}\)
\(=\frac{-4}{13}\)
\(4\frac{1}{3}.\left(\frac{1}{6}-\frac{1}{2}\right)\le x\le\frac{2}{3}.\left(\frac{1}{3}-\frac{1}{2}-\frac{3}{4}\right)\)
\(\Leftrightarrow\frac{13}{3}.\frac{-1}{3}\le x\le\frac{2}{3}.\frac{-11}{12}\)
\(\Leftrightarrow\frac{-13}{9}\le x\le\frac{-11}{18}\)
\(\Leftrightarrow x=-1\)
a) 47.(-147) - 47.(-47)
= 47.(-147 + 47)
= 47 . (-100)
= -4700
b) 24.(15 - 4) + 4.(24 - 15)
= 24.15 - 25.4 + 4.25 - 4.15
= (24 - 4).15 - (25.4 - 4.25)
= 20 . 15
= 300
47(-147) - 47(-47)
= 47(-147 + 47)
= 47.(-100)
= -4700