\(\dfrac{4}{7}x8=\)
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\(=\dfrac{2}{7}\cdot\dfrac{3}{2}\cdot8-\dfrac{4}{5}=\dfrac{24}{7}-\dfrac{4}{5}=\dfrac{120-28}{35}=\dfrac{92}{35}\)
Bài 7: Tính
a) \(4\dfrac{2}{5}\times8\dfrac{3}{4}-2\dfrac{3}{4}\)
\(=\dfrac{22}{5}\times\dfrac{35}{4}-\dfrac{11}{4}\)
\(=\dfrac{77}{2}-\dfrac{11}{4}\)
\(=\dfrac{143}{4}\)
b) \(2\dfrac{2}{3}+1\dfrac{2}{5}-\dfrac{2}{15}\)
\(=\dfrac{8}{3}+\dfrac{7}{5}-\dfrac{2}{15}\)
\(=\dfrac{47}{15}-\dfrac{2}{15}\)
\(=\dfrac{45}{15}=3\)
c) \(3\dfrac{1}{3}-2\dfrac{2}{3}+1\dfrac{5}{6}\)
\(=\dfrac{10}{3}-\dfrac{8}{3}+\dfrac{11}{6}\)
\(=\dfrac{2}{3}+\dfrac{11}{6}\)
\(=\dfrac{15}{6}\)
a)
x 8 = − 1 4 x .4 = − 1.8 x .4 = − 8 x = − 2
b)
6 x − 3 = 9 2 x − 7 6. 2 x − 7 = 9. x − 3 12 x − 42 = 9 x − 27 3 x = 15 x = 5
c)
x − 1 2 2 = 4 x − 1 2 = ± 2
TH1:
x − 1 2 = 2 x = 5 2
TH2:
x − 1 2 = − 2 x = − 3 2
Vậy x = 5 2 hoặc x = − 3 2
\(B=\dfrac{2\cdot8^4\cdot27^2+4\cdot6^9}{2^7\cdot6^7+2^7\cdot40\cdot9^4}\)
\(B=\dfrac{2\cdot\left(2^3\right)^4\cdot\left(3^3\right)^2+2^2\cdot2^9\cdot3^9}{2^7\cdot2^7\cdot3^7+2^7\cdot2^2\cdot10\cdot\left(3^2\right)^4}\)
\(B=\dfrac{2\cdot2^{12}\cdot3^6+2^{11}\cdot3^9}{2^{14}\cdot3^7+2^9\cdot10\cdot3^8}\)
\(B=\dfrac{2^{11}\cdot2^2\cdot3^6+2^{11}\cdot3^6\cdot3^3}{2^{11}\cdot2^3\cdot3^6\cdot3+\dfrac{2^{11}}{2^2}\cdot10\cdot3^6\cdot3^2}\)
\(B=\dfrac{\left(2^{11}\cdot3^6\right)\left(2^2+3^3\right)}{\left(2^{11}\cdot3^6\right)\left(2^3\cdot3\right)+2^{11}\cdot\dfrac{1}{2^2}\cdot10\cdot3^6\cdot3^2}\)
\(B=\dfrac{\left(2^{11}\cdot3^6\right)\left(2^2+3^3\right)}{\left(2^{11}\cdot3^6\right)\left(2^3\cdot3+\dfrac{1}{2^2}\cdot10\cdot3^2\right)}\)
\(B=\dfrac{2^2+3^3}{2^3\cdot3+\dfrac{1}{2^2}\cdot10\cdot3^2}\)
\(B=\dfrac{4+27}{8\cdot3+\dfrac{1}{4}\cdot10\cdot9}\)
\(B=\dfrac{31}{24+\dfrac{1}{4}\cdot90}\)
\(B=\dfrac{31}{24+\dfrac{45}{2}}\)
\(B=\dfrac{31}{\dfrac{48}{2}+\dfrac{45}{2}}\)
\(B=\dfrac{31}{\dfrac{93}{2}}\)
\(B=31\div\dfrac{93}{2}\)
\(B=31\times\dfrac{2}{93}\)
\(B=\dfrac{2}{3}\)
(-4).8.25.(-125).7
=[(-4x.25]x[8.(-125)]
=-100 x -1000
=100000
32/7
hello @acc4 :3
\(\dfrac{4}{7}\times8=\dfrac{4}{7}\times\dfrac{8}{1}=\dfrac{32}{7}\)