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)

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

2a) pt <=> (x + 6)^2 = 0

<=> x = -6

b) pt <=> (4x - 1)^2 = 0

<=> x = 1/4

c) pt<=> (x + 1)^3 = 0

<=> x = -1

Bài 1:

a: Ta có: \(A=\left(4x+3y\right)^2+\left(4x-3y\right)^2\)

\(=16x^2+24xy+9y^2+16x^2-24xy+9y^2\)

\(=32x^2+18y^2\)

b: Ta có: \(B=\left(x-2\right)^3-\left(x+2\right)^3\)

\(=x^3-6x^2+12x-8-x^3-6x^2-12x-8\)

\(=-12x^2-24\)

Bài 2: 

a: Ta có: \(x^2+12x+36=0\)

\(\Leftrightarrow x+6=0\)

hay x=-6

b: Ta có: \(16x^2-8x+1=0\)

\(\Leftrightarrow4x-1=0\)

hay \(x=\dfrac{1}{4}\)

Bài 1: 

a: Ta có: \(A=\left(4x+3y\right)^2+\left(4x-3y\right)^2\)

\(=16x^2+24xy+9y^2+16x^2-24xy+9y^2\)

\(=32x^2+18y^2\)

b: Ta có: \(B=\left(x-2\right)^3-\left(x+2\right)^3\)

\(=x^3-6x^2+12x-8-x^3-6x^2-12x-8\)

\(=-12x^2-24\)

c: Ta có: \(C=\left(x+2y\right)^2+2\left(x+2y\right)\left(x-2y\right)+\left(x-2y\right)^2\)

\(=\left(x+2y+x-2y\right)^2\)

\(=4x^2\)

câu 2: c,4x^4 y^4

\(2,B\\ 3,D\\ 4,D\\ 5,B,C\\ 6,A\\ 7,C\\ 8,A\)

a: \(3x\left(x-3\right)+4x-12=0\)

=>\(3x\left(x-3\right)+\left(4x-12\right)=0\)

=>\(3x\left(x-3\right)+4\left(x-3\right)=0\)

=>\(\left(x-3\right)\left(3x+4\right)=0\)

=>\(\left[{}\begin{matrix}x-3=0\\3x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{4}{3}\end{matrix}\right.\)

b: Sửa đề:\(\left(x+1\right)\left(x^2-x+1\right)-x^3+2x=17\)

\(\Leftrightarrow x^3+1-x^3+2x=17\)

=>2x+1=17

=>2x=17-1=16

=>\(x=\dfrac{16}{2}=8\)

c: \(\left(x-3\right)\left(x+5\right)+\left(x-1\right)^2-6x^4y^2:3x^2y^2=15x\)

=>\(x^2+2x-15+x^2-2x+1-2x^2=15x\)

=>\(15x=-14\)

=>\(x=-\dfrac{14}{15}\)

Bài 1: 

a, (\(x\) - 4).(\(x\) + 4) - (5 - \(x\)).(\(x\) + 1)

= \(x^2\) -  16 - 5\(x\) - 5 + \(x^2\) + \(x\) 

= (\(x^2\) + \(x^2\)) - (5\(x\) - \(x\)) - (16 + 5)

= 2\(x^2\) - 4\(x\) - 21

b, (3\(x^2\) - 2\(xy\) + 4) + (5\(xy\) - 6\(x^2\) - 7)

=  3\(x^2\) - 2\(xy\) + 4 + 5\(xy\) - 6\(x^2\) - 7

= (3\(x^2\) - 6\(x^2\)) + (5\(xy\) - 2\(xy\)) - (7 - 4)

= - 3\(x^2\) + 3\(xy\) - 3

Bài 2:

a, 3\(x^2\).(2\(x\) + y) - 2y(4\(x^2\) - y)

= 6\(x^3\) + 3\(x^2\).y - 8y\(x^2\) + 2y²

= 6\(x^3\) - (8\(x^2\)y - 3\(x^2\)y) + 2y²

= 6\(x^3\) - 5\(x^2\)y + 2y²

a: \(\dfrac{-6x^3y^4+4x^4y^3}{2x^3y^3}\)

\(=\dfrac{-6x^3y^4}{2x^3y^3}+\dfrac{4x^4y^3}{2x^3y^3}\)

\(=-3y+2x\)

b: \(\dfrac{5x^4y^2-x^3y^2}{x^3y^2}=\dfrac{5x^4y^2}{x^3y^2}-\dfrac{x^3y^2}{x^3y^2}\)

\(=5x-1\)

c: \(\dfrac{27x^3y^5+9x^2y^4-6x^3y^3}{-3x^2y^3}\)

\(=-\dfrac{27x^3y^5}{3x^2y^3}-\dfrac{9x^2y^4}{3x^2y^3}+\dfrac{6x^3y^3}{3x^2y^3}\)

\(=-9xy^2-3y+2x\)

a) \(\dfrac{-6x^3y^4+4x^4y^3}{2x^3y^3}\)

\(=\dfrac{2x^3y^3\cdot\left(-3y+2x\right)}{2x^3y^3}\)

\(=-3y+2x\)

\(=2x-3y\)

b) \(\dfrac{5x^4y^2-x^3y^2}{x^3y^2}\)

\(=\dfrac{5x\cdot x^3y^2-x^3y^2\cdot1}{x^3y^2}\)

\(=\dfrac{x^3y^2\cdot\left(5x-1\right)}{x^3y^2}\)

\(=5x-1\)

c) \(\dfrac{27x^3y^5+9x^2y^4-6x^3y^3}{-3x^2y^3}\)

\(=\dfrac{-3x^2y^3\cdot-9xy^2+-3x^2y^3\cdot-3y+-3x^2y^3\cdot2x}{-3x^2y^3}\)

\(=\dfrac{-3x^2y^3\cdot\left(-9xy^2-3y+2x\right)}{-3x^2y^3}\)

\(=-9xy^2-3x+2x\)