\(a,\left(2x-1\right)^2-\left(x-3\right)\left(x+3\right)-1969\\ =4x^2-4x+1-x^2+9-1969\\ =3x^2-4x-1959\)

\(b,\left(2x-3y\right)\left(2x+3y\right)-\left(2x-y\right)^2\\ =4x^2-9y^2-4x^2+4xy-y^2\\ =8y^2+4xy=4y\left(2y+x\right)\)

\(c,\left(x+3y\right)^2+\left(x+y\right)\left(x-y\right)+280\\ =x^2+6xy+9y^2+x^2-y^2+280\\ =2x^2+8y^2+6xy+280\)

a: \(\left(2x-1\right)^2-\left(x-3\right)\cdot\left(x+3\right)-1969\)

\(=4x^2-4x+1-x^2+9-1969\)

\(=3x^2-4x-1959\)

b: \(\left(2x-3y\right)\left(2x+3y\right)-\left(2x-y\right)^2\)

\(=4x^2-9y^2-4x^2+4xy-y^2\)

\(=-10y^2+4xy\)

a)\(\text{( 2 x − 1 )^2− ( x − 3 ) ( x + 3 ) − 1969}\)

\(\text{= 4x^2 − 4x + 1 − x^2 + 9 − 1969}\)

\(\text{=3x^2− 4 x − 1959}\)

b) \(\text{( 2 x − 3 y ) ( 2 x + 3 y ) − ( 2 x − y )^2}\)

=\(\text{= 4 x^2− 9 y^2− 4 x^2 + 4 x y − y^2}\)

\(\text{= -10 y^2+ 4 x y = -2 y ( 5 y -2 x )}\)

c)\(\text{( x + 3 y )^2 + ( x + y ) ( x − y ) + 280}\)

\(\text{= x^2 + 6 x y + 9 y^2 + x^2 − y^2 + 280}\)

\(\text{= 2 x^2 + 8 y^2 + 6 x y + 280}\)

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) A = [(2x + y) - (2x - y)] . [(2x +y) + (2x - y)]

b) B = [(x - 2y) - 2y]²

\(a,A=\left(2x+y\right)^2-\left(2x-y\right)^2\\ =\left(2x+y-2x+y\right)\left(2x+y+2x-y\right)\\ =2y\cdot4x\\ =8xy\\ b,B=\left(x-2y\right)^2-4y\left(x-2y\right)+4y^2\\ =x^2-4xy+4y^2-4xy+8y^2+4y^2\\ =x^2+16y^2-8xy\\ =\left(x-4y\right)^2\)

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

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

\(=2y.4x=8xy\)

Vậy \(A=8xy\)

\(----------\)

\(b,B=\left(x-2y\right)^2-4\left(x-2y\right)y+4y^2\)

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

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

\(=\left(x-4y\right)^2\)

Vậy \(B=\left(x-4y\right)^2\)

B) Ta có: 2x-2y-x²+2xy-y²

⇔ 2(x-y)-(x²-2xy+y²)

⇔ 2(x-y)-(x-y)²

⇔ (x-y)(2-x+y)

Đúng thì tick nhé

a) 2x(x-3y)+3y(2x+5y)

=2x²-6xy+6xy+15y²

=2x²+15y²

b)(5x-3y)(2x+y)-x(10x-y)

=10x²+5xy-6xy-3y²-10x²+xy

=0

c)(x-y)(x²+xy+y²)-(x+y)(x²-xy+y²)

=x³-y³-(x³+y³)

=x³-y³-x³-y³

=-2y³

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)

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