bài 1 :
a) 90^2/15^2 ; b) 790^4/79^4
bài 2 :
so sánh 2^24 và 3^16
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\(1,A=\dfrac{2}{3\cdot7}+\dfrac{2}{7\cdot11}+\dfrac{2}{11\cdot15}+...+\dfrac{2}{99\cdot103}\\ 2A=\dfrac{4}{3\cdot7}+\dfrac{4}{7\cdot11}+\dfrac{4}{11\cdot15}+...+\dfrac{4}{99\cdot103}\\ 2A=\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+...+\dfrac{1}{99}-\dfrac{1}{103}\\ 2A=\dfrac{1}{3}-\dfrac{1}{103}=\dfrac{100}{309}\\ A=\dfrac{100}{309}\cdot\dfrac{1}{2}=\dfrac{50}{309}\)
\(2,A=\dfrac{7}{2}+\dfrac{7}{6}+\dfrac{7}{12}+\dfrac{7}{20}+\dfrac{7}{30}+\dfrac{7}{42}+\dfrac{7}{56}+\dfrac{7}{72}+\dfrac{7}{90}\\ A=7\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{9\cdot10}\right)\\ A=7\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)\\ A=7\left(1-\dfrac{1}{10}\right)=7\cdot\dfrac{9}{10}=\dfrac{63}{10}\)
\(A=\frac{1}{25.27}+\frac{1}{27.29}+\frac{1}{29.31}+...+\frac{1}{83.85}\)
\(\Rightarrow2A=\frac{2}{25.27}+\frac{2}{27.29}+\frac{2}{29.31}+...+\frac{2}{83.85}\)
\(2A=\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+\frac{1}{29}-\frac{1}{31}+...+\frac{1}{83}-\frac{1}{85}\)
\(2A=\frac{1}{25}-\frac{1}{85}\)
\(2A=\frac{12}{425}\)
\(A=\frac{12}{425}:2\)
\(A=\frac{6}{425}\)
\(C=\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+...+\frac{2}{399}\)
\(C=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{19.21}\)
\(C=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{19}-\frac{1}{21}\)
\(C=\frac{1}{3}-\frac{1}{21}\)
\(C=\frac{2}{7}\)
CÂU B LÀM TƯƠNG TỰ NHA
HOK TOT
b1/A=25/1.6+25/6.11+25/11.16+....+25/41.46
=5.(5/1.6+5/6.11+5/11.16+...+5/41.46)
=5.(1/1-1/6+1/6-1/11+1/11-1/16+....+1/41-1/46)
=5.(1/1-1/46)
=5.45/46
=225/46
\(\dfrac{1}{2};\dfrac{5}{10};\dfrac{45}{90};\dfrac{90}{180}\)
Ta có : 224 = (26)4. =64 4 (1). Ta có : 316 = (34)4= =81 4 (2). Từ (1) và (2),suy ra : 2 24<3 16.