4/5.7+7/7.9+...+4/57.59
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Đặt : A = \(\frac{4}{5.7}+\frac{4}{7.9}+...+\frac{4}{59.61}\)
A = \(2.\)\(\left(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{59.61}\right)\)
A = 2 . ( \(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{59}-\frac{1}{61}\))
A = 2 . \(\left(\frac{1}{5}-\frac{1}{61}\right)\)
A = 2 . \(\frac{56}{305}\)= \(\frac{112}{305}\)
\(=4\left(\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+...+\frac{1}{53.55}\right)\)
\(=4\left(\frac{1}{5}-\frac{1}{5}+\frac{1}{7}-\frac{1}{7}+...+\frac{1}{53}-\frac{1}{55}\right)\)
\(=4\left(\frac{1}{5}-\frac{1}{55}\right)\)
\(=4.\frac{2}{11}\)
\(=\frac{8}{11}\)
\(\frac{4}{3\cdot5}+\frac{4}{5\cdot7}+\frac{4}{7\cdot9}+...+\frac{4}{97\cdot99}\)
\(=\frac{2\cdot2}{3\cdot5}+\frac{2\cdot2}{5\cdot7}+\frac{2\cdot2}{7\cdot9}+...+\frac{2\cdot2}{97\cdot99}\)
\(=\frac{2}{3}+\frac{2}{5}-\frac{2}{5}+\frac{2}{7}-\frac{2}{7}+\frac{2}{9}-...+\frac{2}{97}-\frac{2}{99}\)
\(=\frac{2}{3}-\frac{2}{99}\)
\(=\frac{64}{99}\)
Ta có :
4/5 . 7 + 4/7 . 9 + ...+ 4/59 . 61
= 2 . ( 2/5 . 7 + 2/7 . 9 + ...+ 2/59 . 61 )
= 2 . ( 1/5 - 1/7 + 1/7 - 1/9 + ...+ 1/59 - 1/61 )
= 2 . ( 1/5 - 1/61 )
= 2 . 56/305
= 112/305
Tham khảo nha !!!
\(\dfrac{4}{3.5}+\dfrac{4}{5.7}+\dfrac{4}{7.9}+...+\dfrac{4}{97.95}\)
\(=2\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{95.97}\right)\)
\(=2\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{95}-\dfrac{1}{97}\right)\)
\(=2\left(\dfrac{1}{3}-\dfrac{1}{97}\right)\)
\(=2.\dfrac{94}{291}=\dfrac{188}{291}\)
\(=2\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{95\cdot97}\right)\\ =2\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{95}-\dfrac{1}{97}\right)\\ =2\left(\dfrac{1}{3}-\dfrac{1}{97}\right)=2\cdot\dfrac{94}{291}=\dfrac{188}{291}\)
a/
\(A=3^2+3^2.2^2+3^2.3^2+3^2.4^2+...+3^2.30^2=\)
\(=3^2\left(1^2+2^2+3^2+...+30^2\right)\)
Đăt biểu thức trong dấu ngoặc là B
\(B=1\left(2-1\right)+2\left(3-1\right)+3\left(4-1\right)+...+30\left(31-1\right)=\)
\(=1.2+2.3+3.4+30.31-\left(1+2+3+...+30\right)=\)
\(C=1+2+3+...+30=\dfrac{30\left(1+30\right)}{2}=465\)
\(D=1.2+2.3+3.4+...+30.31\)
\(3D=1.2.3+2.3.3+3.4.3+...+30.31.3=\)
\(=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+30.31.\left(32-29\right)=\)
\(=1.2.3-1.2.3+2.3.4-2.3.4+3.4.5-...-29.30.31+30.31.32=\)
\(=30.31.32\Rightarrow D=\dfrac{30.31.32}{3}=10.31.32\)
\(\Rightarrow A=3^2\left(C-D\right)=3^2\left(10.31.32-465\right)\)
b/
Đặt biểu thức là A
\(\dfrac{A}{2}=\dfrac{5-3}{3.5}+\dfrac{7-5}{5.7}+\dfrac{9-7}{7.9}+...+\dfrac{39-37}{37.39}=\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{37}-\dfrac{1}{39}=\)
\(=\dfrac{1}{3}-\dfrac{1}{39}=\dfrac{12}{39}\Rightarrow A=\dfrac{2.12}{39}=\dfrac{24}{39}=\dfrac{8}{13}\)
\(A=3.3+6.6+9.9+...+90.90\)
\(A=3^2\left(1+2^2+3^2+...+10^2\right)\)
\(A=9.\dfrac{10.\left(10+1\right)\left(2.10+1\right)}{6}\)
\(A=3.\dfrac{10.11.21}{2}\)
\(A=3465\)
4/5.7+4/7.9+...+4/59.61
=2.(2/5.7+2/7.9+...+2/59.61)
=2.(1/5-1/7+1/7-1/9+...+1/59-1/61)
=2.(1/5-1/61)
=2.56/305
=112/203
A=( 2/5.7+2/7.9+.........+2/59.61).2
A = (1/5-1/7+1/7-1/9+.......+1/59-1/61).2
A= ( 1/5-1/61)2