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3) C thiếu đề
4) \(D=\frac{1}{9}-\left|\frac{-5}{23}\right|-\left(\frac{-5}{23}+\frac{1}{9}+\frac{25}{7}\right)+\frac{50}{4}-\frac{7}{30}\)
\(D=\frac{1}{9}-\frac{5}{23}+\frac{5}{23}-\frac{1}{9}-\frac{25}{7}+\frac{50}{4}-\frac{7}{30}\)
\(D=\frac{1}{9}-\frac{1}{9}-\frac{5}{23}+\frac{5}{23}+\frac{-25}{7}+\frac{50}{4}-\frac{7}{30}\)
\(D=0+0+\frac{125}{14}-\frac{7}{30}\)
\(D=\frac{913}{105}\)
a) \(\frac{11}{24}-\frac{5}{41}+\frac{13}{24}+0,5-\frac{36}{41}\)
= \(\frac{11}{24}-\frac{5}{41}+\frac{13}{24}+\frac{1}{2}-\frac{36}{41}\)
= \(\frac{1}{2}-\left\{\frac{11}{24}+\frac{13}{24}\right\}-\left\{\frac{5}{41}+\frac{36}{41}\right\}\)
=\(\frac{1}{2}-\frac{24}{24}-\frac{41}{41}\)
=\(\frac{1}{2}-1-1\)
=\(\frac{-3}{2}\)
b) \(-12:\left\{\frac{3}{4}-\frac{5}{6}\right\}^2\)
= \(-12:\left\{\frac{9}{12}-\frac{10}{12}\right\}^2\)
= \(-12:\left\{\frac{-1}{12}\right\}^2\)
= \(-12:\frac{1}{144}\)
= \(-12.144\)
= -1728
c) \(\frac{7}{23}.\left[\left(\frac{-8}{6}\right)-\frac{45}{18}\right]\)
= \(\frac{7}{23}.\left[\left(\frac{-24}{18}\right)-\frac{45}{18}\right]\)
= \(\frac{7}{23}.\left(\frac{-23}{6}\right)\)
= \(\frac{-7}{6}\)
d) \(23\frac{1}{4}.\frac{7}{5}-13\frac{1}{4}:\frac{5}{7}\)
= \(23\frac{1}{4}.\frac{7}{5}-13\frac{1}{4}.\frac{7}{5}\)
= \(\left\{23\frac{1}{4}-13\frac{1}{4}\right\}.\frac{7}{5}\)
= \(10.\frac{7}{5}\)
= 14
e) (1+23−14).(0,8−34)2
= (1+23−14).(\(\frac{4}{5}\)−34)2
= \(\left(\frac{12}{12}+\frac{8}{12}-\frac{3}{12}\right).\left(\frac{16}{20}-\frac{15}{20}\right)^2\)
= \(\frac{17}{12}.\left(\frac{1}{20}\right)^2\)
= \(\frac{17}{20}.\frac{1}{400}\)
= \(\frac{17}{8000}\)
a) \(\frac{17}{9}-\frac{17}{9}:\left(\frac{7}{3}+\frac{1}{2}\right)\)
= \(\frac{17}{9}-\frac{17}{9}:\frac{17}{6}\)
= \(\frac{17}{9}-\frac{2}{3}\)
= \(\frac{11}{9}\)
b) \(\frac{4}{3}.\frac{2}{5}-\frac{3}{4}.\frac{2}{5}\)
= \(\frac{2}{5}.\left(\frac{4}{3}-\frac{3}{4}\right)\)
= \(\frac{2}{5}.\frac{7}{12}\)
= \(\frac{7}{30}\)
Mình lười làm quá, hay mình nói kết quả cho bn thôi nha
c) -6
d) 3
e) 3
g) 12
h) \(\frac{23}{18}\)
i) \(\frac{-69}{20}\)
k) \(\frac{-1}{2}\)
l) \(\frac{49}{5}\)
\(a,\frac{-8}{15}.\left(-30\right).\frac{15}{-8}.\frac{9}{10}\)
\(=-\left(\frac{8}{15}.\frac{15}{8}\right).\left(30.\frac{9}{10}\right)\)
\(=-1.27
=-27\)
\(b,2\frac{1}{18}.\frac{23}{24}.\frac{9}{37}.\frac{48}{-15}\)
\(=\frac{-37.23.9.48}{18.24.37.15}=\frac{23}{15}\)
c, chịu rồi
a) \(4\frac{5}{9}:\left(-\frac{5}{7}\right)+\frac{49}{9}:\left(-\frac{5}{7}\right)=\frac{41}{9}:\left(-\frac{5}{7}\right)+\frac{49}{9}:\left(-\frac{5}{7}\right)\)
\(=\frac{41}{9}\cdot\left(-\frac{7}{5}\right)+\frac{49}{9}\cdot\left(-\frac{7}{5}\right)=\left(\frac{41}{9}+\frac{49}{9}\right)\cdot\left(-\frac{7}{5}\right)=10\cdot\left(-\frac{7}{5}\right)=-14\)
b) \(\left(\frac{-3}{5}+\frac{4}{9}\right):\frac{7}{11}+\left(\frac{-2}{5}+\frac{5}{9}\right):\frac{7}{11}\)
