Question

 giúp mik câu a s ạ

Answer 1

\(a,\Rightarrow\left(x-2\right)\left(3x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2=0\\3x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\\ b,\Rightarrow\left(x-2\right)\left(x^4-1\right)=0\\ \Rightarrow\left(x-2\right)\left(x^2+1\right)\left(x-1\right)\left(x+1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=1\\x=-1\\x^2=-1\left(vô.lí\right)\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=1\\x=-1\end{matrix}\right.\)

\(e,\Rightarrow5\left(x+3\right)-2x\left(x+3\right)=0\\ \Rightarrow\left(x+3\right)\left(5-2x\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-3\\x=\dfrac{5}{2}\end{matrix}\right.\)

Answer 2

a. 3x(x - 2) - x + 2 = 0

<=> 3x(x - 2) - (x - 2) = 0

<=> (3x - 1)(x - 2) = 0

<=> \(\left[{}\begin{matrix}3x-1=0\\x-2=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=2\end{matrix}\right.\)

c. x⁴(x - 2) - 2 + x = 0

<=> x⁴(x - 2) + x - 2 = 0

<=> x⁴(x - 2) + (x - 2) = 0

<=> (x⁴ + 1)(x - 2) = 0

<=> \(\left[{}\begin{matrix}x^4+1=0\left(VLí\right)\\x-2=0\end{matrix}\right.\)

<=> x - 2 = 0

<=> x = 2

e. 5(x + 3) = 2x(3 + x)

<=> 5(x + 3) - 2x(x + 3) = 0

<=> (5 - 2x)(x + 3) = 0

<=> \(\left[{}\begin{matrix}5-2x=0\\x+3=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-3\end{matrix}\right.\)

f. Thiếu đề

Answer 3

a: Ta có: \(3x\left(x-2\right)-x+2=0\)

\(\Leftrightarrow\left(x-2\right)\left(3x-1\right)=0\)

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

b: Ta có: \(x^2\left(x+1\right)+2x\left(x+1\right)=0\)

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

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