\(\text{Δ}=\left(-450\right)^2-4\cdot1\cdot1800=195300>0\)

Do đó: Phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}a_1=\dfrac{450-30\sqrt{217}}{2}=225-15\sqrt{217}\\a_2=225+15\sqrt{217}\end{matrix}\right.\)

a=0 or a=450

a² nha ...

hình như sai đề

ak là 19a² nha

\(=\left(x^2+5\right)\left(x^2-5\right)-4\left(x^2+5\right)\)

\(=\left(x^2+5\right)\left(x^2-5-4\right)\)

\(=\left(x^2+5\right)\left(x^2-9\right)\)

\(=\left(x^2+5\right)\left(x^2-3^2\right)=\left(x^2+5\right)\left(x-3\right)\left(x+3\right)\)

EZ :))

MAI CÔ KIỂM TRA HẢ? CHÚC BẠN MAY MẮN LẦN SAU

\(x^3-64x=x\left(x^2-64\right)=x\left(x-8\right)\left(x+8\right)\)

𝑥(𝑥−8)(𝑥+8)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)

\(8x\left(x^2-9\right)=0\Rightarrow8x\left(x-3\right)\left(x+3\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x+3=0\\x-3=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\pm3\end{matrix}\right.\)

a) \(8x\left(x-3\right)+x-3=0\)

\(\Rightarrow8x\left(x-3\right)+\left(x-3\right)=0\)

\(\Rightarrow\left(x-3\right)\left(8x+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{1}{8}\end{matrix}\right.\)

b) \(x^2+36=12x\)

\(\Rightarrow x^2-12x+36=0\)

\(\Rightarrow\left(x-6\right)^2=0\)

\(\Rightarrow x=6\)