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Bài giải
Ta có :
\(\frac{16,2\times3,7+5,7\times16,2-7,8\times4,8+4,6\times7,8}{11,2+12,3+13,4+12,6+11,5+10,4}\)
\(=\frac{16,2\times\left(3,7+5,7\right)-7,8\times\left(4,8+4,6\right)}{\left(13,4-12,6\right)+\left(12,3-11,5\right)+\left(11,2-10,4\right)}\)
\(=\frac{16,2\times9,4-7,8\times9,4}{0,8+0,8+0,8}\)
\(=32,9\)
Vậy ............
\(=\frac{16,2\times\left(5,7+3,7\right)-6,2\times\left(4,8+4,6\right)}{\left(13,4+12,3+11,2\right)-\left(10,4+11,5+12,6+0,4\right)}=\frac{16,2\times9,4-6,2\times9,4}{\left(13+12+11+0,4+0,3+0,2\right)-\left(10+11+0,4+0,5+13\right)}\)
\(=\frac{9,4\times\left(16,2-6,2\right)}{\left(13+12+11+0,9\right)-\left(13+11+10+0,9\right)}=\frac{9,4\times10}{12-10}=\frac{94}{2}=47\)
(Những số giống nhau trong hai ngoặc trừ hết cho nhau)
a: 891+(359+109)
=891+109+359
=1000+359=1359
b: \(\dfrac{19}{11}+\left(\dfrac{5}{13}+\dfrac{3}{11}\right)\)
\(=\dfrac{19}{11}+\dfrac{3}{11}+\dfrac{5}{13}\)
\(=2+\dfrac{5}{13}=\dfrac{31}{13}\)
c: \(\dfrac{17,8\cdot3,7-7,8\cdot4,8+5,7\cdot17,8-4,6\cdot7,8}{11,2+12,3+13,4-12,6-11,5-10,4}\)
\(=\dfrac{17,8\left(3,7+5,7\right)-7,8\left(4,8+4,6\right)}{11,2-10,4+12,3-11,5+13,4-12,6}\)
\(=\dfrac{17,8\cdot9,4-7,8\cdot9,4}{0,8+0,8+0,8}=\dfrac{9,4\cdot10}{2,4}=\dfrac{94}{2,4}=\dfrac{940}{24}=\dfrac{235}{6}\)
\(a,891+\left(359+109\right)\\ =\left(891+109\right)+359\\ =1000+359\\ =1359\\ b,\dfrac{19}{11}+\left(\dfrac{5}{13}+\dfrac{3}{11}\right)\\ =\left(\dfrac{19}{11}+\dfrac{3}{11}\right)+\dfrac{5}{13}\\ =2+\dfrac{5}{13}\\ =\dfrac{26}{13}+\dfrac{5}{13}\\ =\dfrac{31}{13}\\ c,\dfrac{17,8\times3,7-7,8\times4,8+5,7\cdot17,8-4,6\times7,8}{11,2+12,3+13,4-12,6-11,5-10,4}\\ =\dfrac{17,8\times\left(3,7+5,7\right)-7,8\times\left(4,8+4,6\right)}{\left(11,2+12,3-11,5\right)+\left(13,4-10,4\right)-12,6}\\ =\dfrac{17,8\times9,4-7,8\times9,4}{12+3-12,6}\\ =\dfrac{9,4\times\left(17,8-7,8\right)}{2,4}\\ =\dfrac{94}{2,4}\\ =\dfrac{940}{24}=\dfrac{235}{6}\)