tính nhanh tổng sau:(4/3.6)+(4/6.9)+...+(4/12.15)
NH
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4/3.6+ 4/6.9+ 4/9.12 + 4/12.15
1/3.6+1/6.9+1/9.12+1/12.15+1/15.18
\(=\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+...+\dfrac{1}{15}-\dfrac{1}{18}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{18}\right)=\dfrac{1}{3}\cdot\dfrac{5}{18}=\dfrac{5}{54}\)
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frac{4}{3.6}+ frac{4}{6.9}+ frac{4}{9.12}+frac{4}{12.15} 4(frac{1}{3.6}+frac{1}{6.9}+frac{1}{9.12}+frac{1}{12.15})frac{4}{3}( frac{1}{3}-frac{1}{6}+frac{1}{6}-frac{1}{9}+frac{1}{9}-frac{1}{12}+frac{1}{12}-frac{1}{15} )frac{4}{3}(frac{1}{3}-frac{1}{15})frac{4}{3}.frac{4}{15}frac{16}{45}mk làm đúng chưa
Đọc tiếp
\(\frac{4}{3.6}\)+ \(\frac{4}{6.9}\)+ \(\frac{4}{9.12}\)+\(\frac{4}{12.15}\)
= 4(\(\frac{1}{3.6}\)+\(\frac{1}{6.9}\)+\(\frac{1}{9.12}\)+\(\frac{1}{12.15}\))
=\(\frac{4}{3}\)( \(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+\frac{1}{12}-\frac{1}{15}\) )
=\(\frac{4}{3}\)(\(\frac{1}{3}-\frac{1}{15}\))
=\(\frac{4}{3}\).\(\frac{4}{15}\)
=\(\frac{16}{45}\)
mk làm đúng chưa
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Tính Nhanh : a) \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{99.101}\)
b) \(\frac{1}{3.6}+\frac{1}{6.9}+\frac{1}{9.12}+\frac{1}{12.15}\)
a) Đặt \(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)
\(2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(2A=1-\frac{1}{101}=\frac{100}{101}\)
\(A=\frac{100}{101}\div2=\frac{50}{101}\)
b) Đặt \(B=\frac{1}{3.6}+\frac{1}{6.9}+\frac{1}{9.12}+\frac{1}{12.15}\)
\(3B=\frac{3}{3.6}+\frac{3}{6.9}+\frac{3}{9.12}+\frac{3}{12.15}\)
\(3B=\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+\frac{1}{12}-\frac{1}{15}\)
\(3B=\frac{1}{3}-\frac{1}{15}=\frac{4}{15}\)
\(B=\frac{4}{15}\div3=\frac{4}{45}\)
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Đặt \(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)
\(2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(2A=1-\frac{1}{101}=\frac{100}{101}\)
\(A=\frac{100}{101}\div2=\frac{50}{101}\)
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\(\left(\frac{-1}{3.6}-\frac{1}{6.9}-\frac{1}{9.12}-\frac{1}{12.15}:\left|x\right|=\frac{-8}{15}\right)\)
TH
2 tháng 10 2021 lúc 19:39
BÀI 2 TÍNH :
A = \(\dfrac{6^3+3.6^2+3^3}{13}\)
B = \(\dfrac{4^6.9^5+6^9.120}{8^4-3^{12}.6^{11}}\)
\(A=\dfrac{6^3+3\cdot6^2+3^3}{13}\)
\(=\dfrac{3^3\cdot8+3^3\cdot4+3^3}{13}\)
=27
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\(\frac{4}{3.6}+\frac{4}{6.9}+.........\frac{4}{27.30}\)
\(\frac{4}{3.6}+\frac{4}{6.9}+...+\frac{4}{27.30}=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+...+\frac{1}{27}-\frac{1}{30}\right)\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{30}\right)=\frac{1}{2}.\frac{3}{10}=\frac{3}{20}\)
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\(\frac{4}{3.6}+\frac{4}{6.9}+....+\frac{4}{27.30}\)
\(=\frac{4}{3}\left(\frac{3}{3.6}+\frac{3}{6.9}+...+\frac{3}{27.30}\right)\)
\(=\frac{4}{3}\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+...+\frac{1}{27}-\frac{1}{30}\right)\)
\(=\frac{4}{3}\left(\frac{1}{3}-\frac{1}{30}\right)\)
\(=\frac{4}{3}.\frac{3}{10}\)
\(=\frac{2}{5}\)
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Lập các tỉ lệ thức có thể đucợ từ các đẳng thức sau:
a)3.18 = 6.9
b) (-15).4 = (-2).30
c) 12.15 = 4.45
d) 0,3 . 6,3 = 6.(-2,1)
a) 3 . 18 = 6 . 9
\(\Rightarrow\frac{3}{6}=\frac{9}{18}\)
\(\frac{3}{9}=\frac{6}{18}\)
\(\frac{9}{3}=\frac{18}{6}\)
\(\frac{18}{9}=\frac{6}{3}\)
mấy bài còn lại tương tự
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a) So sánh \(1995^n.1997^n\)với \(1996^{2n}\)
b) Rút gọn \(A=\frac{1}{3.6}+\frac{1}{6.9}+\frac{1}{9.12}+\frac{1}{12.15}+\frac{1}{15.18}+\frac{1}{18.21}+\frac{1}{21.24}\)
a/
\(1995^n.1997^n=\left(1995.1997\right)^n\)
\(1996^{2n}=\left(1996^2\right)^n\)
\(1995.1997=\left(1996-1\right).\left(1996+1\right)=1996^2-1\)
\(\Rightarrow1995.1997< 1996^2\Rightarrow1995^n.1997^n< 1996^{2n}\)
b/
\(A=\frac{1}{2.9}+\frac{1}{6.9}+\frac{1}{9.12}+\frac{1}{9.20}+\frac{1}{9.30}+\frac{1}{9.42}+\frac{1}{9.56}\)
\(A=\frac{1}{9}\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\right)\)
\(A=\frac{1}{9}\left(\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{8-7}{7.8}\right)\)
\(A=\frac{1}{9}\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{7}-\frac{1}{8}\right)\)
\(A=\frac{1}{9}\left(1-\frac{1}{9}\right)=\frac{1}{9}.\frac{8}{9}=\frac{8}{81}\)
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