pạn ơi pạn đã lm đk chưa? nếu lm đk oy cho mk xem cách lm bài 2 nhé. cảm ơn pạn nhìu lắm

a) \(\left(6x^3y^2-4x^2y^3-10x^2y^2\right):2xy\)

=\(\left(6x^3y^2:2xy\right)-\left(4x^2y^3:2xy\right)-\left(10x^2y^2:2xy\right)\)

\(=3x^2y-2xy^2-5xy\)

b) \(\dfrac{2y}{x-2}+\dfrac{5y}{x-2}\)

=\(\dfrac{2y+5y}{x-2}\)

=\(\dfrac{7y}{x-2}\)

c)\(\dfrac{xy}{3x-y}+\dfrac{3x^2}{y-3x}\)

\(=\dfrac{xy}{3x-y}-\dfrac{3x^2}{3x-y}\)

=\(\dfrac{x\left(y-3x\right)}{3x-y}\)

=\(\dfrac{-x\left(3x-y\right)}{3x-y}\)

=-x

d)\(\dfrac{x-1}{6x+12}.\dfrac{x+2}{x-1}\)

=\(\dfrac{\left(x-1\right)\left(x+2\right)}{6\left(x+2\right)\left(x-1\right)}\)

=\(\dfrac{1}{6}\)

a ) \(\left(5x+2y\right)^2=25x^2+20xy+4y^2\)

b ) \(\left(-3x+2\right)^2=9x^2-12x+4\)

c ) \(\left(\dfrac{2}{3}x+\dfrac{1}{3}y\right)^2=\dfrac{4}{9}x^2+\dfrac{4}{9}xy+\dfrac{1}{9}y^2\)

d ) \(\left(2x-\dfrac{5}{2}y\right)^2=4x^2-10xy+\dfrac{25}{4}y^2\)

e ) \(\left(x+\dfrac{4}{3}y^2\right)^2=x^2+\dfrac{8}{3}xy^2+\dfrac{16}{9}y^4\)

f ) \(\left(2x^2+\dfrac{5}{3}y\right)^2=4x^4+\dfrac{20}{3}x^2y+\dfrac{25}{9}y^2\)

a) (5x-y)² = (5x)² - 2.5x.y + y² = 25x² - 10xy +y²

b) (2x + y² )³ = (2x)³ - 3.(2x)².y² + 3.2x.(y²)² - (y²)³ = 8x³ - 12x²y² + 6xy⁴ - y⁶

c) (x + \(\dfrac{1}{4}\))² = x² + 2.x.\(\dfrac{1}{4}\) + (\(\dfrac{1}{4}\))² = x² + \(\dfrac{1}{2}\).x + \(\dfrac{1}{16}\)

d) (\(\dfrac{2}{3}\)x² - \(\dfrac{1}{2}\)y)³ = (\(\dfrac{2}{3}\)x²)³ - 3.(\(\dfrac{2}{3}\)x²)². \(\dfrac{1}{2}\)y + 3.\(\dfrac{2}{3}\)x². ( \(\dfrac{1}{2}\)y)² - (\(\dfrac{1}{2}\)y)³

^{= \(\dfrac{8}{27}\)}x⁶ - \(\dfrac{2}{3}\)x⁴y + \(\dfrac{1}{2}\)x²y² - \(\dfrac{1}{8}\)y³

Đăng ít thôi.

Bài 2:

a) \(\dfrac{x}{x-3}+\dfrac{9-6x}{x^2-3x}\)

\(=\dfrac{x}{x-3}+\dfrac{9-6x}{x\left(x-3\right)}\)

\(=\dfrac{x^2-6x+9}{x\left(x-3\right)}\)

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

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

b) \(\dfrac{6x-3}{x}:\dfrac{4x^2-1}{3x^2}\)

\(=\dfrac{6x-3}{x}.\dfrac{3x^2}{4x^2-1}\)

\(=\dfrac{3\left(2x-1\right).3x^2}{x\left(2x-1\right)\left(2x+1\right)}\)

\(=\dfrac{9x}{2x+1}\)

c) \(\dfrac{x+2}{3x}+\dfrac{x-5}{5x}-\dfrac{x+8}{4x}\)

\(=\dfrac{20x\left(x+2\right)+12x\left(x-5\right)-15x\left(x+8\right)}{60x}\)

\(=\dfrac{20x^2+40x+12x^2-60x-15x^2-120x}{60x}\)

\(=\dfrac{17x^2-140x}{60x}\)

d) \(\dfrac{x^2-x+1}{x^2+x}.\dfrac{x+1}{3x-2}.\dfrac{9x-6}{x^2-x+1}\)

\(=\dfrac{x^2-x+1}{x\left(x+1\right)}.\dfrac{x+1}{3x-2}.\dfrac{3\left(3x-2\right)}{x^2-x+1}\)

\(=\dfrac{3\left(x^2-x+1\right)\left(x+1\right)\left(3x-2\right)}{x\left(x+1\right)\left(3x-2\right)\left(x^2-x+1\right)}\)

\(=\dfrac{3}{x}\).

$a)$ \(x^{12}:\left(-x\right)^6\)

\(=x^{12}:x^6\)

\(=x^{12-6}\)

\(=x^6\)

$b) $ \(\left(-x\right)^7:\left(-x\right)^5\)

\(=\left(-x\right)^{7-5}\)

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

\(=x^2\)

$c)$ \(5x^2y^4:10x^2y\)

\(=\dfrac{1}{2}y^3\)

$e)$ \(\left(-xy\right)^{14}:\left(-xy\right)^7\)

\(=\left(-xy\right)^{14-7}\)

\(=\left(-xy\right)^7\)

Các câu còn lại tương tự nha bạn!

a; \(=x^5-2x^4-x^3-x^3-x^2=x^5-2x^4-2x^3-x^2\)

b: \(=2x^3-6x^2+x^2-3x+x-3\)

\(=2x^3-5x^2-2x-3\)

c: \(=6x^3y^2-3x^3+3x^2-2x^2y^3+x^2y-xy\)

d: \(=x^3-x^3y+x^3y-x^2y^2+xy^3-y^4\)

\(=x^3-x^2y^2+xy^3-y^4\)

Bài 2:

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

Bài 1:

a: \(=\left(\dfrac{2}{3}:\dfrac{-1}{9}\right)\cdot x^4y^2z^6=-6x^4y^2z^6\)

b: \(=-12x^8-21x^5\)

c: =x^3+8

d: \(=125x^3-75x^2+15x-1\)