Question

a) (x - 4)(x - 2)(2x + 1)(2x + 2) - 6x² = 0

b) 2(x² + x + 1)² - 7(x - 1)² = 13(x³ - 1)

Mình cần gấp nhaaa

Answer 1

a, \(\left(x-4\right)\left(x-2\right)\left(2x+1\right)\left(2x+2\right)-6x^2=0\)

\(\Leftrightarrow\left(x^2-6x+8\right)\left(4x^2+6x+2\right)-6x^2=0\)

\(\Leftrightarrow4x^4+6x^3+2x^2-24x^3-36x^2-12x+32x^2+48x+16-6x^2=0\) \(\Leftrightarrow4x^4-18x^3-8x^2+36x+16=0\)

Answer 2

a/ \(\Leftrightarrow\left(x-4\right)\left(2x+1\right)\left(x-2\right)\left(2x+2\right)-6x^2=0\)

\(\Leftrightarrow\left(2x^2-4-7x\right)\left(2x^2-4-2x\right)-6x^2=0\)

Nhận thấy \(x=0\) không phải nghiệm, pt tương đương:

\(\left(2x-\frac{4}{x}-7\right)\left(2x-\frac{4}{x}-2\right)-6=0\)

Đặt \(2x-\frac{4}{x}-7=t\) ta được:

\(t\left(t+5\right)-6=0\)

\(\Leftrightarrow t^2+5t-6=0\Rightarrow\left[{}\begin{matrix}t=1\\t=-6\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x-\frac{4}{x}-7=1\\2x-\frac{4}{x}-7=-6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x^2-8x-4=0\\2x^2-x-4=0\end{matrix}\right.\) (bấm máy)

Answer 3

b/

\(\Leftrightarrow2\left(x^2+x+1\right)^2-7\left(x-1\right)^2=13\left(x-1\right)\left(x^2+x+1\right)\)

Đặt \(\left\{{}\begin{matrix}x^2+x+1=a\\x-1=b\end{matrix}\right.\)

\(\Rightarrow2a^2-7b^2=13ab\)

\(\Leftrightarrow2a^2-13ab-7b^2=0\)

\(\Leftrightarrow\left(a-7b\right)\left(2a+b\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}a-7b=0\\2a+b=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+x+1-7\left(x-1\right)=0\\2\left(x^2+x+1\right)+x-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-6x+8=0\\2x^2+3x+1=0\end{matrix}\right.\) (bấm máy)