a: ĐKXD: x<>0

\(\dfrac{14x^3+12x^2-14x}{2x}=\left(x+2\right)\left(3x-4\right)\)

=>\(\dfrac{2x\left(7x^2+6x-7\right)}{2x}=\left(x+2\right)\left(3x-4\right)\)

=>\(7x^2+6x-7=3x^2-4x+6x-8\)

=>\(7x^2+6x-7=3x^2+2x-8\)

=>\(4x^2+4x+1=0\)

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

=>2x+1=0

=>x=-1/2(nhận)

b: \(\left(4x-5\right)\left(6x+1\right)-\left(8x+3\right)\left(3x-4\right)=15\)

=>\(24x^2+4x-30x-5-\left(24x^2-32x+9x-12\right)=15\)

=>\(24x^2-26x-5-24x^2+23x+12=15\)

=>-3x+7=15

=>-3x=8

=>\(x=-\dfrac{8}{3}\)

PT <=> (x⁴ - 2x³ + 3x²) + (- 4x³ + 8x² - 12x) + (x² - 2x + 3) = 0

<=> (x² - 2x + 3)(x² - 4x + 1) = 0

bạn ơi nhầm rồi đề bài là cộng 2 chứ có phải cộng 3 đâu

Đề sai đó bạn. Nếu là 2 thì nó ra nghiệm vô tỷ phức tạp lắm.

Bài 1.

a)

\((x-2)(2x-1)-(2x-3)(x-1)-2\\=2x^2-x-4x+2-(2x^2-2x-3x+3)-2\\=2x^2-5x+2-(2x^2-5x+3)-2\\=2x^2-5x+2-2x^2+5x-3-2\\=(2x^2-2x^2)+(-5x+5x)+(2-3-2)\\=-3\)

b)

\(x(x+3y+1)-2y(x-1)-(y+x+1)x\\=x^2+3xy+x-2xy+2y-xy-x^2-x\\=(x^2-x^2)+(3xy-2xy-xy)+(x-x)+2y\\=2y\)

Bài 2.

a)

\((14x^3+12x^2-14x):2x=(x+2)(3x-4)\\\Leftrightarrow 14x^3:2x+12x^2:2x-14x:2x=3x^2-4x+6x-8\\ \Leftrightarrow 7x^2+6x-7=3x^2+2x-8\\\Leftrightarrow (7x^2-3x^2)+(6x-2x)+(-7+8)=0\\\Leftrightarrow 4x^2+4x+1=0\\\Leftrightarrow (2x)^2+2\cdot 2x\cdot 1+1^2=0\\\Leftrightarrow (2x+1)^2=0\\\Leftrightarrow 2x+1=0\\\Leftrightarrow 2x=-1\\\Leftrightarrow x=\frac{-1}2\)

b)

\((4x-5)(6x+1)-(8x+3)(3x-4)=15\\\Leftrightarrow 24x^2+4x-30x-5-(24x^2-32x+9x-12)=15\\\Leftrightarrow 24x^2-26x-5-(24x^2-23x-12)=15\\\Leftrightarrow 24x^2-26x-5-24x^2+23x+12=15\\\Leftrightarrow -3x+7=15\\\Leftrightarrow -3x=8\\\Leftrightarrow x=\frac{-8}3\\Toru\)

\(=\dfrac{x^4-4x^3+x^2-2x^3+8x^2-2x+3x^2-12x+3}{x^2-4x+1}\)

=x^2-2x+3

\(x^4-6x^3+12x^2-14x+3\) 

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

= \(x^2\left(x^2-4x+1\right)-2x\left(x^2-4x+1\right)+3\left(x^2-4x+1\right)\) 

= \(\left(x^2-4x+1\right)\left(x^2-2x+3\right)\)

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

\(=x^2\left(x^2-4x-1\right)-2x\left(x^2-4x-1\right)+3\left(x^2-4x+1\right)\)

\(=\left(x^2-2x+3\right)\left(x^2-4x-1\right)\)

12x² hay 12x

a. x² - 6x = -9

<=> x² - 6x + 9 = 0

<=> (x - 3)² = 0

<=> x - 3 = 0

<=> x = 3

b. 2(x + 3) - x² + 3x = 0

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

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

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

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

=\(\left(x^2-2x+3\right)\left(x^2-4x+1\right)\)

x⁴-6x³+12x²-14x+3 = (x² - 4x + 1) (x² - 2x + 3) 

x⁴-6x³+12x²-14x+3=x⁴-2x³+3x²-4x³+8x²-12x+x²-2x+3

=x²(x²-2x+3)-4x(x²-2x+3)+(x²-2x+3)

=(x²-2x+3)(x²-4x+1)

văn lớp 6 nêu bài học rút ra từ truyện cười và truyện ngụ ngôn

\(x^4-6x^2+12x^2-14x+3\)

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

\(=x^2\left(x^2-2x+3\right)-4x\left(x^2-2x+3\right)+\left(x^2-2x+3\right)\)

\(=\left(x^2-2x+3\right)\left(x^2-4x+1\right)\)