Bạn sửa lại đề bài câu 2) nhé ^^

2) \(a+b+c+d=0\Leftrightarrow a+b=-c-d\Leftrightarrow\left(a+b\right)^3=-\left(c+d\right)^3\)

\(\Leftrightarrow a^3+b^3+3ab\left(a+b\right)=-\left[c^3+d^3+3cd\left(c+d\right)\right]\)

\(\Leftrightarrow a^3+b^3+c^3+d^3=-3cd\left(c+d\right)-3ab\left(a+b\right)\)

\(\Leftrightarrow a^3+b^3+c^3+d^3=3ab\left(c+d\right)-3cd\left(c+d\right)\)

\(\Leftrightarrow a^3+b^3+c^3+d^3=3\left(c+d\right)\left(ab-cd\right)\)

đề đúng ak bạn

a)nhân S với 3² ta dc:

9S=3^2+3^4+...+3^2002+3^2004

=>9S-S=(3^2+3^4+...+3^2004)-(3^0+3^4+...+2^2002)

=>8S=3^2004-1

=>S=3^2004-1/8

Ai trả lời nhanh mà đúng mình

Đặt A=0-1+2-3+4-5+6-7+...+2004-2005

=(-1)+(-1)+.....+(-1)

Số số hạng -1 là:

\(\frac{\left(2005-0\right)\div1+1}{2}=1003\left(số\right)\)

Do đó A=1003(-1)

=-1003

a/ \(5\left(x+3\right)-2x\left(x+3\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(5-2x\right)=0\) \(\Leftrightarrow\left[\begin{array}{nghiempt}x=-3\\x=\frac{5}{2}\end{array}\right.\)

b/ \(4x\left(x-2004\right)-x+2004=0\)

\(\Leftrightarrow4x\left(x-2004\right)-\left(x-2004\right)=0\)

\(\Leftrightarrow\left(x-2007\right)\left(4x-1\right)=0\) \(\Leftrightarrow\left[\begin{array}{nghiempt}x=2007\\x=\frac{1}{4}\end{array}\right.\)

c/ \(\left(x+1\right)^2=x+1\Leftrightarrow\left(x+1\right)\left(x+1-1\right)=0\Leftrightarrow x\left(x+1\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=-1\end{array}\right.\)

A) 5(x+3)-2x(3+x)=0

=> 5(x+3)-2x(x+3)=0

=> (5-2x)(x+3)=0

\(\Rightarrow\left[\begin{array}{nghiempt}5-2x=0\\x+3=0\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{5}{2}\\x=-3\end{array}\right.\)

b thiếu vế phải