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Giusp m với. Tìm số A biết. 121:A dư 10, 61 : A dư 10. giải thích cách làm theo lớp 6 b nhé.Thank
\(\frac{9}{27}+\frac{8}{24}+\frac{18}{27}+\frac{16}{24}+\frac{2}{3}\)
=\(\frac{1}{3}+\frac{1}{3}+\frac{2}{3}+\frac{2}{3}+\frac{2}{3}\)
\(=\frac{8}{3}\)
\(\frac{9}{27}+\frac{8}{24}+\frac{18}{27}-\frac{-16}{24}+\frac{2}{3}\)
\(=\frac{9}{27}+\frac{8}{24}+\frac{18}{27}+\frac{16}{24}+\frac{2}{3}\)
\(=\frac{1}{3}+\frac{1}{3}+\frac{2}{3}+\frac{2}{3}+\frac{2}{3}\)
\(=\left[\frac{1}{3}+\frac{2}{3}\right]+\left[\frac{1}{3}+\frac{2}{3}\right]+\frac{2}{3}\)
\(=1+1+\frac{2}{3}=2+\frac{2}{3}=\frac{8}{3}\)
a) \(\frac{x}{5}-\frac{x}{6}=\frac{3}{10}\\ \frac{6x}{30}-\frac{5x}{30}=\frac{3\cdot3}{10\cdot3}\\ \frac{x}{30}=\frac{9}{30}\\ \Rightarrow x=9\) Vậy x = 9
b) \(-1\frac{5}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\\ \frac{-32}{27}+\frac{24}{27}=\left(3x-\frac{7}{9}\right)^3\\ \left(3x-\frac{7}{9}\right)^3=\frac{-8}{27}\\ \left(3x-\frac{7}{9}\right)^3=\left(\frac{-2}{3}\right)^3\\ \Rightarrow3x-\frac{7}{9}=\frac{-2}{3}\\ 3x=\frac{-2}{3}+\frac{7}{9}\\ 3x=\frac{-6}{9}+\frac{7}{9}\\ 3x=\frac{1}{9}\\ x=\frac{1}{9}:3\\ x=\frac{1}{9\cdot3}\\ x=\frac{1}{27}\)Vậy \(x=\frac{1}{27}\)
a)1.2.3.4...9-1.2.3.4...8-1.2.3.4...8.8
=1.2.3.4...8(9-1-8)
=1.2.3.4...8.0
=0
b)(3.4.216)2/11.123.411-169=(3.22.216)2/11.213.222-236=32.24.232/11.235-236=32.226/235.(11-2)
=32.236/235.9=32.236/235.32=2
c)70.(131313/565656+131313/727272+131313/909090
=70.(13/56+13/72+13/90)
=70.39/70=39
d)1/4.9+1/9.14+1/14.19+...+1/64.69
=4/4.9.4+4/9.4.14+4/14.19.4+...+4/64.69.4.
=1/4.(4/4.9+4/9.14+4/14.19+...+4/64.69)
=1/4.(1/4-1/9+1/9-1/14+1/14-1/19+...+1/64-1/69)
=1/4.(1/4-1/69)
=1/4.65/276=65/1104
~~~~~~~~Chúc bạn học giỏi nhé !~~~~~~~~