Điều kiện: { x + 2 ≥ 0 y + 7 ≥ 0 ⇔ { x ≥ − 2 y ≥ − 7 \left\{\begin{matrix} x+2\geq 0\\ y+7\geq 0 \end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq -2\\ y\geq -7 \end{matrix}\right. { x + 2 ≥ 0 y + 7 ≥ 0 ⇔ { x ≥ − 2 y ≥ − 7
Từ phương trình (1) ta có: ( x − 1 ) 3 + 5 ( x − 1 ) = ( y − 1 ) 3 + 5 ( y − 1 ) ( 3 ) (x-1)^3+5(x-1)=(y-1)^3+5(y-1) \ \ \ (3) ( x − 1 ) 3 + 5 ( x − 1 ) = ( y − 1 ) 3 + 5 ( y − 1 ) ( 3 ) Xét hàm số f ( t ) = t 3 + 5 t f(t)=t^3+5t f ( t ) = t 3 + 5 t , trên tập R f ′ ( t ) = 3 t 2 + 5 > 0 ∀ t ∈ R ⇒ f'(t)=3t^2+5>0\forall t\in R\Rightarrow f ′ ( t ) = 3 t 2 + 5 > 0 ∀ t ∈ R ⇒ hàm số f(t) đồng biến trên R. Từ (3): f ( x − 1 ) = f ( y − 1 ) ⇔ x = y ( 4 ) f(x-1)=f(y-1)\Leftrightarrow x=y \ \ (4) f ( x − 1 ) = f ( y − 1 ) ⇔ x = y ( 4 ) Thay (4) vào (2) ta được pt: ( 5 x 2 − 5 x + 10 ) x + 7 + ( 2 x + 6 ) x + 2 = x 3 + 13 x 2 − 6 x + 32 ( 5 ) (5x^2-5x+10)\sqrt{x+7}+(2x+6)\sqrt{x+2}=x^3+13x^2-6x+32 \ \ (5) ( 5 x 2 − 5 x + 1 0 ) x + 7 + ( 2 x + 6 ) x + 2 = x 3 + 1 3 x 2 − 6 x + 3 2 ( 5 ) D K : x ≥ 2 DK : x\geq 2 D K : x ≥ 2
( 5 x 2 − 5 x + 10 ) ( x + 7 − 3 ) + ( 2 x + 6 ) ( x + 2 − 2 ) = x 3 − 2 x 2 + 5 x − 10 ( 5 ) (5x^2-5x+10)(\sqrt{x+7}-3)+(2x+6)(\sqrt{x+2}-2)=x^3-2x^2+5x-10 \ \ (5) ( 5 x 2 − 5 x + 1 0 ) ( x + 7 − 3 ) + ( 2 x + 6 ) ( x + 2 − 2 ) = x 3 − 2 x 2 + 5 x − 1 0 ( 5 ) ( x − 2 ) ( 5 x 2 − 5 x + 10 x + 7 + 3 + 2 x + 6 x + 2 + 2 ) = ( x − 2 ) ( x 2 + 5 ) (x-2)\left ( \frac{5x^2-5x+10}{\sqrt{x+7}+3}+\frac{2x+6}{\sqrt{x+2}+2} \right )=(x-2)(x^2+5) ( x − 2 ) ( x + 7 + 3 5 x 2 − 5 x + 1 0 + x + 2 + 2 2 x + 6 ) = ( x − 2 ) ( x 2 + 5 ) ( x − 2 ) ( 5 x 2 − 5 x + 10 x + 7 + 3 + 2 x + 6 x + 2 + 2 − ( x 2 + 5 ) ) = 0 (x-2)\left ( \frac{5x^2-5x+10}{\sqrt{x+7}+3}+\frac{2x+6}{\sqrt{x+2}+2}-(x^2+5) \right )=0 ( x − 2 ) ( x + 7 + 3 5 x 2 − 5 x + 1 0 + x + 2 + 2 2 x + 6 − ( x 2 + 5 ) ) = 0 x = 2 → ( 4 ) y = 2 ⇒ ( x ; y ) = ( 2 ; 2 ) x=2\overset{(4)}{\rightarrow}y=2\Rightarrow (x;y)=(2;2) x = 2 → ( 4 ) y = 2 ⇒ ( x ; y ) = ( 2 ; 2 ) (thỏa mãn đ/k) 5 x 2 − 5 x + 10 x + 7 + 3 + 2 x + 6 x + 2 + 2 − ( 5 x 2 − 5 x + 10 5 + 2 x + 6 2 ) = 0 \frac{5x^2-5x+10}{\sqrt{x+7}+3}+\frac{2x+6}{\sqrt{x+2}+2}-\left (\frac{5x^2-5x+10}{5} +\frac{2x+6}{2} \right )=0 x + 7 + 3 5 x 2 − 5 x + 1 0 + x + 2 + 2 2 x + 6 − ( 5 5 x 2 − 5 x + 1 0 + 2 2 x + 6 ) = 0
( 5 x 2 − 5 x + 10 ⎵ > 0 , ∀ x ≥ − 2 ) ( 1 x + 7 + 3 − 1 5 ⎵ > 0 , ∀ x ≥ − 2 ) + ( 2 x + 6 ⎵ > 0 , ∀ x ≥ − 2 ) ( 1 x + 2 + 2 − 1 2 ⎵ > 0 , ∀ x ≥ − 2 ) = 0 \small \left (\underbrace{ 5x^2-5x+10 }_{>0,\forall x\geq -2} \right )\left (\underbrace{\frac{1}{\sqrt{x+7}+3} -\frac{1}{5}}_{>0,\forall x\geq -2} \right )+\left (\underbrace{ 2x+6 }_{>0,\forall x\geq -2} \right )\left (\underbrace{ \frac{1}{\sqrt{x+2}+2} -\frac{1}{2}}_{>0,\forall x\geq -2} \right )=0 ( > 0 , ∀ x ≥ − 2 5 x 2 − 5 x + 1 0 ) ⎝ ⎜ ⎛ > 0 , ∀ x ≥ − 2 x + 7 + 3 1 − 5 1 ⎠ ⎟ ⎞ + ( > 0 , ∀ x ≥ − 2 2 x + 6 ) ⎝ ⎜ ⎛ > 0 , ∀ x ≥ − 2 x + 2 + 2 1 − 2 1 ⎠ ⎟ ⎞ = 0 (pt vô nghiệm)
Vậy hệ phương trình có một nghiệm duy nhất: ( x ; y ) = ( 2 ; 2 ) \small (x;y)=(2;2) ( x ; y ) = ( 2 ; 2 )