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co 99 so hang
A=99x+1.2+2.3+..+99.100
tinh tong day
3.B=1.2.3+2.3.3+3.4.3+...+99.100.3
3.B=1.2.3+2.3(4-1)+3.4(5-2)+..+99.100.(101-98)
=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+..+99.100.101-99.98.100=99.100.101
B=99.100.101/3=33.100.101
A=99x+33.100.101=334290
99.x=330000+4290-330000+3300=4290+3300=7590
x=7590/99
Đặt \(A=1\cdot2+2\cdot3+\cdots+99\cdot100\)
\(=1\left(1+1\right)+2\left(2+1\right)+\cdots+99\left(99+1\right)\)
\(=\left(1^2+2^2+\cdots+99^2\right)+\left(1+2+\cdots+99\right)\)
\(=\frac{99\left(99+1\right)\left(2\cdot99+1\right)}{6}+\frac{99\cdot\left(99+1\right)}{2}\)
\(=\frac{99\cdot100\cdot199}{6}+\frac{99\cdot100}{2}=\frac{99\cdot100\cdot199+3\cdot99\cdot100}{6}\)
\(=\frac{99\cdot100\cdot202}{6}=33\cdot100\cdot101=333300\)
Ta có: \(x^2+\left(x^2+1\right)+\left(x^2+2\right)+\cdots+\left(x^2+99\right)\)
\(=\left(x^2+0\right)+\left(x^2+1\right)+\cdots+\left(x^2+99\right)\)
\(=\left(x^2+x^2+\cdots+x^2\right)+\left(0+1+\cdots+99\right)=100x^2+\frac{99\cdot100}{2}=100x^2+99\cdot50=100x^2+4950\)
Ta có: \(\frac{1\cdot2+2\cdot3+\cdots+99\cdot100}{x^2+\left(x^2+1\right)+\left(x^2+2\right)+\cdots+\left(x^2+99\right)}=50\frac{116}{131}\)
=>\(\frac{333300}{100x^2+4950}=50+\frac{116}{131}\)
=>\(\frac{6666}{2x^2+99}=\frac{6666}{131}\)
=>\(2x^2+99=131\)
=>\(2x^2=32\)
=>\(x^2=16\)
=>\(\left[\begin{array}{l}x=4\\ x=-4\end{array}\right.\)
Cai phan 1+3+5+...+99 chac em biet lam roi phai ko? Con 3/1.2+3/2.3+3/3.4+...+3/99.100 thi em cu tach lam sao cho tro thanh dang ban dau thi lam . Anh phai nghi roi !~ Neu chieu anh ranh ranh thi len giai tiep . BYE BYE
\(\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+...+\frac{3}{99.100}+4x=1+3+5+...+99\)
\(3\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)+4x=\left(1+99\right)+\left(3+97\right)+\left(5+95\right)+...+\left(49+51\right)\)\(3\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)+4x=100+100+100+...+100\)\(3\left(1-\frac{1}{100}\right)+4x=100.25\)
\(3.\frac{99}{100}+4x=2500\)
\(\frac{297}{100}+4x=2500\)
\(4x=2500-\frac{297}{100}\)
\(4x=2500-2,97\)
\(4x=2497,03\)
\(x=624,2575\)
\(x=2497,03:4\)
\(\Leftrightarrow x-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}\right)=\frac{1}{100}+\frac{1}{99}-\frac{1}{100}\)
\(\Leftrightarrow x-\frac{98}{99}=\frac{1}{99}\Leftrightarrow x=1\)