Mong mọi người giúp đỡ mình với!!!

\(2.\left(x-4\right)-x+3=0\)

\(2x-8-x+3=0\)

\(x-5=0\)

\(x=5\)
\(x^2-25-x-5=0\)

\(\left(x-5\right)\left(x+5\right)-\left(x+5\right)=0\)

\(\left(x+5\right)\left(x-5-1\right)=0\)

\(\left(x+5\right)\left(x-6\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+5=0\\x-6=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-5\\x=6\end{cases}}}\)

Vậy \(\orbr{\begin{cases}x=-5\\x=6\end{cases}}\)

\(x^3+6x^2+9x=0\)

\(x\left(x^2+6x+9\right)=0\)

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

\(\Rightarrow\hept{\begin{cases}x=0\\x=-3\end{cases}}\)

x³-5x²-9x+45=0

=>(x³-5x²)-(9x-45)=0

=>x²(x-5)-9(x-5)=0

=>(x²-9)(x-5)=0

=>x²-9=0  =>x²=9  => x=3;-3

     x-5=0  =>x=5

\(5x^2-x+y-5y^2\)

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

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

\(=5\left(x-y\right)\left(x+y\right)-\left(x-y\right)\)

\(=\left(x-y\right)\left[5\left(x+y\right)-1\right]\)

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

\(1,\left(x-2\right)\left(x+2\right)\left(x^2+4\right)-\left(x^2-3\right)\left(x^2+3\right)\)

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

\(=x^2-16-x^2+9\)

\(=-7\)

\(2,\left(6x+1\right)^2+\left(6x-1\right)^2-2\left(1+6x\right)\left(6x-1\right)\)

\(=\left(6x+1-6x+1\right)^2\)

\(=2^2=4\)

a) Đặt \(A=x\left(x+1\right)\left(x+2\right)\left(x+3\right)+1\)

\(A=\left[x\left(x+3\right)\right]\left[\left(x+1\right)\left(x+2\right)\right]\)

\(A=\left(x^2+3x\right)\left(x^2+3x+1\right)+1\)

Đặt \(a=x^2+3x+1\)

\(\Leftrightarrow A=\left(a-1\right)\left(a+1\right)+1\)

\(\Leftrightarrow A=a^2-1^2+1\)

\(\Leftrightarrow A=a^2\)

Thay \(a=x^2+3x+1\)vào A ta có :

\(A=\left(x^2+3x+1\right)^2\)

Dòng 2 ghi thiếu +1 bổ sung hộ :))