\(=\left(\frac{-3}{5}+\frac{4}{9}+\frac{-2}{5}+\frac{5}{9}\right):\frac{7}{11}\)
\(=\left(\frac{-3}{5}+\frac{-2}{5}+\frac{4}{9}+\frac{5}{9}\right):\frac{7}{11}\)
\(=\left(-1+1\right):\frac{7}{11}=0\cdot\frac{11}{7}=0\)
c) \(\left(\frac{3}{4}\right)^4\cdot\left(\frac{8}{9}\right)^2=\left(\frac{3}{4}\right)^2\cdot\left(\frac{3}{4}\right)^2\cdot\left(\frac{8}{9}\right)^2=\left(\frac{3}{4}\cdot\frac{3}{4}\cdot\frac{8}{9}\right)^2\)
\(=\left(\frac{1}{2}\right)^2=\frac{1}{4}\)
d) \(\left(-\frac{3}{5}\right)^6\cdot\left(-\frac{5}{3}\right)^5=\left(-\frac{3}{5}\right)^5\cdot\left(-\frac{3}{5}\right)\cdot\left(-\frac{5}{3}\right)^5=\left[\left(-\frac{3}{5}\right)\cdot\left(-\frac{5}{3}\right)\right]^5\cdot\left(-\frac{3}{5}\right)\)
\(=1^5\cdot\left(-\frac{3}{5}\right)=1\cdot\left(-\frac{3}{5}\right)=-\frac{3}{5}\)
e) \(\frac{8^{14}}{4^4\cdot64^5}=\frac{\left(2^3\right)^{14}}{\left(2^2\right)^4\cdot\left(2^6\right)^5}=\frac{2^{42}}{2^8\cdot2^{30}}=\frac{2^{42}}{2^{38}}=2^4=16\)
f) \(\frac{9^{10}\cdot27^7}{81^7\cdot3^{15}}=\frac{\left(3^2\right)^{10}\cdot\left(3^3\right)^7}{\left(3^4\right)^7\cdot3^{15}}=\frac{3^{20}\cdot3^{21}}{3^{28}\cdot3^{15}}=\frac{3^{41}}{3^{43}}=3^{-2}=\frac{1}{3^2}=\frac{1}{9}\)
\(\left(5+\frac{1}{5}-\frac{2}{9}\right)-\left(2-\frac{1}{23}-2\frac{3}{35}+\frac{5}{6}\right)-\left(8+\frac{2}{7}-\frac{1}{18}\right).\)
\(=5+\frac{1}{5}-\frac{2}{9}-2+\frac{1}{23}+\frac{73}{35}-\frac{5}{6}-8-\frac{2}{7}+\frac{1}{18}\)
\(=\left(5-2-8\right)+\left(\frac{1}{5}+\frac{73}{35}-\frac{2}{7}\right)+\left(-\frac{2}{9}-\frac{5}{6}+\frac{1}{18}\right)+\frac{1}{23}\)
\(=-5+\left(\frac{7}{35}+\frac{73}{35}-\frac{10}{35}\right)+\left(-\frac{4}{18}-\frac{15}{18}+\frac{1}{18}\right)+\frac{1}{23}\)
\(=-5+\frac{70}{35}+\frac{-18}{18}+\frac{1}{23}\)
\(=-5+2+\left(-1\right)+\frac{1}{23}\)
\(=-4+\frac{1}{23}=-\frac{91}{23}\)
Tính nhanh nếu có thể:
\(\left(5+\frac{1}{5}-\frac{2}{9}\right)-\left(2-\frac{1}{23}-2\frac{3}{35}+\frac{5}{6}\right)-\left(8+\frac{2}{7}-\frac{1}{18}\right)\)
\(=\)\(\left(5+\frac{1}{5}-\frac{2}{9}\right)-\left(2-\frac{1}{23}-\frac{73}{35}+\frac{5}{6}\right)-\left(8+\frac{2}{7}-\frac{1}{18}\right)\)
\(=\)\(5+\frac{1}{5}-\frac{2}{9}-2+\frac{1}{23}+\frac{73}{35}-\frac{5}{6}-8-\frac{2}{7}+\frac{1}{18}\)
\(=\)\(\left(5-2-8\right)+\left(\frac{1}{5}+\frac{73}{35}-\frac{2}{7}\right)+\left(\frac{2}{9}-\frac{5}{6}+\frac{1}{18}\right)+\frac{1}{23}\)
\(=\)\(\left(-5\right)+\left(\frac{7}{35}+\frac{73}{35}-\frac{10}{35}\right)+\left(\frac{4}{18}-\frac{15}{18}+\frac{1}{18}\right)+\frac{1}{23}\)
\(=\)\(\left(-5\right)+2+-\frac{5}{9}+\frac{1}{23}\)
\(=\)\(\left(-3\right)+-\frac{5}{9}+\frac{1}{23}\)
\(=\)\(-\frac{727}{207}\)
\(A=\frac{1135}{23}-\left(\frac{167}{32}+\frac{330}{23}\right)=\frac{1135}{23}-\frac{330}{23}+\frac{167}{32}=35+\frac{167}{32}=\frac{1287}{32}\)