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\(=\dfrac{2}{3}+\dfrac{1}{3}.\left(\dfrac{7}{18}\right):\dfrac{7}{12}\)
\(=\dfrac{2}{3}+\dfrac{7}{54}:\dfrac{7}{12}\)
\(=\dfrac{2}{3}+\dfrac{2}{9}\)
\(=\dfrac{8}{9}\)
B=1+2-3-4+5+6-7-8+9+10-11-12+...-299-300+301+302
B=1+2+2*150=303
\(A=\frac{-7}{21}+\left(1+\frac{1}{3}\right)\)
\(A=\frac{-7}{21}+1+\frac{1}{3}\)
\(A=\left(\frac{-7}{21}+\frac{1}{3}\right)+1\)
\(A=0+1\\ A=1\)
\(B=\frac{2}{15}+\left(\frac{5}{9}+-\frac{6}{9}\right)\)
\(B=\frac{2}{15}+\frac{-1}{9}\)
\(B=\frac{1}{45}\)
\(C=\left(\frac{-1}{5}+\frac{3}{12}\right)+\frac{-3}{4}\)
\(C=\frac{-1}{5}+\left(\frac{3}{12}+\frac{-3}{4}\right)\)
\(C=\frac{-1}{5}+\frac{-1}{2}\)
\(C=\frac{-7}{10}\)
\(A=\frac{-7}{21}+\left(1+\frac{1}{3}\right)\)
\(A=\frac{-1}{3}+1+\frac{1}{3}\)
\(A=1\)
\(B=\frac{2}{15}+\left(\frac{5}{9}+\frac{-6}{9}\right)\)
\(B=\frac{2}{15}+\frac{5}{9}+\frac{-6}{9}\)
\(B=\frac{2}{15}+\frac{5}{9}+\frac{-10}{15}\)
\(B=\frac{-8}{15}+\frac{5}{9}\)
\(B=\frac{1}{45}\)
\(C=\left(\frac{-1}{5}+\frac{3}{12}\right)+\frac{-3}{4}\)
\(C=\frac{-1}{5}+\frac{3}{4}+\frac{-3}{4}\)
\(C=\frac{-1}{5}\)
\(\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}.\left(-2\right)^2=\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}.4=\dfrac{6}{7}+\dfrac{1}{8}-\dfrac{3}{4}=\dfrac{5}{56}\)
\(\dfrac{2}{3}+\dfrac{1}{3}.\left(-\dfrac{4}{9}+\dfrac{5}{6}\right):\dfrac{7}{12}=\dfrac{2}{3}+\dfrac{1}{3}.\dfrac{7}{18}:\dfrac{7}{12}=\dfrac{2}{3}+\dfrac{2}{9}=\dfrac{8}{9}\)
a) \(A=\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right).\left(1-\dfrac{1}{4}\right).\left(1-\dfrac{1}{5}\right)...\left(1-\dfrac{1}{2003}\right).\left(1-\dfrac{1}{2004}\right)\)
\(=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}...\dfrac{2002}{2003}.\dfrac{2003}{2004}\)
\(=\dfrac{1}{2004}\)
b) \(B=5\dfrac{9}{10}:\dfrac{3}{2}-\left(2\dfrac{1}{3}.4\dfrac{1}{2}-2.2\dfrac{1}{3}\right):\dfrac{7}{4}\)
\(=\dfrac{59}{10}:\dfrac{3}{2}-\left(\dfrac{7}{3}.\dfrac{9}{2}-2.\dfrac{7}{3}\right).\dfrac{4}{7}\)
\(=\dfrac{59}{15}-\left(\dfrac{21}{2}-\dfrac{14}{3}\right).\dfrac{4}{7}\)
\(=\dfrac{59}{15}-\dfrac{35}{6}.\dfrac{4}{7}\)
\(=\dfrac{59}{15}-\dfrac{10}{3}\)
\(=\dfrac{3}{5}\)
\(\frac{-5}{7}.\frac{2}{11}+\frac{-5}{7}.\frac{9}{11}\) \(\frac{3}{5}.\frac{2}{8}+\frac{-6}{16}.\frac{2}{5}+\frac{-6}{15}:\left(-16\right)\)
\(=\frac{-5}{7}\left(\frac{2}{11}+\frac{9}{11}\right)\) \(=\frac{3}{20}+\frac{-3}{20}+\frac{1}{40}\)
\(=\frac{-5}{7}.1=\frac{-5}{7}\) \(=0+\frac{1}{40}=\frac{1}{40}\)
\(x-\frac{2}{5}=0,24\) \(\left(\frac{7}{3}x-0,6\right):3\frac{2}{5}=1\)
\(\Rightarrow x=0,24+\frac{2}{5}=\frac{16}{25}\) \(\Rightarrow\left(\frac{7}{3}x-0,6\right):\frac{17}{5}=1\)
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
a)2+4+6+...+998 b)1+5+9+...+1001 c)1+3+5+...+997
Số số hạng là:(998-2):2+1=499 Số số hạng:(1001-1):4+1=251 số số hạng:(997-1):2+1=499
tổng là: (998+2)x499:2=249500 tổng là:(1001+1)x251:2=125751 tổng là:(997+1)x499:2=249001
d)2+9+16+...+7352
số số hạng:(7352-2):7+1=1051
tổng là:(7352+2)x1051:2=3864527
Cho cậu công thức nè: Số số hạng: (số cuối trừ số đầu)chia khỏang cách rồi cộng 1
Tổng:(số cuối cộng số đầu) nhân với số số hạng vừa tính ra rồi chia cho 2
Cảm ơn cậu nhé T^T