\(A=\left(x-2\right)^2+\left(x-2\right)^2-2\left(x+3\right)\left(x-2\right)\)

\(A=x^2-4x^2+4+x^2-4x^2+4-2x^2-2x+6\)

\(A=-10x+14\)

\(A=\left(x-2\right)^2+\left(x-2\right)^2-2\left(x+3\right)\left(x-2\right)\)

\(=\left(x-2\right)\left[\left(x-2\right)+\left(x-2\right)-2\left(x+3\right)\right]\)

\(=\left(x-2\right)\left[x-2+x-2x-6\right]\)

\(=\left(x-2\right)\left(-8\right)\)

\(=-8x+16\)

a,ta có:

DM // AB=>ABDM  là hình thang

AH=DH => ABDM là hbh mà AD vuông góc với BC 

=> ABDM là hình thoi

còn câu b,c thì sao bạn

\(\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}+\frac{1}{x^2+13x+42}=\frac{1}{18}\)

\(\Leftrightarrow\frac{1}{x\left(x+4\right)+5\left(x+4\right)}+\frac{1}{x\left(x+5\right)+6\left(x+5\right)}+\frac{1}{x\left(x+6\right)+7\left(x+6\right)}=\frac{1}{18}\)(điều kiện: \(x\ne\left\{-4;-5;-6;-7\right\}\) )

\(\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+7}=\frac{1}{18}\)

\(\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+7}=\frac{1}{18}\)

\(\Leftrightarrow\frac{3}{\left(x+4\right)\left(x+7\right)}=\frac{1}{18}\)

\(\Rightarrow54=\left(x+4\right)\left(x+7\right)\)

\(\Leftrightarrow x^2+11x-26=0\)

\(\Leftrightarrow x\left(x+13\right)-2\left(x+13\right)=0\Leftrightarrow\left(x+13\right)\left(x-2\right)=0\Leftrightarrow\orbr{\begin{cases}x=-13\\x=2\end{cases}}\)(thỏa mãn ĐKXĐ)

Vậy tập nghiệm của pt là: \(S=\left\{-13;2\right\}\)

Lâu lắm không làm nhể

\(\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}+\frac{1}{x^2+13x+42}=\frac{1}{18}\)

\(\Rightarrow\frac{1}{x^2+4x+5x+20}+\frac{1}{x^2+5x+6x+30}+\frac{1}{x^2+6x+7x+42}=\frac{1}{18}\)

\(\Rightarrow\frac{1}{x.\left(x+4\right)+5.\left(x+4\right)}+\frac{1}{x.\left(x+5\right)+6.\left(x+5\right)}+\frac{1}{x.\left(x+6\right)+7.\left(x+6\right)}=\frac{1}{18}\)

\(\Rightarrow\frac{1}{\left(x+4\right).\left(x+5\right)}+\frac{1}{\left(x+5\right).\left(x+6\right)}+\frac{1}{\left(x+6\right).\left(x+7\right)}=\frac{1}{18}\)

Dùng công thứ \(\frac{1}{x.\left(x+1\right)}=\frac{1}{x}-\frac{1}{x+1}\)

Khi đó \(\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+7}=\frac{1}{18}\)

\(\Rightarrow\frac{1}{x+4}-\frac{1}{x+7}=\frac{1}{18}\)

\(\Rightarrow\frac{x+7}{\left(x+4\right).\left(x+7\right)}-\frac{\left(x+4\right)}{\left(x+4\right).\left(x+7\right)}=\frac{1}{18}\)

\(\Rightarrow\frac{3}{\left(x+4\right).\left(x+7\right)}=\frac{1}{18}\Rightarrow\left(x+4\right).\left(x+7\right)=54\)

\(\Rightarrow\hept{\begin{cases}x+4=6\\x+7=9\end{cases}}\)hoặc \(\hept{\begin{cases}x+4=-6\\x+7=-9\end{cases}}\)

Suy ra \(x=3\)hoặc \(x=-3\)

\(f\left(1\right)=\left(1^2+1-1\right)^{2014}+\left(1^2-1-1\right)^{2014}-2=1+1-2=0\)

Nên \(f\left(x\right)⋮\left(x-1\right)\)

\(f\left(-1\right)=\left[\left(-1\right)^2+\left(-1\right)-1\right]^{2014}.\left[\left(-1\right)^2-\left(-1\right)-1\right]^{2014}-2=1+1-2=0\)

Nên \(f\left(x\right)⋮\left(x+1\right)\)

Vậy \(f\left(x\right)⋮\left[\left(x-1\right)\left(x+1\right)\right]\Rightarrow f\left(x\right)⋮\left(x^2-1\right)\)

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

\(=2x^3-4x^2-4x^2+8x\)

\(=\left(2x^3-4x^2\right)-\left(4x^2-8x\right)\)

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

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

\(b,2x^2-3x-5=2x^2-5x+2x-5\)

\(=\left(2x^2-5x\right)+\left(2x-5\right)=x\left(2x-5\right)+\left(2x-5\right)\)

\(=\left(x+1\right)\left(2x-5\right)\)

\(c,x^2y-x^3-9y+9x\)

\(=\left(x^2y-x^3\right)-\left(9y-9x\right)\)

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

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