(Mình giải theo cách lớp 8 nhé)

\(A=1^2-2^2+3^2-4^2+...+2015^2\)

    \(=1+\left(3^2-2^2\right)+\left(5^2-4^2\right)+...+\left(2015^2-2014^2\right)\)

    \(=1+\left(3-2\right)\left(3+2\right)+\left(5-4\right)\left(5+4\right)+...+\left(2015-2014\right)\left(2015+2014\right)\)

    \(=1+\left(2+3\right)+\left(4+5\right)+...+\left(2014+2015\right)\)

    \(=1+2+3+...+2015=B\)

             \(\Leftrightarrow A=B\)

1.(x+2)^2 + 3.(x+2) =0

\(\Leftrightarrow\left(x+2\right)\left(x+2+3\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\x+5=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-5\end{cases}}\)

2. 3x^2 +6x = 0

\(\Leftrightarrow3x\left(x+2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x=0\\x+2=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-2\end{cases}}\)

3. x. (x-3) + 2 .(3-x) = 0

\(\Leftrightarrow x\left(x-3\right)-2\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x-2=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=2\end{cases}}\)

1) (x + 2)² + 3(x + 2) = 0

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

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

=> \(\orbr{\begin{cases}x+2=0\\x+5=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=-2\\x=-5\end{cases}}\)

2) 3x² + 6x = 0

=> 3x(x + 2) = 0

=> \(\orbr{\begin{cases}3x=0\\x+2=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=0\\x=-2\end{cases}}\)

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

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

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

=> \(\orbr{\begin{cases}x-2=0\\x-3=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=2\\x=3\end{cases}}\)

a³+b³+ab=(a+b)³-3ab(a+b)+ab=(a+b)³-ab(3a+3b-1)

=(a+b)³-ab(2a+4b)

=(a+b)³-2ab(a+2b)             (đề bài sai phải không????)

Ta có:   \(\left(x+y+z\right)^2=3\left(xy+yz+xz\right)\)

\(\Leftrightarrow x^2+y^2+z^2+2\left(xy+yz+xz\right)=3\left(xy+yz+xz\right)\)

\(\Leftrightarrow x^2+y^2+z^2=xy+yz+xz\)

\(\Leftrightarrow2x^2+2y^2+2z^2=2xy+2yz+2xz\)

\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2xz-2yz=0\)

\(\Leftrightarrow x^2+x^2+y^2+y^2+z^2+z^2-2xy-2xz-2yz=0\)

\(\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(x^2-2xz+z^2\right)=0\)

\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2=0\)

Vì \(\left(x-y\right)^2;\left(y-z\right)^2\)và \(\left(x-z\right)^2\)luôn luôn \(\ge0\)

để phương trình trên bằng 0 thì \(\hept{\begin{cases}\left(x-y\right)^2=0\\\left(y-z\right)^2=0\\\left(x-z\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}x-y=0\\y-z=0\\x-z=0\end{cases}\Rightarrow}\hept{\begin{cases}x=y\\y=z\\x=z\end{cases}}}\)

\(\Rightarrow x=y=z\left(ĐPCM\right)\)

A=(2005² -1) . 2008

B=2005² . (2008² - 1)

SUY RA A>B

sao 2018 vs 2005 lại bỏ mũ 2 vậy bn 

bạn search gg có bài này đó

bạn xem lại đề đc k

hạng tử cuối là \(2^{2019}\)

đúng ko bạn

a)999x1001=(1000-1)(1000+1)=1000²-1²=1000000-1=999999

b)bạn viết đúng đề câu b k thế?

https://olm.vn/hoi-dap/detail/740021926146.html?auto=1