Question

[Châu Kim]

Giúp mk bài 2; bài 3 nha

Mk cần gấp lắm!!!!!

Thanks trước😊😊😊

Answer 1

[T.Thùy Ninh]

\(a,\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=6\)

\(\Leftrightarrow x^2-4x+4-x^2+9=6\)

\(\Leftrightarrow-4x=-7\)

\(\Leftrightarrow x=\dfrac{7}{4}\)

\(b,9x^2-4-\left(3x-2\right)\left(4x-5\right)=0\)

\(\Leftrightarrow9x^2-4-12x^2+23x-10=0\)

\(\Leftrightarrow-3x^2+23x-14=0\)

\(\Leftrightarrow-3x^2+21x+2x-14=0\)

\(\Leftrightarrow-3x\left(x-7\right)+2\left(x-7\right)=0\)

\(\Leftrightarrow\left(x-7\right)\left(2-3x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-7=0\\2-3x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=7\\3x=2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=7\\x=\dfrac{2}{3}\end{matrix}\right.\)

\(c,4x^2\left(x-1\right)-x+1=0\)

\(\Leftrightarrow4x^2\left(x-1\right)-\left(x-1\right)=0\)

\(\Leftrightarrow\left(4x^2-1\right)\left(x-1\right)=0\)

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

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

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

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

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

Answer 2

[Châu Kim]

bài 3 thì sao