\(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+\frac{3}{7.9}\)
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Thuc hien phep tinh : A= \(\frac{-1}{54}-\frac{3}{1.3}-\frac{3}{3.5}-\frac{3}{5.7}-\frac{3}{7.9}-...-\frac{3}{79.81}\)
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a) (frac{1}{2}+frac{1}{6}+frac{1}{12}+frac{1}{20}+...+frac{1}{132}) . x frac{1}{3}
b) (frac{2}{1.3}+frac{2}{3.5}+frac{2}{5.7}+frac{2}{7.9}+frac{2}{9.11}) : x frac{2}{3}
c) (frac{1}{1.3}+frac{1}{3.5}+frac{1}{5.7}+frac{1}{7.9}+frac{1}{9.11}) . x frac{2}{3}
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a) (\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{132}\)) . x =\(\frac{1}{3}\)
b) (\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)) : x = \(\frac{2}{3}\)
c) (\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)) . x = \(\frac{2}{3}\)
Mik đang cần gấp
a)(\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{11.12}\)). x=\(\frac{1}{3}\)
(1-\(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{11}_{ }+\frac{1}{12}\)).x=\(\frac{1}{3}\)
(1+\(\frac{1}{12}\)).x=\(\frac{1}{3}\)
x=\(\frac{1}{3}:\frac{13}{12}\)
x=\(\frac{4}{13}\)
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b)( \(2-\frac{2}{3}+\frac{2}{3}-\frac{2}{5}+...+\frac{2}{9}-\frac{2}{11}_{ }\)):x =\(\frac{2}{3}\)
Giống câu a
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Tính B=\(\frac{1.3}{3.5}+\frac{2.4}{5.7}+\frac{3.5}{7.9}+.....+\frac{49.51}{99.101}\)
tínhA frac{11}{1.3}+ frac{47}{3.5}+ frac{107}{5.7}+ frac{191}{7.9}+...+ frac{971}{17.19}B frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}- frac{5^{10}.7^3-5^{10}.7^4}{5^9.7^3-5^9.7^3.2^3}C 1.3+3.5+5.7+...+ (2n-1)(2n+1)Giúp mình vs nhanh nhanh nha các bạn
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tính
A =\(\frac{11}{1.3}\)+ \(\frac{47}{3.5}\)+ \(\frac{107}{5.7}\)+ \(\frac{191}{7.9}\)+...+ \(\frac{971}{17.19}\)
B = \(\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}\)- \(\frac{5^{10}.7^3-5^{10}.7^4}{5^9.7^3-5^9.7^3.2^3}\)
C = 1.3+3.5+5.7+...+ (2n-1)(2n+1)
Giúp mình vs nhanh nhanh nha các bạn
\(F=\frac{11}{1.3}+\frac{47}{3.5}+\frac{107}{5.7}+\frac{191}{7.9}+...+\frac{971}{17.19}\)
\(B=\frac{2^3}{3.5}+\frac{2^3}{5.7}+\frac{2^3}{7.9}+......+\frac{2^3}{101.103}\)
\(B=\frac{2^3}{3.5}+\frac{2^3}{5.7}+....+\frac{2^3}{101.103}\)
\(\Rightarrow\frac{1}{2^2}.B=\frac{2}{3.5}+\frac{2}{4.7}+....+\frac{2}{101.103}\)
\(\Rightarrow\frac{1}{4}.B=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{101}-\frac{1}{103}\)
\(\Rightarrow\frac{1}{4}.B=\frac{1}{3}-\frac{1}{103}=\frac{100}{309}\)
\(\Rightarrow B=\frac{100}{309}:\frac{1}{4}=\frac{400}{309}\)
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\(=2^2\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{101.103}\right)\)
\(=4\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{101}-\frac{1}{103}\right)\)
\(=4\left(\frac{1}{3}-\frac{1}{103}\right)\)
\(=4\cdot\frac{100}{309}=\frac{400}{309}\)
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\(A=\frac{3}{3.5}+\frac{3}{5.7}+\frac{3}{7.9}+...+\frac{3}{49.51}=?\)
NHẦM GIẢI LẠI :
\(A=\frac{3}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)=\frac{3}{2}.\left(\frac{1}{3}-\frac{1}{51}\right)=\frac{3}{2}.\frac{16}{51}=\frac{8}{17}\)
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D=\(\frac{1}{54}-\frac{3}{1.3}-\frac{3}{3.5}-\frac{3}{5.7}-....-\frac{3}{79.81}\)
D = \(\frac{1}{54}-\frac{3}{1.3}-\frac{3}{3.5}-\frac{3}{5.7}-...-\frac{1}{79.81}\)
\(=\frac{1}{54}-\left(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{79.81}\right)\)
\(=\frac{1}{54}-\frac{3}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{79.81}\right)\)
\(=\frac{1}{54}-\frac{3}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{79}-\frac{1}{81}\right)\)
\(=\frac{1}{54}-\frac{3}{2}.\left(1-\frac{1}{81}\right)\)
\(=\frac{1}{54}-\frac{3}{2}.\frac{80}{81}\)
\(=\frac{1}{54}-\frac{40}{27}\)
\(=\frac{1}{54}-\frac{80}{54}\)
\(=\frac{79}{54}\)
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C = \(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{99.101}\)
\(C=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{99.101}\)
\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-...-\frac{1}{99}+\frac{1}{99}-\frac{1}{101}\)
\(=\frac{1}{1}-\frac{1}{101}\)
\(=\frac{100}{101}\)
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\(C=\frac{3}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+......+\frac{1}{99}-\frac{1}{101}\right)\)
\(C=\frac{3}{2}\left(1-\frac{1}{101}\right)\)
\(C=\frac{3}{2}.\frac{100}{101}=\frac{150}{101}\)
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C = 31.3+33.5+35.7+...+399.101
C = 3\(\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\right)\)
C =3\(\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
C =3\(\left(1-\frac{1}{101}\right)\)
C=3 . \(\frac{100}{101}\)
C=\(\frac{300}{101}\)
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