B=1.98+2.97+3.96+......+98.1/1.2+2.3+3.4+....+98.99 = ?
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B=\(\frac{1.\left(100-2\right)+2.\left(100-3\right)+3.\left(100-4\right)+...+98.\left(100-99\right)}{1.2+2.3+3.4+...+98.99}\)
B=\(\frac{100.\left(1+2+3+...+98\right)-\left(1.2+2.3+3.4+...+98.99\right)}{1.2+2.3+3.4+...+98.99}\)
B=\(\frac{100.\left(1+98\right).98:2}{1.2+2.3+3.4+...+98.99}-\frac{1.2+2.3+3.4+...+98.99}{1.2+2.3+3.4+...+98.99}\)
B=\(\frac{50.98.99}{1.2+2.3+3.4+...+98.99}\)
Đặt M = 1.2+2.3+3.4+....+98.99
=> 3M=3.(1.2+2.3+3.4+...+98.99)
=> 3M = 1.2.3+2.3.(4-1)+...+098.99.(100-97)
3M= 1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+98.99.100-97.98.100
3M=98.99.100
=> M = 98.33.100
=> B = \(\frac{50.98.99}{98.33.100}-1=\frac{3}{2}-1=\frac{1}{2}\)
Đặt A=1.98+2.97+3.96+...+96.3+97.2+98.1
B=1.2+2,3+3.4+...+96.97+97.98+98.99
Ta có: A=1+(1+2)+...+(1+2+3+...+97+98)
=\(\dfrac{1.2}{2}+\dfrac{2.3}{2}+...+\dfrac{98.99}{3}\)
=\(\dfrac{1.2+2.3+3.4+4.5+...+98.99}{2}\)=\(\dfrac{B}{2}\)
=>E=\(\dfrac{B}{2}\):2=\(\dfrac{1}{2}\)
a)\(\frac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+98\right)}{1.98+2.97+3.96+....+98.1}\)
\(=\frac{\left(1+1+....+1\right)+\left(2+2+...2\right)+....+\left(97+97\right)+98}{ }\)
\(=\frac{1.98+2.97+3.96+....+97.2+98.1}{1.98+2.97+3.96+....+98.1}=1\)
\(\text{Đặt C = 1.2 + 2.3 + 3.4 + ..... +98.99 }\)
\(\text{ Và A = 1.98 + 2.97 + 3.96 + .... + 98.1 }\)
\(\text{Khi đó : }A=1+\left(1+2\right)+....+\left(1+2+...+98\right)\)
\(=\frac{1.2}{2}+\frac{2.3}{2}+\frac{3.4}{2}+....+\frac{98.99}{2}\)
\(=\frac{1.2+2.3+3.4+....+98.99}{2}=\frac{C}{2}\)
\(\Rightarrow B=\frac{B}{\frac{2}{B}}=\frac{1}{2}\)
ko thiếu bạn ạ
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\(\text{Đặt C = 1.2 + 2.3 + 3.4 + ..... +98.99 }\)
\(\text{ Và A = 1.98 + 2.97 + 3.96 + .... + 98.1 }\)
\(\text{Khi đó : }A=1+\left(1+2\right)+....+\left(1+2+...+98\right)\)
\(=\frac{1.2}{2}+\frac{2.3}{2}+\frac{3.4}{2}+....+\frac{98.99}{2}\)
\(=\frac{1.2+2.3+3.4+....+98.99}{2}=\frac{C}{2}\)
\(\Rightarrow B=\frac{C}{\frac{2}{C}}=\frac{1}{2}\)