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\(\frac{\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)}{\left(1-\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)}=\frac{\left(\frac{16}{16}+\frac{8}{16}+\frac{4}{16}+\frac{2}{16}+\frac{1}{16}\right)}{\left(\frac{16}{16}-\frac{8}{16}+\frac{4}{16}+\frac{2}{16}+\frac{1}{16}\right)}=\frac{\frac{31}{16}}{\frac{15}{16}}=\frac{31}{16}:\frac{15}{16}=\frac{31}{16}\times\frac{16}{15}=\frac{31}{15}\)
b)Ta có:\(A=1+\frac{1}{2.\left(1+2\right)}+\frac{1}{3.\left(1+2+3\right)}+...+\frac{1}{16.\left(1+2+3+...+16\right)}\)
\(=1+\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)+...+\frac{1}{16}.\left(1+2+3+...+16\right)\)
\(=1+\frac{1}{2}.3+\frac{1}{3}.6+...+\frac{1}{16}.136\)
\(=1+1,5+2+...+8,5\)
\(=\frac{\left(8,5+1\right).\left[\left(8,5-1\right):0,5+1\right]}{2}=76\)
B=\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\frac{1}{8^2}<\)
B=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)
B=\(1-\frac{1}{8}=\frac{8}{8}-\frac{7}{8}=\frac{1}{8}<2\)
Vậy 1/8<2 hay 1/8<16/8
Cách 1:
B=1/2+1/4+1/8+1/16+1/32+1/64
B=1-1/2 + 1/2-1/4 + 1/4-1/8 +1/8-1/16 + 1/16-1/32 + 1/32-1/64
B=1-1/64
B=63/64
Cách 2:
B=1/2+1/4+1/8+1/16+1/32+1/64
B=1/21+1/22+1/23+1/24+1/25+1/26
2B=1+1/21+1/2^2+1/2^3+1/2^4+1/2^5
2B-B=1-1/2^6
B=1-1/64
B=63/64
\(a,\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)
\(=1-\frac{1}{7}\)
\(=\frac{6}{7}\)
\(b,\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)
Ta có :
\(\frac{1}{2}=1-\frac{1}{2}\)
\(\frac{1}{4}=\frac{1}{2}-\frac{1}{4}\)
\(\frac{1}{8}=\frac{1}{4}-\frac{1}{8}\)
\(\frac{1}{16}=\frac{1}{8}-\frac{1}{16}\)
\(\frac{1}{32}=\frac{1}{16}-\frac{1}{32}\)
Thay vào ta có :
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+\frac{1}{16}-\frac{1}{32}\)
\(=1-\frac{1}{32}\)
\(=\frac{31}{32}\)
\(c,\)\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{256}\)
Ta có :
\(\frac{1}{2}=1-\frac{1}{2}\)
\(\frac{1}{4}=\frac{1}{2}-\frac{1}{4}\)
...................
\(\frac{1}{256}=\frac{1}{128}-\frac{1}{256}\)
Thay vào ta có :
\(=\)\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+...+\frac{1}{128}-\frac{1}{256}\)
\(=1-\frac{1}{256}\)
\(=\frac{255}{256}\)
\(a,\dfrac{4}{5}+\dfrac{2}{3}+\dfrac{1}{9}=\dfrac{12}{15}+\dfrac{10}{15}+\dfrac{1}{9}=\dfrac{22}{15}+\dfrac{1}{9}=\dfrac{66}{45}+\dfrac{5}{45}=\dfrac{71}{45}\)
\(b,\dfrac{3}{7}+\dfrac{11}{14}+\dfrac{19}{28}=\dfrac{12}{28}+\dfrac{22}{28}+\dfrac{19}{28}=\dfrac{53}{28}\)
\(c,\dfrac{1}{2}+\dfrac{1}{7}+\dfrac{-1}{5}=\dfrac{7}{14}+\dfrac{2}{14}+\dfrac{-1}{5}=\dfrac{9}{14}+\dfrac{-1}{5}=\dfrac{45}{70}+\dfrac{-14}{70}=\dfrac{31}{70}\)
\(d,\dfrac{7}{8}+\dfrac{5}{16}+\dfrac{-3}{4}=\dfrac{14}{16}+\dfrac{5}{16}+\dfrac{-12}{16}=\dfrac{7}{16}\)
\(e,\dfrac{1}{4}+\dfrac{5}{12}+\dfrac{-1}{13}=\dfrac{3}{12}+\dfrac{5}{12}+\dfrac{-1}{13}=\dfrac{8}{12}+\dfrac{-1}{13}=\dfrac{2}{3}+\dfrac{-1}{13}=\dfrac{26}{39}+\dfrac{-3}{39}=\dfrac{23}{39}\)
\(g,\dfrac{2}{3}+\dfrac{3}{8}+\dfrac{-5}{12}=\dfrac{16}{24}+\dfrac{9}{24}+\dfrac{-5}{12}=\dfrac{25}{24}+\dfrac{-5}{12}=\dfrac{25}{24}+\dfrac{-10}{24}=\dfrac{15}{24}\)
A= 13;21;34
B= 37;70;135
C= 64;128;256
D= 22;29;37
E= 53;68;75
F= 127;255;511
G= 49;64;81
H= 324;841;2209
I= chịu
k cho mk nha!
a, A={x thuộc các số nguyên tố |2<hoặc bằng x<hoặc bằng 7}
oặc A={x thuộc R |(x^2-5*x+6)*(x^2-12*x+35)=0}
b,B={x thuộc Z | -3<hoặc bằng x<hoặc bằng 3}
c,C={5*x thuộc Z |-1<hoặc bằng x<hoặc bằng 3}
\(a)\)
\(2^{2x-1}+6.2^8=14.2^8\)
\(\Leftrightarrow2^{2x-1}=14.2^8-6.2^8\)
\(\Leftrightarrow2^{2x-1}=8.2^8\)
\(\Leftrightarrow2^{2x-1}=2^3.2^8\)
\(\Leftrightarrow2x-1=11\)
\(\Leftrightarrow2x=11+1\)
\(\Leftrightarrow x=\frac{12}{2}\)
\(\Leftrightarrow x=6\)
\(b)\)
\(A=\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)\left(1-\frac{1}{16}\right)\left(1-\frac{1}{25}\right)...\left(1-\frac{1}{196}\right)\left(1-\frac{1}{225}\right)\)
\(\Leftrightarrow A=\frac{3}{4}.\frac{8}{9}...\frac{224}{225}\)
\(\Leftrightarrow A=\frac{1.3}{2.2}.\frac{2.4}{3.3}...\frac{14.16}{15.15}\)
\(\Leftrightarrow A=\frac{1.2.3.4...14.16}{2.2.3.3...25.25}\)
\(\Leftrightarrow A=\frac{1.2.3...14}{2.3.4...15}.\frac{3.4.5...16}{2.3.4...15}\)
\(\Leftrightarrow A=\frac{1}{15}.\frac{16}{2}\)
\(\Leftrightarrow A=\frac{8}{15}\)