bn ơi sao y/2=z/3 => y/10=z/25 đc v ạ.

Hoa is interested in playing badminton in her free time.

Ta thấy: (2x - 5)²⁰²⁴≥ 0 ∀ x ∈ R

              (3y + 4)²⁶ ≥ 0 ∀ y ∈ R

=> (2x - 5)²⁰²⁴ + (3y + 4)²⁶ ≥ 0 

Mặt khác:  (2x - 5)²⁰²⁴ + (3y + 4)²⁶ ≤ 0

Suy ra:  (2x - 5)²⁰²⁴ + (3y + 4)²⁶ = 0

\(\Leftrightarrow\left\{{}\begin{matrix}2x-5=0\\3y+4=0\end{matrix}\right.\)                          \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{2}\\y=-\dfrac{4}{3}\end{matrix}\right.\)

Vậy...

= \(\left(\dfrac{7}{3}+\dfrac{7}{2}\right):\left(-\dfrac{25}{6}+\dfrac{22}{7}\right)+\dfrac{15}{2}\)

= \(\dfrac{35}{6}:\left(-\dfrac{43}{42}\right)+\dfrac{15}{2}\)

= \(-\dfrac{245}{43}+\dfrac{15}{2}\) = \(\dfrac{155}{86}\)

G = x² - x + 10

= \(\left[x^2-2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\right]+\dfrac{39}{4}\)

= \(\left(x-\dfrac{1}{2}\right)^2+\dfrac{39}{4}\)

Ta thấy: \(\left(x-\dfrac{1}{2}\right)^2\ge0\forall x\in R\)

=> \(\left(x-\dfrac{1}{2}\right)^2+\dfrac{39}{4}\ge\dfrac{39}{4}\)

=> Min G = \(\dfrac{39}{4}\)\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2=0\)

\(\Leftrightarrow x=\dfrac{1}{2}\)

Vậy Min G = 39/4 khi x = 1/2

Có: (2x + y)(4x² - 2xy + y²) = 8x³ + y³

Lời giải:

Ta có: (2x + y)(4x² - 2xy + y²) 

= 2x(4x² - 2xy + y²) + y(4x² - 2xy + y²)

= 8x³ - 4x²y + 2xy² + 4x²y - 2xy² + y³

= 8x³ + y^3.

Có: A = (2,4x² + 1,7y² + 2xy) - (0,4x² - 1,3y² + xy)

= (2,4x² - 0,4x²) + (1,7y² + 1,3y²) + (2xy - xy)

= 2x² + 3y² + xy

Bậc của đa thức A = 2