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

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

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

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

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

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

\(D=\dfrac{2}{-x^2-7}=\dfrac{2}{-\left(x^2+7\right)}=\dfrac{-2}{x^2+7}\)

Ta có: x² \(\ge0\forall x\)

Dấu ''='' xảy ra khi x² = 0 ⇔ x = 0

Do đó: x² + 7 \(\ge7>0\)

⇔ \(\dfrac{2}{x^2+7}\le\dfrac{2}{7}\)

⇔ \(\dfrac{-2}{x^2+7}\ge\dfrac{-2}{7}\)

Dấu ''='' xảy ra khi x = 0

Vậy Min D = \(\dfrac{-2}{7}\) tại x = 0

\(C=\dfrac{5}{x^2+6}\)

Ta có: \(x^2\ge0\forall x\)

Dấu ''='' xảy ra khi x² = 0 ⇔ x = 0

Do đó: x² + 6 \(\ge6>0\)

⇔ \(\dfrac{5}{x^2+6}\le\dfrac{5}{6}\)

Dấu ''='' xảy ra khi x = 0

Vậy Max C = \(\dfrac{5}{6}\) tại x = 0

a) -y² + 2xy - x² + 3x - 3y

= (3x - 3y) - (x² - 2xy + y²)

= 3(x - y) - (x - y)²

= (x - y)(3 - x + y)

b) x³ - 2x² - x + 2

= (x³ - x) - (2x² - 2)

= x(x² - 1) - 2(x² - 1)

= (x² - 1)(x - 2)

= (x - 2)(x - 1)(x + 1)

c) x²(x + 1) - 2x(x + 1) + x + 1

= (x + 1)(x² - 2x + 1)

= (x + 1)(x - 1)²

d) a² + b² + 2a - 2b - 2ab

= (a² - 2ab + b²) + (2a - 2b)

= (a - b)² + 2(a - b)

= (a - b)(a - b + 2)

e) 4x² - 8x + 3

= (4x² - 2x) - (6x - 3)

= 2x(2x - 1) - 3(2x - 1)

= (2x - 1)(2x - 3)

f) 25 - 16x²

= 5² - (4x)²

= (5 - 4x)(5 + 4x)

a) (x³ + 8y³) : (2y + x)

= (x + 2y)(x² - 2xy + 4y²) : (2y + x)

= x² - 2xy + 4y²

b) (x³ + 3x²y + 3xy² + y³) : (2x + 2y)

= (x + y)³ : 2(x + y)

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

c) (6x⁵y² - 9x⁴y³ + 15x³y⁴) : 3x³y²

= 3x³y²(2x² - 3xy + 5y²) : 3x³y²

= 2x² - 3xy + 5y²

10x(x - y) - 8 (y - x)

= 10x(x - y) + 8(x - y)

= (x - y)(10x + 8)

\(\dfrac{2x^3}{x^2-1}-\dfrac{6}{3-x}-\dfrac{2x-6}{\left(x^2-1\right)\left(x-3\right)}\)

\(=\dfrac{2x^3\left(x-3\right)}{\left(x^2-1\right)\left(x-3\right)}+\dfrac{6\left(x^2-1\right)}{\left(x^2-1\right)\left(x-3\right)}-\dfrac{2x-6}{\left(x^2-1\right)\left(x-3\right)}\)

\(=\dfrac{2x^4-6x^3+6x^2-6-2x+6}{\left(x^2-1\right)\left(x-3\right)}\)

\(=\dfrac{2x^4-6x^3+6x^2-2x}{\left(x^2-1\right)\left(x-3\right)}\)

n_Mg = \(\dfrac{2,4}{24}=0,1\left(mol\right)\)

PT: 2Mg + O₂ ➝ 2MgO

mol 0,1 0,05 0,1

a) \(V_{O_2\left(đktc\right)}=0,05.22,4=1,12\left(l\right)\)

b) C1: \(m_{O_2}=0,05.32=1,6\left(g\right)\)

m_MgO = 2,4 + 1,6 = 4 (g)

C2: m_MgO = 0,1.40 = 4 (g)

pt vô số nghiệm \(\left(x;y\right)=\left(x;\frac{3x-35}{5}\right)\)

giải chi tiết cko mink đi. mink cần lắm