a, Đặt \(\left\{{}\begin{matrix}a+b-c=x\\b+c-a=y\\c+a-b=z\end{matrix}\right.\Rightarrow x+y+z=a+b+c\)

\(\Rightarrow A=\left(x+y+z\right)^3-x^3-y^3-z^3\)

\(=\left(y+z\right)\left[\left(x+y+z\right)^2+x\left(x+y+z\right)+x^2\right]-\left(y+z\right)\left(y^2-yz+z^2\right)\)

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

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

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

\(=3\left(y+z\right)\left[x\left(x+z\right)+y\left(x+z\right)\right]\)

\(=3\left(y+z\right)\left(x+y\right)\left(x+z\right)\)

\(=24abc\)

Vậy A = 24abc

a, \(x^3+8-4x^2-9x\)

\(=\left(x+2\right)\left(x^2-2x+4\right)-4x\left(x+2\right)\)

\(=\left(x+2\right)\left(x^2-6x+4\right)\)

b, \(x^2+2x+1-y^2\)

\(=\left(x+1\right)^2-y^2=\left(x-y+1\right)\left(x+y+1\right)\)

c, \(x^3-64=\left(x-4\right)\left(x^2+4x+16\right)\)

\(\left(2x+1\right)\left(3x-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}2x+1=0\\3x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{-1}{2}\\x=\dfrac{2}{3}\end{matrix}\right.\)

Vậy...

Bài 3:

\(a,\dfrac{x-1}{10}+\dfrac{x-1}{11}=\dfrac{x-1}{12}+\dfrac{x-1}{13}\)

\(\Rightarrow\dfrac{x-1}{10}+\dfrac{x-1}{11}-\dfrac{x-1}{12}-\dfrac{x-1}{13}=0\)

\(\Rightarrow\left(x-1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}-\dfrac{1}{12}-\dfrac{1}{13}\right)=0\)

Mà \(\dfrac{1}{10}+\dfrac{1}{11}-\dfrac{1}{12}-\dfrac{1}{13}\ne0\)

\(\Rightarrow x-1=0\Rightarrow x=1\)

Vậy x = 1

b, \(\dfrac{x-2000}{10}+\dfrac{x-1999}{9}=\dfrac{x-1998}{8}+\dfrac{x-1997}{7}\)

\(\Rightarrow\dfrac{x-2000}{10}+1+\dfrac{x-1999}{9}+1=\dfrac{x-1998}{8}+\dfrac{x-1997}{7}+1\)

\(\Rightarrow\dfrac{x-1990}{10}+\dfrac{x-1990}{9}-\dfrac{x-1990}{8}-\dfrac{x-1990}{7}=0\)

\(\Rightarrow\left(x-1990\right)\left(\dfrac{1}{10}+\dfrac{1}{9}-\dfrac{1}{8}-\dfrac{1}{7}\right)=0\)

Mà \(\dfrac{1}{10}+\dfrac{1}{9}-\dfrac{1}{8}-\dfrac{1}{7}\ne0\)

\(\Rightarrow x-1990=0\Rightarrow x=1990\)

\(4x^4+4x^2y^2+y^4-9y^4\)

\(=\left(2x^2+y^2\right)^2-9y^4\)

\(=\left(2x^2+4y^2\right)\left(2x^2-2y^2\right)\)

\(=4\left(x^2+2y^2\right)\left(x^2-y^2\right)\)

\(=4\left(x^2+2y^2\right)\left(x-y\right)\left(x+y\right)\)

\(\dfrac{1}{2.5}+\dfrac{1}{5.8}+...+\dfrac{1}{90.93}\)

\(=\dfrac{1}{3}\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+...+\dfrac{3}{90.93}\right)\)

\(=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{90}-\dfrac{1}{93}\right)\)

\(=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{93}\right)\)

\(=\dfrac{91}{558}\)

a, Áp dụng tính chất dãy tỉ số bằng nhau:

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{x+y}{2+3}=\dfrac{-15}{5}=-3\)

\(\Rightarrow\left\{{}\begin{matrix}x=-6\\y=-9\end{matrix}\right.\)

Vậy...

b, tương tự

c, Đặt \(\dfrac{x}{3}=\dfrac{y}{4}=k\Rightarrow\left\{{}\begin{matrix}x=3k\\y=4k\end{matrix}\right.\)

Ta có: \(xy=12\)

\(\Rightarrow3k.4k=12\)

\(\Rightarrow k^2=1\)

\(\Rightarrow\left[{}\begin{matrix}k=1\\k=-1\end{matrix}\right.\)

+) \(k=1\Rightarrow x=3,y=4\)

+) \(k=-1\Rightarrow x=-3,y=-4\)

Vậy cặp số \(\left(x;y\right)\) là \(\left(3;4\right);\left(-3;-4\right)\)

\(\left(x-2\right)\left(x+\dfrac{2}{3}\right)>0\)

\(\Rightarrow\left\{{}\begin{matrix}x-2>0\\x+\dfrac{2}{3}>0\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}x-2< 0\\x+\dfrac{2}{3}< 0\end{matrix}\right.\)

\(\Rightarrow x>2\) hoặc \(x< \dfrac{-2}{3}\)

Vậy \(\left[{}\begin{matrix}x>2\\x< \dfrac{-2}{3}\end{matrix}\right.\)

2 + 2 + 2 + 2 + 2 = 12  là sai nha bạn

tick cho mk đi bạn phan nguyễn trung thuận mk sẽ tick lại cho bạn ngay

\(M=-1^2+2^2-3^2+4^2-...-17^2+18^2-19^2+20^2\)

\(=\left(2-1\right)\left(2+1\right)+\left(4-3\right)\left(4+3\right)+...+\left(18-17\right)\left(18+17\right)+\left(20-19\right)\left(20+19\right)\)

\(=3+7+...+35+39\)

\(=210\)