`a)6x(x-1999)-x+1999=0`

`<=>6x(x-1999)-(x-1999)=0`

`<=>(x-1999)(6x-1)=0`

`<=>` \(\left[ \begin{array}{l}x-1999=0\\6x-1=0\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=1999\\x=\dfrac16\end{array} \right.\) 

`b)x^2-9-4(x+3)=0`

`<=>(x-3)(x+3)-4(x+3)=0`

`<=>(x+3)(x-3-4)=0`

`<=>(x+3)(x-7)=0`

`<=>` \(\left[ \begin{array}{l}x+3=0\\x-7=0\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=7\\x=-3\end{array} \right.\) 

Câu b sai rồi :v

=2666666000

Có công thức như sau

1x2+2x3+3x4+...+nx(n+1)=nx(n+1)x(n+2):3

phần sau thì sao

là 4000000 đó nha

Đặt A=\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2000}}{\frac{1999}{1}+\frac{1998}{2}+....+\frac{1}{1999}}\)

Xét mẫu số:

\(\frac{1999}{1}+\frac{1998}{2}+\frac{1997}{3}+\frac{1996}{4}+....+\frac{1}{1999}\)

=\(\left(\frac{1998}{2}+1\right)+\left(\frac{1997}{3}+1\right)+\left(\frac{1996}{4}+1\right)+....+\left(\frac{1}{1999}+1\right)+1\)

=\(\frac{2000}{2}+\frac{2000}{3}+\frac{2000}{4}+....+\frac{2000}{1999}+\frac{2000}{2000}\)

= 2000\(\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{1999}+\frac{1}{2000}\right)\)

=> A = \(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2000}}{2000\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2000}\right)}\)

=> A = \(\frac{1}{2000}\)

\(x^4+\sqrt{x^2+1999}=1999\)

\(\Leftrightarrow4x^4+4x^2+1=4\left(x^2+1999\right)-4\sqrt{x^2+1999}+1\)

\(\Leftrightarrow\left(2x^2+1\right)^2=\left(2\sqrt{x^2+1999}-1\right)^2\)

TH1:Đặt \(t=x^2\left(t>0\right)\)

\(\Leftrightarrow2t+1=2\sqrt{t+1999}-1\)

\(\Leftrightarrow t+1=\sqrt{t+1999}\)(tiếp theo bình phương rồi giải)

TH còn lại tương tự

hello you

Dat A=(-1).2(-3).4....(-1999)

Ta co : 1,3,5....1999 co tat ca : (1999-1):2+1=1000(so)

=> (-1),(-3),(-5)....(-1999) cung co 1000 so ma 1000 la so chan nen

(-1).(-3)...(-1999)=1.3.5...1999 ( so am nhan voi so am thi ra so duong )

hay A=(-1).2.(-3)...(-1999)=1.2.3....1999>1.3.5.1999