Question

[Diệp Hằng]

Giải các phương trình sau bằng cách đưa về phương trình tích

a. x² - 3x + 2 = 0 b.-x² + 5x - 6 = 0 c.4x² - 12x + 5 = 0 d. 2x² + 5x + 3 = 0 e. x³ + 2x² - x -2 = 0

g. (3x-1)² - 5(2x +1)² + (6x-3)(2x+1) = (x-1)² h. 2x³ - 7x² + 7x -2 = 0 i.x⁴ + 2x³ + 5x² + 4x - 12 = 0

Mọi người giúp em với ạ em cảm ơn

Answer 1

[Nguyễn Lê Phước Thịnh]

a) Ta có: \(x^2-3x+2=0\)

\(\Leftrightarrow x^2-x-2x+2=0\)

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

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

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

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

Vậy: \(x\in\left\{1;2\right\}\)

b) Ta có: \(-x^2+5x-6=0\)

\(\Leftrightarrow-\left(x^2-5x+6\right)=0\)

\(\Leftrightarrow-\left(x^2-2x-3x+6\right)=0\)

\(\Leftrightarrow-\left[\left(x^2-2x\right)-\left(3x-6\right)\right]=0\)

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

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

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

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

Vậy: x∈{2;3}

c) Ta có: \(4x^2-12x+5=0\)

\(\Leftrightarrow4x^2-10x-2x+5=0\)

⇔(4x²-10x)-(2x-5)=0

\(\Leftrightarrow2x\left(2x-5\right)-\left(2x-5\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(2x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\2x-1=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}2x=5\\2x=1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{5}{2}\\x=\frac{1}{2}\end{matrix}\right.\)

Vậy: \(x\in\left\{\frac{1}{2};\frac{5}{2}\right\}\)

d) Ta có: \(2x^2+5x+3=0\)

\(\Leftrightarrow2x^2+2x+3x+3=0\)

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

\(\Leftrightarrow2x\left(x+1\right)+3\left(x+1\right)=0\)

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

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

Vậy: \(x\in\left\{-1;\frac{-3}{2}\right\}\)

e) Ta có: \(x^3+2x^2-x-2=0\)

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

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

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

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

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

Vậy: \(x\in\left\{-2;1;-1\right\}\)

g) Ta có: \(\left(3x-1\right)^2-5\left(2x+1\right)^2+\left(6x-3\right)\left(2x+1\right)=\left(x-1\right)^2\)

\(\Leftrightarrow9x^2-6x+1-20x^2-20x-5+12x^2-3-x^2+2x-1=0\)

\(\Leftrightarrow-24x-8=0\)

\(\Leftrightarrow-8\left(3x+1\right)=0\)

⇔3x+1=0

\(\Leftrightarrow3x=-1\)

\(\Leftrightarrow x=-\frac{1}{3}\)

Vậy: \(x=-\frac{1}{3}\)

Answer 2

[Ngọc Hưng]

h) \(2x^3-7x^2+7x-2=0\)

\(\Leftrightarrow2x^3-4x^2-3x^2+6x+x-2=0\)

\(\Leftrightarrow2x^2\left(x-2\right)-3x\left(x-2\right)+\left(x-2\right)=0\)

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

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

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

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

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

Vậy S = {2; 1; \(\frac{1}{2}\)}

i) \(x^4+2x^3+5x^2+4x-12=0\)

\(\Leftrightarrow x^4-x^3+3x^3-3x^2+8x^2-8x+12x-12=0\)

\(\Leftrightarrow x^3\left(x-1\right)+3x^2\left(x-1\right)+8x\left(x-1\right)+12\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^3+3x^2+8x+12\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^3+2x^2+x^2+2x+6x+12\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left[x^2\left(x+2\right)+x\left(x+2\right)+6\left(x+2\right)\right]=0\)

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

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left[\left(x+\frac{1}{2}\right)^2+\frac{23}{4}\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\\\left(x+\frac{1}{2}\right)^2+\frac{23}{4}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\\\left(x+\frac{1}{2}\right)^2=\frac{-23}{4}\left(loai\right)\end{matrix}\right.\)

Vậy S = {1;-2}