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\(A=11\dfrac{3}{13}-\left(2\dfrac{4}{7}+5\dfrac{3}{13}\right)\)
\(A=11\dfrac{3}{13}-5\dfrac{3}{13}-2\dfrac{4}{7}\)
\(A=6-2\dfrac{4}{7}\)
\(A=5\dfrac{7}{7}-2\dfrac{4}{7}\)
\(A=3\dfrac{3}{7}\)
\(B=\left(6\dfrac{4}{9}+3\dfrac{7}{11}\right)-4\dfrac{4}{9}\)
\(B=\left(6\dfrac{4}{9}-4\dfrac{4}{9}\right)+3\dfrac{7}{11}\)
\(B=2+3\dfrac{7}{11}\)
\(B=5\dfrac{7}{11}\)
\(C=\dfrac{-5}{7}.\dfrac{2}{11}+\dfrac{-5}{7}-\dfrac{9}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{-5}{7}.\left(\dfrac{2}{11}+1\right)-\dfrac{9}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{-5}{7}.\dfrac{13}{11}-\dfrac{9}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{-65}{77}-\dfrac{9}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{4}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{160}{11}\)
\(D=0,7.2\dfrac{2}{3}.20.0,375.\dfrac{5}{28}\)
\(D=\dfrac{7}{10}.\dfrac{8}{3}.20.\dfrac{375}{1000}.\dfrac{5}{28}\)
\(D=\dfrac{7}{28}=\dfrac{5}{2}\)
\(E=\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right)\left(\dfrac{1}{3}-0,25-\dfrac{1}{12}\right)\)
\(E=\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right)\left(\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{1}{12}\right)\)
\(E=\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right)\left(\dfrac{1}{12}-\dfrac{1}{12}\right)\)
\(E=\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right).0\)
\(\Rightarrow E=0\)
\(C=1-2+2^2-2^3+...-2^{2011}+2^{2012}\)
\(\Rightarrow2C=2-2^2+2^3-2^4+...-2^{2012}+2^{2013}\)
\(\Rightarrow3C=1+2^{2013}\)
\(\Rightarrow C=\frac{1+2^{2013}}{3}\)
Vậy
\(D=-2+2^2-2^3+2^4-...-2^{2019}+2^{2020}\)
\(\Rightarrow-2D=2^2-2^3+2^4-2^5+...+2^{2020}-2^{2021}\)
\(\Rightarrow-3D=-2^{2021}+2\)
\(\Leftrightarrow D=\frac{2^{2021}-2}{3}\)
\(1)\)\(\frac{3}{4}\cdot2+\frac{5}{2}\cdot\frac{1}{3}=\frac{3}{2}+\frac{5}{6}=\frac{9+5}{6}=\frac{14}{6}=\frac{7}{3}\)
\(2)\)\(\frac{5}{2}+\frac{3}{11}\cdot\frac{7}{26}\left(19-6\right)=\frac{5}{2}+\frac{3\cdot7}{11\cdot2}=\frac{5}{2}+\frac{21}{22}==\frac{38}{11}\)
A=(27+53)+(46+34)+79
A=80+80+79
A=160+79
A=239
B = -377 - (98 - 277)
B=-377-98-277
B=[(-377)+277]-98
B=(-100)-98
B=-198
C= -1,7.2,3 + 1,7.(-3,7) - 1,7.3 - 0,17 : 0,1
C=1,7.(-2.3)+1,7.(-3,7)+1,7.(-3)+1,7
C=1,7.[(-2,3+(-3,7)+(-3)+1]
C=1,7.(-8)
C=-13,6
Còn mái câu kia mk hk pít lm pn th/cảm cho mk nhae
Ta có : \(\frac{3^{10}.11+3^{10}.5}{3^9.2^4}=\frac{3^{10}.\left(11+5\right)}{3^9.16}=\frac{3^{10}.16}{3^9.16}=3\)3
\(\frac{3^{10}.11+3^{10}.5}{3^9.2^4}=\frac{3^{10}.\left(11+5\right)}{3^9.2^4}\) \(=\frac{3^{10}.16}{3^9.2^4}=\frac{3^{10}.2^4}{3^9.2^4}=3\)
a)\(A=\frac{3^{10}.11+3^{10}.5}{3^9.2^4}=\frac{3^{10}\left(11+5\right)}{3^9.2^4}=\frac{3.16}{2^4}=\frac{3.2^4}{2^4}=3\)
b)\(B=\frac{2^{10}.13+2^{10}.65}{2^8.104}=\frac{2^{10}\left(13+65\right)}{2^8.2^3.13}=\frac{2^{10}.78}{2^{11}.13}=3\)
c)\(C=\frac{4^9.36+64^4}{16^4.100}=\frac{2^{18}.2^2.3^2+2^{24}}{2^{16}.2^2.5^2}=\frac{2^{20}\left(3^2+2^4\right)}{2^{18}.5^2}=\frac{2^2.25}{25}=4\)
\(A=\frac{3}{4}-\left(\frac{5}{2}+\frac{5}{3}\right)+\left(\frac{-1}{3}\right)^2\)
\(\Rightarrow A=\frac{3}{4}-\left(\frac{15}{6}+\frac{10}{6}\right)+\frac{1}{9}\)
\(\Rightarrow A=\frac{3}{4}-\frac{25}{6}+\frac{1}{9}\)
\(\Rightarrow A=\frac{18}{24}-\frac{100}{24}+\frac{1}{9}\)
\(\Rightarrow A=\frac{-82}{24}+\frac{1}{9}\)
\(\Rightarrow A=\frac{-41}{12}+\frac{1}{9}=\frac{-123}{36}+\frac{4}{36}=\frac{-119}{36}\)
Vậy A=\(\frac{-119}{36}\)
b)\(B=\frac{-11}{19}.\frac{4}{13}+\frac{-11}{13}.\frac{15}{19}+\frac{11}{13}\)
\(\Rightarrow B=\frac{-11}{19}.\frac{4}{13}+\frac{11}{13}.\left(\frac{-15}{19}+1\right)\)
\(\Rightarrow B=\frac{-11}{19}.\frac{4}{13}+\frac{11}{13}.\frac{4}{19}\)
\(\Rightarrow B=\frac{-11}{13}.\frac{4}{19}+\frac{11}{13}.\frac{4}{19}=\frac{4}{19}.\left(\frac{-11}{13}+\frac{11}{13}\right)=\frac{4}{19}.\frac{0}{13}=\frac{4}{19}.0=0\)
Vậy B=0
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