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câu a :
\(\dfrac{-8}{24}+\dfrac{-4}{12}=\dfrac{-1}{3}+\dfrac{-1}{3}=\dfrac{-2}{3}\)
câu b :
\(\dfrac{-20}{35}+\dfrac{16}{24}=\dfrac{-4}{7}+\dfrac{2}{3}=\dfrac{2}{21}\)
câu c :
\(\dfrac{-3}{9}+\dfrac{-6}{15}=\dfrac{-1}{3}+\dfrac{-2}{5}=\dfrac{-11}{15}\)
câu d :
\(\dfrac{3}{13}-\dfrac{4}{10}=\dfrac{3}{13}-\dfrac{2}{5}=\dfrac{-11}{65}\)
câu e :
\(\dfrac{5}{17}-\dfrac{9}{15}=\dfrac{5}{17}-\dfrac{3}{5}=\dfrac{-26}{85}\)
câu g :
\(\dfrac{9}{18}-\dfrac{6}{15}+\dfrac{3}{-9}=\dfrac{9}{18}-\dfrac{6}{15}+\dfrac{-3}{9}\\ =\dfrac{1}{2}-\dfrac{2}{5}+\dfrac{-1}{3}=\dfrac{-7}{30}\)
câu h :
\(\dfrac{5}{4}-\dfrac{1}{2}+\dfrac{-7}{8}=\dfrac{10}{8}-\dfrac{4}{8}+\dfrac{-7}{8}=\dfrac{-1}{8}\)
b)=3^1+(3^2+3^3+3^4)+(3^5+3^6+3^7)+....+(3^58+3^59+3^60)
=3^1+(3^2.1+3^2.3+3^2.9)+(3^5.1+3^5.3+3^5.9)+......+(3^58.1+3^58.3+3^58.9)
=3^1+3^2.(1+3+9)+3^5.(1+3+9)+.....+3^58.(1+3+9)
=3+3^2.13+3^5.13+.........+3^58.13
=3.13.(3^2+3^5+....+3^58)
vi tich tren co thua so 13 nen tich do chia het cho 13
=
bai1
a) A=(31+32)+(33+34)+...+(359+360)
=(3^1.1+3^1.3)+...+(3^59.1+3^59.2)
=3^1.(1+3)+...+3^59.(1+3)
=3^1.4+....+3^59.4
=4.(3^1+...+3^59)
vi tich tren co thua so 4 nen tich do chia het cho 4
a; \(\dfrac{3}{11}\) + \(\dfrac{5}{-9}\) + \(\dfrac{4}{11}\) - \(\dfrac{4}{9}\) + \(\dfrac{3}{17}\) + \(\dfrac{15}{11}\)
= (\(\dfrac{3}{11}\) + \(\dfrac{4}{11}\) + \(\dfrac{15}{11}\)) - (\(\dfrac{5}{9}\) + \(\dfrac{4}{9}\)) + \(\dfrac{3}{17}\)
= 2 - 1 + \(\dfrac{3}{17}\)
= 1 + \(\dfrac{3}{17}\)
= \(\dfrac{20}{17}\)
c; N = \(\dfrac{\dfrac{5}{7}-\dfrac{5}{9}-\dfrac{5}{11}}{\dfrac{15}{7}+\dfrac{15}{9}+\dfrac{15}{11}}\)
Phải là - \(\dfrac{5}{7}\) chỗ tử số mới đúng em nhé!
up từng bài thôi,nhiều thế ko thánh nào làm cho đâu.thách nhau ak
\(A=\frac{1\cdot2+2\cdot4+3\cdot6+4\cdot8+5\cdot10}{3\cdot4+6\cdot8+9\cdot12+12\cdot16+15\cdot20}\)
\(=>A=\frac{1\cdot2+4\cdot1\cdot2+9\cdot1\cdot2+16\cdot1\cdot2+25\cdot1\cdot2}{3\cdot4+4\cdot3\cdot4+9\cdot3\cdot4+16\cdot3\cdot4+25\cdot3\cdot4}\)
\(=>A=\frac{\left(1+4+9+16+25\right)\cdot1\cdot2}{\left(1+4+9+16+25\right)\cdot3\cdot4}=\frac{1}{6}=\frac{111111}{666666}\)
Mà \(\frac{111111}{666666}< \frac{111111}{666665}\)
\(=>A< B\)
\(B=\frac{1.2+2.4+3.6+4.8+5.10}{3.4+6.8+9.12+12.16+15.20}\)
\(B=\frac{1.2+2^2.1.2+3^21.2+4^2.1.2+5^2.1.2}{3.4+2^23.4+3^23.4+4^23.4+5^23.4}\)
\(B=\frac{2.\left(1+2^2+3^2+4^2+5^2\right)}{12\left(1+2^2+3^2+4^2+5^2\right)}\)\(\Rightarrow B=\frac{2}{12}=\frac{1}{6}\)