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B=4*13/9*3-4/3*40/9
B=4/3*13/9-4/3*40/9
B=4/3*(13/9-40/9)
B=4/3*(-27)/9
B=4*(-3)/9
B=-4
A=6/7 + 1/7.(2/7+5/7)
A=6/7 + 1/7.7/7=6/7+1/7.1
A=6/7+1/7=7/7=1
A=(2+3+...+13)-(1+2+...+12)=2+3+...+13-1-2-...-12=(13-1)+(2-2)+(3-3)+...+(12-12)=12
\(\frac{-5}{7}.\frac{2}{11}+\frac{-5}{7}.\frac{9}{11}\) \(\frac{3}{5}.\frac{2}{8}+\frac{-6}{16}.\frac{2}{5}+\frac{-6}{15}:\left(-16\right)\)
\(=\frac{-5}{7}\left(\frac{2}{11}+\frac{9}{11}\right)\) \(=\frac{3}{20}+\frac{-3}{20}+\frac{1}{40}\)
\(=\frac{-5}{7}.1=\frac{-5}{7}\) \(=0+\frac{1}{40}=\frac{1}{40}\)
\(x-\frac{2}{5}=0,24\) \(\left(\frac{7}{3}x-0,6\right):3\frac{2}{5}=1\)
\(\Rightarrow x=0,24+\frac{2}{5}=\frac{16}{25}\) \(\Rightarrow\left(\frac{7}{3}x-0,6\right):\frac{17}{5}=1\)
vậy x = 16/25 \(\Rightarrow\frac{7}{3}x-0,6=\frac{17}{5}\)
\(\Rightarrow\frac{7}{3}x=\frac{17}{5}+0,6=4\)
\(\Rightarrow x=4:\frac{7}{3}=\frac{12}{7}\)
vậy x = 12/7
d)\(\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{97.99}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{97}-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}\)
\(=\frac{32}{99}\)
a) 11/4x(-0,4)x-1,6x11/4
= -1,1 x-1,6x11/4
= 1,76x11/4
= 4,48
b) 6/19x-7/11+6/19x-4/19x-4/11+13/19
= -42/209+96/3971+13/19
= -702/3971+13/19
= 2015/3971
c) 2/5x3/7+-3/7x3/5+3/7
= 3/7x(2/5+3/5+1)
= 3/7x(1+1)
= 3/7x2
= 6/7
d) 2/3x5+2x5/7+2/7x9+....+2/97x99
= 1/3-1/5+1/5-1/7+1/7-1/9+......+1/97-1/99
= 1/3-1/99
= 32/99
Mình nghĩ đề là : 2/8 sẽ hay hơn.
\(B=\dfrac{5}{2}+\dfrac{6}{11}+\dfrac{2}{8}+\dfrac{7}{2}+\dfrac{6}{8}+\dfrac{5}{11}\)
\(=\left(\dfrac{5}{2}+\dfrac{7}{2}\right)+\left(\dfrac{6}{11}+\dfrac{5}{11}\right)+\left(\dfrac{2}{8}+\dfrac{6}{8}\right)\)
\(=6+1+1=8\)
b: \(B=\dfrac{5}{2}-\dfrac{7}{2}+\dfrac{3}{8}+\dfrac{6}{8}+\dfrac{-6}{11}-\dfrac{5}{11}=-2-1+\dfrac{9}{8}=\dfrac{9}{8}-3=-\dfrac{15}{8}\)
c: \(C=\left(\dfrac{4}{3}+\dfrac{7}{3}+\dfrac{1}{3}\right)+\left(\dfrac{2}{5}+\dfrac{3}{5}\right)=4+1=5\)
d: \(D=\dfrac{4}{19}\left(\dfrac{-5}{6}-\dfrac{7}{12}\right)-\dfrac{40}{57}\)
\(=\dfrac{4}{19}\cdot\dfrac{-17}{12}-\dfrac{40}{57}=-1\)
e: \(E=\dfrac{1}{3}\left(\dfrac{4}{5}-\dfrac{9}{5}\right)+\dfrac{2}{3}=\dfrac{2}{3}-\dfrac{1}{3}=\dfrac{1}{3}\)
N = 1 3