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x^2 + 3xy + 2y^2 = 0
=> x^2 + xy + 2xy + 2y^2 = 0
=> x(x+y) + 2y ( x+ y ) = 0 =
=> ( x+ 2y)( x + y ) = 0
=> x = -2y hoặc x = -y
(+) x = -2y thay vào ta có :
8y^2 + 6y + 5 = 0 giải ra y => x
(+) thay x = -y ta có :
2y^2 - 3y + 5 = 0 tương tự
Ta có:
\(x^2+x^2y^2-2y=0\)
\(\Leftrightarrow x^2=\frac{2y}{y^2+1}\le1\)(cái này chứng minh đơn giản b tự làm lấy nhé)
\(\Leftrightarrow-1\le x\le1\left(1\right)\)
Ta lại có:
\(x^3+2y^2-4y+3=0\)
\(\Leftrightarrow x^3=-1-2\left(y-1\right)^2\le-1\left(2\right)\)
Từ (1) và (2) \(\Rightarrow x=-1\)
\(\Rightarrow y=1\)
\(\Rightarrow x^2+y^2=1+1=2\)
Bài 1 :
Ta có :
\(x^7+\frac{1}{x^7}=\left(x^3+\frac{1}{x^3}\right)\left(x^4+\frac{1}{x^4}\right)-\left(x+\frac{1}{x}\right)\)
\(\left(x+\frac{1}{x}\right)=a\Leftrightarrow\left(x+\frac{1}{x}\right)^2=a^2\)
\(\Leftrightarrow x^2+\frac{1}{x^2}+2.x.\frac{1}{x}=a^2\)
\(\Leftrightarrow x^2+\frac{1}{x^2}=a^2-2\)
\(x^3+\frac{1}{x^3}=\left(x+\frac{1}{x}\right)\left(x^2-x.\frac{1}{x}+\frac{1}{x^2}\right)\)
\(=a\left(x^2+\frac{1}{x^2}-1\right)=a\left(a^2-3\right)\)
\(x^4+\frac{1}{x^4}=\left(x^2+\frac{1}{x^2}\right)^2-2.x^2.\frac{1}{x^2}\)
\(=\left(a^2-2\right)^2-2=a^4-4a^2+4-2\)
\(=a^4-4a^2+2\)
\(\Rightarrow x^7+\frac{1}{x^7}=a.\left(a^2-3\right).\left(a^4-4a^2+2\right)-a\)
\(=\left(a^3-3a\right)\left(a^4-4a^2+2\right)-a\)
\(=a^7-4a^5+2a^3-3a^5+12a^3-6a-a\)
\(=a^7-7a^5+14a^3-7a\)
Bài 2 :
Ta có :
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2\)
\(\Rightarrow\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2=2^2\)
\(\Rightarrow\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{2}{xy}+\frac{2}{yz}+\frac{2}{zx}=4\)
\(\Rightarrow\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{2}{xy}+\frac{2}{yz}+\frac{2}{zx}=\frac{2}{xy}-\frac{1}{z^2}\)
\(\Rightarrow\frac{1}{x^2}+\frac{1}{y^2}+\frac{2}{z^2}+\frac{2}{yz}+\frac{2}{zx}=0\)
\(\Rightarrow\left(\frac{1}{x^2}+\frac{2}{xz}+\frac{1}{z^2}\right)+\left(\frac{1}{y^2}+\frac{2}{yz}+\frac{1}{z^2}\right)=0\)
\(\Rightarrow\left(\frac{1}{x}+\frac{1}{z}\right)^2+\left(\frac{1}{y}+\frac{1}{z}\right)^2=0\)
\(\Rightarrow\frac{1}{x}+\frac{1}{z}=\frac{1}{y}+\frac{1}{z}=0\) vì \(\left(\frac{1}{x}+\frac{1}{z}\right)^2,\left(\frac{1}{y}+\frac{1}{z}\right)^2\ge0\)
\(\Rightarrow x=y=-z\)
\(\Rightarrow\frac{1}{-z}+\frac{1}{-z}+\frac{1}{z}=2\Rightarrow-\frac{1}{z}=2\Rightarrow z=-\frac{1}{2}\)
\(\Rightarrow x=y=\frac{1}{2}\)
\(\Rightarrow x+2y+z=\frac{1}{2}+2.\frac{1}{2}-\frac{1}{2}=1\)
\(\Rightarrow P=1\)
Xét \(x^3-x^2+x-5=0\)
\(\Leftrightarrow\left(x-\frac{1}{3}\right)^3+\frac{2}{3}\left(x-\frac{1}{3}\right)=\frac{128}{27}\)
Xét \(y^3-2y^2+2y+4=0\)
\(\Leftrightarrow\left(y-\frac{2}{3}\right)^3+\frac{2}{3}\left(y-\frac{2}{3}\right)=-\frac{128}{27}\)
Cộng theo vế 2 dòng có dấu <=> ta có:
\(\left(x-\frac{1}{3}\right)^3+\left(y-\frac{2}{3}\right)^3+\frac{2}{3}\left(x-\frac{1}{3}+y-\frac{2}{3}\right)=0\)
\(\Leftrightarrow\left(x-\frac{1}{3}+y-\frac{2}{3}\right)\left(\left(x-\frac{1}{3}\right)^2+\left(x-\frac{1}{3}\right)\left(y-\frac{2}{3}\right)+\left(y-\frac{2}{3}\right)^2\right)+\frac{2}{3}\left(x+y-1\right)=0\)
\(\Leftrightarrow\left(x+y-1\right)\left(\left(x-\frac{1}{3}\right)^2+\left(x-\frac{1}{3}\right)\left(y-\frac{2}{3}\right)+\left(y-\frac{2}{3}\right)^2\right)+\frac{2}{3}\left(x+y-1\right)=0\)
\(\Leftrightarrow\left(x+y-1\right)\left(\left(x-\frac{1}{3}\right)^2+\left(x-\frac{1}{3}\right)\left(y-\frac{2}{3}\right)+\left(y-\frac{2}{3}\right)^2+\frac{2}{3}\right)=0\)
Dễ thấy: \(\left(x-\frac{1}{3}\right)^2+\left(x-\frac{1}{3}\right)\left(y-\frac{2}{3}\right)+\left(y-\frac{2}{3}\right)^2+\frac{2}{3}>0\)
\(\Rightarrow x+y-1=0\Rightarrow x+y=1\)
Done !!!
sai hdt roi ban oi