\(x^3-3x^2+3x-1\)

\(=x^3-3.x^2.1+3.x.1^2-1^3\)

\(=\left(x-1\right)^3\)

\(x^3-3x^2+3x-1\)

\(=x^3-3.x^2.1+3.x.1^2-1^3\)

\(=\left(x-1\right)^3\)

1200

Diện tích 1 viên gạch:

60 x 60 = 3600(cm²)=0,36(m²)

Căn phòng đó rộng:

0,36 x 500 = 180(m²)=13,... (em xem lại đề)

GIÚP MÌNH VỚI MN ƠIIII

Xét 3 số TN liên tiếp \(\left(n-1\right);n;\left(n+1\right)\) ta có

\(\left(n-1\right).n.\left(n+1\right)=n.\left(n^2-1\right)=n^3-n< n^3\)

\(\Rightarrow A\le\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{20.21.22}=\)

\(=\dfrac{1}{2}\left(\dfrac{3-1}{1.2.3}+\dfrac{4-2}{2.3.4}+\dfrac{5-3}{3.4.5}+...+\dfrac{22-20}{20.21.22}\right)=\)

\(=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{20.21}-\dfrac{1}{21.22}\right)=\)

\(=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{21.22}\right)=\dfrac{1}{2^2}-\dfrac{1}{2.21.22}< \dfrac{1}{2^2}\)

(2x)⁴ + (2x + y)² = 0 (1)

Do (2x)⁴ ≥ 0 với mọi x ∈ R

(2x + y)² ≥ 0 với mọi x, y ∈ R

(1) ⇒ (2x)⁴ = 0 và (2x + y)² = 0

*) (2x)⁴ = 0

2x = 0

x = 0

*) (2x + y)² = 0

2x + y = 0

y = -2x

y = 0

Vậy x = y = 0