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\(\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+...+\frac{1}{41.45}\)
\(=\frac{1}{4}.\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}\right)\)
\(=\frac{1}{4}.\left(1-\frac{1}{45}\right)\)
\(=\frac{1}{4}.\frac{44}{45}\)
\(=\frac{11}{45}\)
Đặt \(\frac{1}{1\cdot5}+\frac{1}{5\cdot9}+\frac{1}{9\cdot13}+\frac{1}{13\cdot17}+...+\frac{1}{41\cdot45}\) là A.
Ta có:
\(A=\frac{1}{1\cdot5}+\frac{1}{5\cdot9}+\frac{1}{9\cdot13}+\frac{1}{13\cdot17}+...+\frac{1}{41\cdot45}\)
\(4A=4\left(\frac{1}{1\cdot5}+\frac{1}{5\cdot9}+\frac{1}{9\cdot13}+\frac{1}{13\cdot17}+...+\frac{1}{41\cdot45}\right)\)
\(4A=\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+...+\frac{4}{41\cdot45}\)
\(4A=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+...+\frac{1}{41}-\frac{1}{45}\)
\(4A=1-\frac{1}{45}\)
\(4A=\frac{44}{45}\)
\(A=\frac{44}{45}:4\)
\(A=\frac{11}{45}\)
Vậy \(\frac{1}{1\cdot5}+\frac{1}{5\cdot9}+\frac{1}{9\cdot13}+\frac{1}{13\cdot17}+...+\frac{1}{41\cdot45}=\frac{11}{45}\)
\(\dfrac{1}{4}+\dfrac{1}{3}\times X=\dfrac{1}{2}\\ \dfrac{1}{3}\times X=\dfrac{1}{2}-\dfrac{1}{4}=\dfrac{1}{4}\\ X=\dfrac{1}{4}:\dfrac{1}{3}=\dfrac{1}{4}\times3=\dfrac{3}{4}\)
14+13×�=1213×�=12−14=14�=14:13=14×3=3441+31×X=2131×X=21−41=41X=41:31=41×3=43
\(\dfrac{3}{4}\times X-1=\dfrac{1}{2}\times X\\ \dfrac{3}{4}\times X-\dfrac{1}{2}\times X=1\\ \left(\dfrac{3}{4}-\dfrac{1}{2}\right)\times X=1\\ \dfrac{1}{4}\times X=1\\ X=1:\dfrac{1}{4}\\ X=4\)
\(\dfrac{3}{4}\times x-1=\dfrac{1}{2}\times x\\ \dfrac{3}{4}\times x-\dfrac{1}{2}\times x=1\\ x\times\left(\dfrac{3}{4}-\dfrac{1}{2}\right)=1\\ x\times\dfrac{1}{4}=1\\ x=1:\dfrac{1}{4}\\x=4\)
\(A=\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{420}\\ \Rightarrow A=\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+...+\dfrac{1}{20\times21}\\ \Rightarrow A=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{20}-\dfrac{1}{21}\\\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{20}=\dfrac{9}{20}\)
Sau khi rút gọn phải còn:
\(A=\dfrac{1}{2}-\dfrac{1}{21}\) (chứ anh)
\(A=\dfrac{19}{42}\)
Đề?
\(\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+...+\frac{1}{41.45}\)
\(=\frac{1}{4}.\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+...+\frac{1}{41}-\frac{1}{45}\right)\)
\(=\frac{1}{4}.\left(\frac{1}{5}-\frac{1}{45}\right)\)
\(=\frac{1}{4}.\frac{8}{45}=\frac{2}{45}\)
1/1.5 + 1/5.9 + 1/9.13 + ... + 1/97.101
= 1/4.(4/1.5 + 4/5.9 + 4/9.13 + ... + 4/97.101)
= 1/4.(1 - 1/5 + 1/5 - 1/9 + 1/9 - 1/13 + ... + 1/97 - 1/101)
= 1/4.(1 - 1/101)
= 1/4.100/101
= 25/101
\(\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+........+\frac{1}{97.101}\)
\(=\frac{1}{4}\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+........+\frac{1}{97}-\frac{1}{101}\right)\)
\(=\frac{1}{4}\left(1-\frac{1}{101}\right)\)
\(=\frac{1}{4}.\frac{100}{101}=\frac{25}{101}\)
3 x X + X = 20
(3+1) x X =20
4 x X = 20
X= 20:4
X=5
\(X+X:0,5=27\)
\(\Rightarrow X\left(1+\dfrac{1}{0,5}\right)=27\)
\(\Rightarrow X\left(1+2\right)=27\)
\(\Rightarrow3X=27\Rightarrow X=27:3=9\)
a) A = 4/5.9 + 4/9.13 + 4/13.17 + ... + 4/41/45
A = 1/5 - 1/9 + 1/9 - 1/13 + 1/13 - 1/17 + ... + 1/41 - 1/45
A = 1/5 - 1/45
A = 8/45
b) B = ( 1 - 1/2 ) . ( 1 - 1/3 ) . ( 1 - 1/4 ) . ..... . ( 1 - 1/100 )
B = 1/2 . 2/3 . 3/4 . .... . 99/100
B = \(\frac{1.2.3.......99}{2.3.4......100}\)
B = 1/100
B = \(\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{3}\right)+...+\left(1-\frac{1}{100}\right)\)
B = \(\frac{1}{2}.\frac{2}{3}.....\frac{99}{100}\)
B = \(\frac{1}{100}\)
D=1/1.5+1/5.9+...+1/41.45
4D=4/1.5+4/5.9+...+4/41.45
4D=1-1/5+1/5-1/9+...+1/41-1/45
4D=1-1/45
D=44/45:4=11/45