Question

[Hiếu]

Làm hộ e bài 1 bài 2 ạ

Answer 1

[ILoveMath]

Bài 1:

a) \(\Leftrightarrow x^2-8x+16-x^2+4=6\\ \Leftrightarrow-8x=-14\\ \Leftrightarrow x=\dfrac{7}{4}\)

b) \(\Leftrightarrow\left(x^2-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)^2\left(x+1\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

c) \(\Leftrightarrow\left[{}\begin{matrix}3x-1=x+2\\3x-1=-x-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x=3\\4x=-1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{1}{4}\end{matrix}\right.\)

Bài 2:

a) \(=5\left(x^2-2xy+y^2-4z^2\right)=5\left[\left(x-y\right)^2-\left(2z\right)^2\right]=5\left(x-y-2z\right)\left(x-y+2z\right)\)

b) \(=\left(x+1\right)^3-\left(3z\right)^3=\left(x-3z+1\right)\left[\left(x+1\right)^2+3z\left(x+1\right)+\left(3z\right)^2\right]=\left(x-3z+1\right)\left(x^2+2x+1+3xz+3z+9z^2\right)\)

c) \(=x\left(x^2-16\right)-15x\left(x-4\right)=x\left(x+4\right)\left(x-4\right)-15x\left(x-4\right)=\left(x-4\right)\left(x^2+4x-15x\right)=\left(x-4\right)\left(x^2-11x\right)=x\left(x-4\right)\left(x-11\right)\)