9 The house were catching fire while they were sleeping 

Áp dụng dãy tỉ số bằng nhau 

\(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}=\dfrac{2y-4}{6}=\dfrac{3z-9}{12}=\dfrac{\left(x-1\right)-\left(2y-4\right)+\left(3z-9\right)}{4-6+12}\)

\(=\dfrac{x-2y+3z-6}{10}=\dfrac{14-6}{10}=\dfrac{4}{5}\)

Khi đó ta được \(\left\{{}\begin{matrix}\dfrac{x-1}{2}=\dfrac{4}{5}\\\dfrac{y-2}{3}=\dfrac{4}{5}\\\dfrac{z-3}{4}=\dfrac{4}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{13}{5}\\y=\dfrac{22}{5}\\z=\dfrac{31}{5}\end{matrix}\right.\)

Khi đó ta được 

x(y + 1) - (y + 1) = 10

<=> (x - 1)(y + 1) = 10

Với x;y \(\inℤ\Rightarrow x-1\inℤ;y+1\inℤ\)

Lập bảng xét các trường hợp

Vậy (x;y) = (2;9) ; (11;0) ; (-9;-2) ; (0;-11) 

A = 1 + 3 + 3² + 3³ + .... + 3⁴¹

= (1 + 3 + 3²) + (3³ + 3⁴ + 3⁵) + (3³⁹ + 3⁴⁰ + 3⁴¹)

= (1 + 3 + 3²) + 3³.(1 + 3 + 3²) + ... + 3³⁹(1 + 3 + 3²)

= (1 + 3 + 3²)(1 + 3³ + ... + 3³⁹)

= 13(1 + 3³ + ... + 3³⁹) 

=> A là hợp số

ĐKXĐ : \(\left\{{}\begin{matrix}x>-2\\y>-2\end{matrix}\right.\)

Có : x³ + x + 2 = y³ - 3y² + 4y

<=> x³ + x + 2 = (y³ - 3y² + 3y - 1) + y + 1

<=> x³ + x + 2 = (y - 1)³ + y + 1 

<=> x³ - (y - 1)³ + x - y + 1 = 0 

<=> (x - y + 1)[x² + x(y - 1) + (y - 1)²]  + (x - y + 1)  = 0

<=>  (x - y + 1)[x² + x(y - 1) + (y - 1)² + 1] = 0

<=> x - y + 1 = 0 (Vì  x² + x(y - 1) + (y - 1)² + 1 > 0 \(\forall x;y\)  )

<=> y = x + 1

Thay y = x + 1 

\(2\sqrt{x+2}=y+2\)

\(\Leftrightarrow2\sqrt{x+2}=x+3\)

\(\Leftrightarrow x-2\sqrt{x+2}+3=0\)

\(\Leftrightarrow(\sqrt{x+2}-1)^2=0\)

\(\Leftrightarrow\sqrt{x+2}=1\)

\(\Leftrightarrow x=-1\) (tm)

Khi đó y = 0

Vậy (x;y) = (-1;0) 

Mình sửa lại chỗ 

\((\overrightarrow{CA}+k\overrightarrow{AB)}(\overrightarrow{AC}+\dfrac{1}{6}\overrightarrow{CB})=0\)

\(\Leftrightarrow-AC^2+\dfrac{1}{6}.CA.CB.cosC+k.AB.AC.cosA+\dfrac{1}{6}.k.AB.CB.cos\left(180^{\text{o}}-B\right)=0\)

\(\Leftrightarrow-12^2+\dfrac{1}{6}.12.11.\dfrac{5}{8}+k.10.12.\dfrac{41}{80}+\dfrac{1}{6}k.10.11.\dfrac{-7}{20}=0\)

\(\Leftrightarrow k=\dfrac{1563}{661}=\dfrac{x}{y}\)

Vì x;y nguyên ; phân số tối giản nên 

(x;y) = (1563;661) ; (-1563;-661)

Với (x,y)  = (1563;661) 

=> P = \(\sqrt{2022.1563-2023.661+2580888}=\sqrt{4404071}\)

Với (x;y) = (-1563;-661)

=> P = \(\sqrt{-2022.1563+2023.661+2580888}=\sqrt{757705}\)

Từ tam giác ABC có AB = 10 ; BC = 11 ; CA = 12 

=> \(cosC=\dfrac{AC^2+BC^2-AB^2}{2AC.BC}=\dfrac{11^2+12^2-10^2}{2.11.12}=\dfrac{5}{8}\)

tương tự cosB = \(\dfrac{7}{20}\)

cos A = \(\dfrac{41}{80}\)

Lại có \(CN\perp AM\)

nên \(\overrightarrow{CN}.\overrightarrow{AM}=0\)

