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

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

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)

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

\(a,\Leftrightarrow x^2-16-x^2-2x=0\\ \Leftrightarrow2x=-16\Leftrightarrow x=-8\\ b,\Leftrightarrow\left(x-2\right)\left(3x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\\ c,\Leftrightarrow6x\left(1-2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\end{matrix}\right.\\ d,\Leftrightarrow12x-56x^2+x^2-16=0\\ \Leftrightarrow55x^2-12x+16=0\\ \Delta=144-4\cdot55\cdot16< 0\\ \Leftrightarrow x\in\varnothing\)

x=13 nên x+1=14

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

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

=x-1

=13-1=12

x 3 - x - 1 3 4 x + 3 x - 5 = 7 x - 1 4 x + 3 - x x - 5         Đ K X Đ :   x ≠ - 3 4   v à   x ≠ 5 ⇔ x 3 - x - 1 3 4 x + 3 x - 5 = 7 x - 1 x - 5 4 x + 3 x - 5 - x 4 x + 3 4 x + 3 x - 5

⇔ x 3 - x - 1 3  = (7x – 1)(x – 5) – x(4x + 3)

⇔ x 3  –  x 3  + 3 x 2  – 3x + 1 = 7 x 2  – 35x – x + 5 – 4 x 2  – 3x

⇔ 3 x 2 – 7 x 2  + 4 x 2 – 3x + 35x + x + 3x = 5 – 1

⇔ 36x = 4 ⇔ x = 1/9 (thoả mãn)

Vậy phương trình có nghiệm x = 1/9

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