Ta có: \(x^2+4x+y^2-12=0\Rightarrow x^2+4x+4+y^2-16=0\)

\(\Rightarrow\left(x+2\right)^2=16-y^2\)

\(\Rightarrow x+2=\sqrt{16-y^2}\Rightarrow x=\sqrt{16-y^2}-2\)

\(\Rightarrow-4\le y\le4\) (Vì y nguyên và để \(16-y^2\ge0\) hay \(\sqrt{16-y^2}\) có nghĩa).

\(\Rightarrow y=\left\{-4,-3,-2,-1,0,1,2,3,4\right\}\)

Vậy y=4, x=-2

Do đó Pmax = 4 + 16 = 20.

(Đây là chỉ là cách giải mẹo. Không chắc có phải cách làm trong bài không. Cách giải chỉ mang tính chất tham khảo).

2. Ta có: \(\dfrac{a_1}{a_2}+\dfrac{b_1}{b_2}+\dfrac{c_1}{c_2}=0\)

\(\Rightarrow\dfrac{a_1b_2c_2+b_1a_2c_2+c_1a_2b_2}{a_2b_2c_2}=0\)

\(\Rightarrow a_1b_2c_2+b_1a_2c_2+c_1a_2b_2=0\)

Lại có: \(\dfrac{a_2}{a_1}+\dfrac{b_2}{b_1}+\dfrac{c_2}{c_1}=1\)

\(\Rightarrow\left(\dfrac{a_2}{a_1}+\dfrac{b_2}{b_1}+\dfrac{c_2}{c_1}\right)^2=1\)

\(\Rightarrow\dfrac{a_2^2}{a_1^2}+\dfrac{b_2^2}{b_1^2}+\dfrac{c_2^2}{c_1^2}+2\left(\dfrac{a_2b_2}{a_1b_1}+\dfrac{b_2c_2}{b_1c_1}+\dfrac{a_2c_2}{a_1c_1}\right)=1\)

Mặt khác: \(\dfrac{a_2b_2}{a_1b_1}+\dfrac{b_2c_2}{b_1c_1}+\dfrac{a_2c_2}{a_1c_1}=\dfrac{a_1b_2c_2+b_1a_2c_2+c_1a_2b_2}{a_1b_1c_1}=0\)

Vậy \(\dfrac{a_2^2}{a_1^2}+\dfrac{b_2^2}{b_1^2}+\dfrac{c_2^2}{c_1^2}=1\) (đpcm)

(10%)²⁼\(\left(\frac{10}{100}\right)^2\)=\(\left(\frac{100}{10000}\right)\)=\(\frac{1}{100}\)=1%

Ta có: \(x+\dfrac{1}{x}=3\Rightarrow\left(x+\dfrac{1}{x}\right)^2=9\Rightarrow x^2+\dfrac{1}{x^2}+2=9\)

\(\Rightarrow x^2+\dfrac{1}{x^2}=7\)

Lại có: \(x+\dfrac{1}{x}=3\Rightarrow\left(x+\dfrac{1}{x}\right)^3=27\)

\(\Rightarrow x^3+\dfrac{1}{x^3}+3\left(x+\dfrac{1}{x}\right)=27\)

\(\Rightarrow x^3+\dfrac{1}{x^3}+9=27\Rightarrow x^3+\dfrac{1}{x^3}=18\)

Do đó: \(\left(x^2+\dfrac{1}{x^2}\right)\left(x^3+\dfrac{1}{x^3}\right)=x^5+\dfrac{1}{x^5}+x+\dfrac{1}{x}=7.18=126\)

\(\Rightarrow x^5+\dfrac{1}{x^5}+3=126\Rightarrow x^5+\dfrac{1}{x^5}=123\)

Vậy \(x^5+\dfrac{1}{x^5}=123\)