\(\Leftrightarrow(\overrightarrow{CA}+\overrightarrow{AN}).(\overrightarrow{AC}+\overrightarrow{CM})=0\)

\(\Leftrightarrow(\overrightarrow{CA}+k\overrightarrow{AB})(\overrightarrow{AC}+\dfrac{1}{5}\overrightarrow{CB})=0\)

\(\Leftrightarrow-AC^2+\dfrac{1}{5}CA.CB.cosC+kAB.AC.cosA+\dfrac{1}{5}k.AB.BC.cos(180^{\text{o}}-B)=0\)

\(\Leftrightarrow-12^2+\dfrac{1}{5}.12.11.\dfrac{5}{8}+k.10.12.\dfrac{41}{80}+\dfrac{1}{5}k.10.11.(-\dfrac{7}{20})=0\)

\(\Leftrightarrow k=\dfrac{1275}{538}=\dfrac{x}{y}\) tối giản => (x ; y) = (1275;538) ; (-1275,-538) (x;y \(\inℤ\))

Với (x,y) = (1275,538)

=> P = \(\sqrt{2022.1275-2023.538+2580888}=\sqrt{4070564}\)

Với (x;y) = (-1275 ; -538) 

=> P = \(\sqrt{1091212}\)

Ta có bất đẳng thức sau 

a² + b² + c² \(\ge\) ab + bc + ca (1)

Dấu "=" xảy ra <=> a = b = c

Thật vậy (1) <=> 2a² + 2b² + 2c² - 2ab - 2bc - 2ca \(\ge0\) 

  <=> (a - b)² + (b - c)² + (c - a)² \(\ge0\) (bđt này luôn đúng)

Khi đó ta được (1) <=> 2(a² + b² + c²) \(\ge\) 2(ab + bc + ca) 

<=> 3(a² + b² + c²) \(\ge\) 2ab + 2bc + 2ca + a² + b² + c² 

<=> 3(a² + b² + c²) \(\ge\) (a + b + c)² 

=> -(a² + b² + c²) \(\le\dfrac{(a+b+c)^2}{3}\)

Ta có \(P=\dfrac{b+c}{b+c-a}+\dfrac{c+a}{c+a-b}+\dfrac{a+b}{a+b-c}\)

\(=\dfrac{a}{b+c-a}+\dfrac{b}{a+c-b}+\dfrac{c}{a+b-c}+3\)

\(=\dfrac{a^2}{ab+ac-a^2}+\dfrac{b^2}{ab+bc-b^2}+\dfrac{c^2}{ac+bc-c^2}+3\)

\(\ge\dfrac{\left(a+b+c\right)^2}{ab+ac-a^2+ab+bc-b^2+ac+bc-c^2}+3\) (BĐT Schwarz)

\(=\dfrac{\left(a+b+c\right)^2}{2ab+2ac+2bc-a^2-b^2-c^2}+3\)

\(=\dfrac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2-2\left(a^2+b^2+c^2\right)}+3\)

\(\ge\dfrac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2-\dfrac{2}{3}\left(a+b+c\right)^2}+3=\dfrac{1}{1-\dfrac{2}{3}}+3=6\) (đpcm) 

B = x² + 9y² + 2011

= x² + (3y)² + 2011

= x² + (5 - 2x)² + 2011 (do 2x + 3y = 5)

= x² + 4x² - 20x + 25 + 2011

= 5x² - 20x + 2036

= 5x² - 20x + 20 + 2016

= 5(x² - 4x + 4)  + 2016

= 5(x - 2)² + 2016 \(\ge2016\)

=> Min B = 2016 khi x - 2 = 0 <=> x = 2 

khi đó y = \(\dfrac{1}{3}\)

Vậy B_min = 2016 khi x = 2 ; \(y=\dfrac{1}{3}\)

Ta có x² + y² + z² = 6 ; xy - 3x + 2z = 10

Khi đó 4(x² + y² + z²) - 4(xy - 3x + 2z) = 24 - 40

<=> 4x² + 4y² + 4z² - 4xy + 12x - 8z + 16 = 0

<=> (x² - 4xy + 4y²) + (3x² + 12x + 12) + (4z² - 8z + 4) = 0

<=> (x - 2y)² + 3(x + 2)² + 4(z - 1)² = 0

<=> \(\left\{{}\begin{matrix}x-2y=0\\x+2=0\\z-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-1\\x=-2\\z=1\end{matrix}\right.\)

Thay x = -2 ; y = -1 ; z = 1 vào P ta được \(P=\dfrac{1006xy-2019y-x^3+z^5}{x^2+2y^3}\)

\(=\dfrac{1006.(-2).(-1)-2019.(-1)-(-2)^3+1^5}{(-2)^2+2.1^3}\)

\(=\dfrac{2020}{3}\)