37.39 + 62.21 − 11.39 − 21.23
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a: \(=\left(5^2+3\right):7=4\)
b: \(=7^2-3^2+8\cdot25=40+200=240\)
37.39+78+37.39
=37.39 .(1+1)+78
=37.39.2+78
=37.78+78
=78.(37+1)=78.38=2964
\(a,=600+72+18=690\\ b,=25+5-20=10\\ c,=45-28+83=100\\ d,=25+36+1=62\\ e,=151-8+3=146\)
a)600+72+18=690
b)25+5-20=10
c)45-28+83=100
d)25+36+1=62
e)151-8+3=146
\(2B=2\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{23}-\frac{1}{25}\right)\)
\(2B=2\left(\frac{1}{3}-\frac{1}{25}\right)\)
\(2B=2\times\frac{22}{75}\)
\(B=\frac{44}{75}\)
\(=\frac{23^{10}\left(1-23\right)}{23^{10}\left(23+21\right)}=\frac{-22}{44}=\frac{-1}{2}\)
\(\frac{23^{10}-23^{11}}{23^{11}+21.23^{10}}\)
\(=\frac{23^{10}.\left(1-23\right)}{23^{10}.\left(23+21\right)}\)
\(=\frac{-22}{44}=\frac{-22}{22.2}=\frac{-1}{2}\)
\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}\)
\(A=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
\(A=\dfrac{1}{1}-\dfrac{1}{50}=\dfrac{49}{50}\)
\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
\(A=\dfrac{1}{1}-\dfrac{1}{50}=\dfrac{50}{50}-\dfrac{1}{50}=\dfrac{49}{50}\)
A=\(\dfrac{2}{1.3}-\dfrac{2}{3.5}-\dfrac{2}{5.7}-.....-\dfrac{2}{23.25}-\dfrac{1}{27}\)
A=\(\dfrac{2}{3}-\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+....+\dfrac{2}{23.25}\right)-\dfrac{1}{27}\)
A=\(\dfrac{2}{3}-\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+......+\dfrac{1}{23}-\dfrac{1}{25}\right)-\dfrac{1}{27}\)
A=\(\dfrac{2}{3}-\left(\dfrac{1}{3}-\dfrac{1}{25}\right)-\dfrac{1}{27}\)
A=\(\dfrac{2}{3}-\dfrac{22}{75}-\dfrac{1}{27}\)
A=\(\dfrac{227}{675}\)
đề sai rùi bn ơi, đáng ra phải là: 53.39+37.39....... mới đúng
kết quả là 4200
62×21=1302
37×39=1443
11×39=429
21×23=483
=1443+1302−429−483=1833
=1833
xin 1 gp
1833
mik đang gấp nên ko kịp trình bày cho bạn nha!
sorry bạn!!!