Question

5/ (x² – 4) + (x – 2)(4 – 2x) = 0

6/ x(2x – 7) – 4x + 14 = 0 7/ x² – x – (3x – 3) = 0

8/ (x² – 2x + 1) – 4 = 0 9/ 4x² + 4x + 1 = x²

10/ x² – x = - 2x + 2 11/ x² – 5x + 6 = 0

11/ x^2 -- 5x + 6 = 0

Câu 11 là x mũ 2 nha giúp dùm e nha anh chị .

Answer 1

5/ (x² – 4) + (x – 2)(4 – 2x) = 0

⇔(x-2)(x+2)+(x – 2)(4 – 2x)=0

⇔(x-2)(x+2+4-2x)=0

⇔(x-2)(6-x)=0

⇔\(\left[{}\begin{matrix}x-2=0\\6-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)

6/ x(2x – 7) – 4x + 14 = 0

⇔2x²-11x+14=0

⇔(x-\(\frac{7}{2}\))(x-2)=0

⇔\(\left[{}\begin{matrix}x-\frac{7}{2}=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{7}{2}\\x=2\end{matrix}\right.\)

7/ x² – x – (3x–3)= 0

⇔x²-4x+3=0

⇔(x-3)(x-1)=0

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

8/ (x² – 2x + 1) – 4 = 0

⇔(x-1)²-4=0

⇔(x-1-4)(x-1+4)=0

⇔(x-5)(x+3)=0

⇔\(\left[{}\begin{matrix}x-5=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)

9/ 4x² + 4x + 1 = x²

⇔3x²+4x+1=0

⇔(3x+1)(x+1)=0

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

10/ x² – x = - 2x + 2

⇔3x²-x-2=0 (chuyển vế)

⇔(3x+2)(x-1)=0

⇔\(\left[{}\begin{matrix}3x+2=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\frac{2}{3}\\x=1\end{matrix}\right.\)

11/ x² – 5x + 6 = 0

⇔x²-3x-2x+6=0

⇔x(x-3)-2(x-3)=0

⇔(x-3)(x-2)=0

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

Mình làm bài khá tắt nên có gì không hiểu bạn cứ hỏi mình nha!

Answer 2

5/(x²-4)+(x-2)(4-2x)=0⇔x²-4+4x-2x²-8+4x=0

⇔-x²+8x-12=0⇔-x²+2x+6x-12=0⇔-x(x-2)-6(x-2)=0⇔(-x-6)(x-2)=0

⇔-x-6=0 hoặc x-2=0⇔x=-6 hoặc x=2