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Ta có \(63,1.2-21,3.6=0,9.7.10.1,2-21.3,6\)
\(=6,3.1,2-21.3,6\)
\(=0,9.7.4.3-7.3.0,9.4\)
\(=6,3.1,2-6,3.1,2\)
\(=0\)
\(\Rightarrow\dfrac{\left(1+2+......+100\right).\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{9}\right)\left(63.1,2-21.3,6\right)}{1-2+3-4+.....+99-100}=\dfrac{\left(1+2+.....+100\right)\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{9}\right)0}{1-2+3-4+......+99-100}=0\)
\(A=\frac{\left(1+2+3+...+100\right)\left(\frac{1}{4}+\frac{1}{6}-\frac{1}{2}\right)\left(63.1,2-21.3,6+1\right)}{1-2+3-4+....+99-100}\)
\(=\frac{\frac{100\left(100+1\right)}{2}\left(\frac{3+2-6}{12}\right)\left[63\left(1,2-1,2\right)+1\right]}{\left(1-2\right)+\left(3-4\right)+....+\left(99-100\right)}\)
\(=\frac{5050.\left(-\frac{1}{12}\right).1}{-1+\left(-1\right)+\left(-1\right)+...+\left(-1\right)}\)
\(=\frac{2525.\left(-\frac{1}{6}\right)}{-50}=\frac{101}{12}\)
Bài 2:
7(x-1)+2x(1-x)=0
=>7(x-1)-2x(x-1)=0
=>(x-1)(7-2x)=0
=>x=1 hoặc x=7/2
\(\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-...-\frac{1}{3.2}-\frac{1}{2.1}.\)
\(=\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)
\(=\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
\(=\frac{1}{100}-\left(1-\frac{1}{100}\right)=\frac{1}{100}-\frac{99}{100}=\frac{98}{100}=\frac{49}{50}\)
Đặt A = \(\dfrac{1}{3}-\dfrac{1}{3^2}+\dfrac{1}{3^3}-...+\dfrac{1}{3^{99}}-\dfrac{1}{3^{100}}\)
\(\Rightarrow3A=\dfrac{1}{3^2}-\dfrac{1}{3^3}+...+\dfrac{1}{3^{100}}-\dfrac{1}{3^{101}}\)
\(\Rightarrow3A+A=\left(\dfrac{1}{3^2}-\dfrac{1}{3^3}+...+\dfrac{1}{3^{100}}-\dfrac{1}{3^{101}}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3^2}+\dfrac{1}{3^3}-...+\dfrac{1}{3^{99}}-\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow4A=\dfrac{1}{3}-\dfrac{1}{3^{101}}\Rightarrow A=\dfrac{\dfrac{1}{3}-\dfrac{1}{3^{101}}}{4}\)
Bạn tham khảo nhé.
Mình gọi A là tổng nhé.
Ta có:
A=13−132+133−134+...+1399−13100A=13−132+133−134+...+1399−13100
=> 3A=1−13+132−133+...+1398−13993A=1−13+132−133+...+1398−1399
=> A+3A=1−13100A+3A=1−13100
=> 4A=3100−131004A=3100−13100
=> A=3100−14.3100