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Lời giải:
Đặt \(\left\{\begin{matrix} (x+y)^2=a\neq 0\\ xy=b\end{matrix}\right.\)
Dùng cách biến đổi tương đương.
Ta có: \(A=x^2+y^2+\left(\frac{xy+1}{x+y}\right)^2=(x+y)^2-2xy+\frac{(xy+1)^2}{(x+y)^2}\)
\(A=a-2b+\frac{(b+1)^2}{a}\)
\(A\geq 2\Leftrightarrow a-2b+\frac{(b+1)^2}{a}\geq 2\)
\(\Leftrightarrow a^2-2ab+(b+1)^2\geq 2a\)
\(\Leftrightarrow a^2+b^2+1-2ab+2b-2a\geq 0\)
\(\Leftrightarrow (-a+b+1)^2\geq 0\) (luôn đúng)
Do đó ta có đpcm.
Dấu bằng xảy ra khi \(-a+b+1=0\Leftrightarrow x^2+y^2+xy=1\)
1/
\(x^2-xy-2y^2=0\Leftrightarrow x^2+xy-2xy-2y^2=0\)
\(\Leftrightarrow x\left(x+y\right)-2y\left(x+y\right)=0\)
\(\Leftrightarrow\left(x+y\right)\left(x-2y\right)=0\Rightarrow x=2y\) (do \(x+y\ne0\))
\(\Rightarrow P=\frac{2y-y}{2y+y}=\frac{y}{3y}=\frac{1}{3}\)
2/
\(x^4-30x^2+31x-30=0\)
\(\Leftrightarrow x^4+x-30x^2+30x-30=0\)
\(\Leftrightarrow x\left(x^3+1\right)-30\left(x^2-x+1\right)=0\)
\(\Leftrightarrow x\left(x+1\right)\left(x^2-x+1\right)-30\left(x^2-x+1\right)=0\)
\(\Leftrightarrow\left(x^2+x-30\right)\left(x^2-x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+x-30=0\\x^2-x+1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left(x-5\right)\left(x+6\right)=0\\\left(x-\frac{1}{2}\right)^2+\frac{3}{4}=0\left(vn\right)\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=5\\x=-6\end{matrix}\right.\)
\(x+y=1\Rightarrow\left\{{}\begin{matrix}y-1=-x\\x-1=-y\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left(y-1\right)^2=x^2\\\left(x-1\right)^2=y^2\end{matrix}\right.\)
\(\frac{x}{y^3-1}-\frac{y}{x^3-1}+\frac{2\left(x-y\right)}{\left(xy\right)^2+3}=\frac{x}{\left(y-1\right)\left(y^2+y+1\right)}-\frac{y}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{2\left(x-y\right)}{\left(xy\right)^2+3}\)
\(=\frac{-1}{y^2+y+1}+\frac{1}{x^2+x+1}+\frac{2\left(x-y\right)}{\left(xy\right)^2+3}=\frac{-1}{x^2+3y}+\frac{1}{y^2+3x}+\frac{2\left(x-y\right)}{\left(xy\right)^2+3}\)
\(=\frac{-y^2-3x+x^2+3y}{\left(xy\right)^2+3x^3+3y^3+9xy}+\frac{2\left(x-y\right)}{\left(xy\right)^2+3}=\frac{\left(x-y\right)\left(x+y\right)-3x+3y}{\left(xy\right)^2+3\left(x+y\right)\left(\left(x+y\right)^2-3xy\right)+9xy}+\frac{2\left(x-y\right)}{\left(xy\right)^2+3}\)
\(=\frac{-2\left(x-y\right)}{\left(xy\right)^2+3}+\frac{2\left(x-y\right)}{\left(xy\right)^2+3}=0\)
\(x^2+y^2+\left(\frac{1+xy}{x+y}\right)^2\ge2\)
\(\Leftrightarrow\left(x+y\right)^2-2xy+\left(\frac{1+xy}{x+y}\right)^2\ge2\)
\(\Leftrightarrow\left(x+y\right)^2-2\left(xy+1\right)+\left(\frac{1+xy}{x+y}\right)^2\ge0\)
\(\Leftrightarrow\left(x+y\right)^2-\frac{2\left(x+y\right)\left(xy+1\right)}{\left(x+y\right)}+\left(\frac{1+xy}{x+y}\right)^2\ge0\)
\(\Leftrightarrow\left(x+y-\frac{xy+1}{x+y}\right)^2\ge0\) (đúng)
Vậy ...
Theo Cauche ta có:
\(\left(x+y\right)^2+\left(\frac{1+xy}{x+y}\right)^2\ge2\left(x+y\right).\frac{1+xy}{x+y}=2\left(1+xy\right)=2+2xy\)
<=> \(x^2+y^2+2xy+\left(\frac{1+xy}{x+y}\right)^2\ge2+2xy\)
<=> \(x^2+y^2+\left(\frac{1+xy}{x+y}\right)^2\ge2+2xy-2xy=2\)=> ĐPCM
Bài 2:
\(a^4+b^4\ge a^3b+b^3a\)
\(\Leftrightarrow a^4-a^3b+b^4-b^3a\ge0\)
\(\Leftrightarrow a^3\left(a-b\right)-b^3\left(a-b\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\left(a^2+ab+b^2\right)\ge0\)
ta thấy : \(\orbr{\orbr{\begin{cases}\left(a-b\right)^2\ge0\\\left(a^2+ab+b^2\right)\ge0\end{cases}}}\Leftrightarrow dpcm\)
Dấu " = " xảy ra khi a = b
tk nka !!!! mk cố giải mấy bài nữa !11
a) \(\text{ }x^4+y^4\ge x^3y+xy^3\)
\(\Leftrightarrow x^4+y^4-x^3y-xy^3\ge0\)
\(\Leftrightarrow x^3\left(x-y\right)-y^3\left(x-y\right)\ge0\)
\(\Leftrightarrow\left(x-y\right)^2\left(x^2+xy+y^2\right)\ge0\)(ĐPCM)
*NOTE: chứng minh đc vì (x-y)^2 >= 0 ; x^2 +xy +y^2 > 0
mình cũng làm đến nơi rồi nhưng sợ x^2+xy+y^2 chưa chắc lớn hơn 0 thanks bạn nhé
\(VT=x^2+y^2+\left(\frac{1+xy}{x+y}\right)^2=\left(x+y\right)^2+\left(\frac{1+xy}{x+y}\right)^2-2xy\)
\(VT\ge2\sqrt{\frac{\left(x+y\right)^2\left(1+xy\right)^2}{\left(x+y\right)^2}}-2xy=2\left|1+xy\right|-2xy\)
\(VT\ge2\left(1+xy\right)-2xy=2\) (đpcm)
Dấu "=" xảy ra khi \(\left(x+y\right)^2=1+xy\)