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Cho các số x , y thỏa mãn x + y \(\ne\)0
Chứng minh : \(x^2+y^2+\left(\frac{1+xy}{x+y}\right)^2\ge2\)
thằng ngu lê anh tú ko biết gì thì im vào
Đặt \(\hept{\begin{cases}S=x+y\\P=xy\end{cases}}\)\(\Rightarrow x^2+y^2=S^2-2P\)
Ta cần chứng minh \(S^2-2P+\left(\frac{P+1}{S}\right)^2\ge2\)
\(\Leftrightarrow S^2-2\left(P+1\right)+\left(\frac{P+1}{S}\right)^2\ge0\)
\(\Leftrightarrow S^2-\frac{2S\left(P+1\right)}{S}+\left(\frac{P+1}{S}\right)^2\ge0\)
\(\Leftrightarrow\left(S-\frac{P+1}{S}\right)^2\ge0\) *luôn đúng*
\(VT=x^2+y^2+\left(\frac{1+xy}{x+y}\right)^2=\left(x+y\right)^2+\left(\frac{1+xy}{x+y}\right)^2-2xy\)
\(VT\ge2\sqrt{\frac{\left(x+y\right)^2\left(1+xy\right)^2}{\left(x+y\right)^2}}-2xy=2\left|1+xy\right|-2xy\)
\(VT\ge2\left(1+xy\right)-2xy=2\) (đpcm)
Dấu "=" xảy ra khi \(\left(x+y\right)^2=1+xy\)
Theo mình nó còn có x,y > 0 nữa nha !
Ta có:
\(x^2+y^2+\left(\dfrac{1+xy}{x+y}\right)^2=\left(x+y\right)^2+\left(\dfrac{1+xy}{x+y}\right)^2-2xy\)
Áp dụng BĐT Cosi ta có:
\(\left(x+y\right)^2+\left(\dfrac{1+xy}{x+y}\right)^2\ge2\sqrt{\left(x+y\right)^2\left(\dfrac{1+xy}{x+y}\right)^2}=2\left(1+xy\right)\)
\(\Leftrightarrow\left(x+y\right)^2+\left(\dfrac{1+xy}{x+y}\right)^2-2xy\ge2\left(1+xy\right)-2xy\)
\(\Leftrightarrow\left(x+y\right)^2+\left(\dfrac{1+xy}{x+y}\right)^2-2xy\ge2+2xy-2xy=2\)
\(\Rightarrow\)đpcm
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/
\(x^4-30x^2+31x-30=0\)
\(\Leftrightarrow x^4+x-30x^2+30x-30=0\)
\(\Leftrightarrow x\left(x^3+1\right)-30\left(x^2-x+1\right)=0\)
\(\Leftrightarrow x\left(x+1\right)\left(x^2-x+1\right)-30\left(x^2-x+1\right)=0\)
\(\Leftrightarrow\left(x^2+x-30\right)\left(x^2-x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+x-30=0\\x^2-x+1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left(x-5\right)\left(x+6\right)=0\\\left(x-\frac{1}{2}\right)^2+\frac{3}{4}=0\left(vn\right)\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=5\\x=-6\end{matrix}\right.\)
\(x+y=1\Rightarrow\left\{{}\begin{matrix}y-1=-x\\x-1=-y\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left(y-1\right)^2=x^2\\\left(x-1\right)^2=y^2\end{matrix}\right.\)
\(\frac{x}{y^3-1}-\frac{y}{x^3-1}+\frac{2\left(x-y\right)}{\left(xy\right)^2+3}=\frac{x}{\left(y-1\right)\left(y^2+y+1\right)}-\frac{y}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{2\left(x-y\right)}{\left(xy\right)^2+3}\)
\(=\frac{-1}{y^2+y+1}+\frac{1}{x^2+x+1}+\frac{2\left(x-y\right)}{\left(xy\right)^2+3}=\frac{-1}{x^2+3y}+\frac{1}{y^2+3x}+\frac{2\left(x-y\right)}{\left(xy\right)^2+3}\)
\(=\frac{-y^2-3x+x^2+3y}{\left(xy\right)^2+3x^3+3y^3+9xy}+\frac{2\left(x-y\right)}{\left(xy\right)^2+3}=\frac{\left(x-y\right)\left(x+y\right)-3x+3y}{\left(xy\right)^2+3\left(x+y\right)\left(\left(x+y\right)^2-3xy\right)+9xy}+\frac{2\left(x-y\right)}{\left(xy\right)^2+3}\)
\(=\frac{-2\left(x-y\right)}{\left(xy\right)^2+3}+\frac{2\left(x-y\right)}{\left(xy\right)^2+3}=0\)
ĐKXĐ: \(...\)
\(P=\dfrac{2}{x}-\left(\dfrac{x^2}{x\left(x+y\right)}-\dfrac{y^2}{y\left(x+y\right)}+\dfrac{y^2-x^2}{xy}\right).\dfrac{x+y}{x^2+xy+y^2}\)
\(P=\dfrac{2}{x}-\left(\dfrac{x-y}{x+y}-\dfrac{\left(x-y\right)\left(x+y\right)}{xy}\right).\dfrac{x+y}{x^2+xy+y^2}\)
\(P=\dfrac{2}{x}-\left(\dfrac{1}{x+y}-\dfrac{x+y}{xy}\right)\dfrac{x^2-y^2}{x^2+xy+y^2}\)
\(P=\dfrac{2}{x}-\dfrac{-\left(x^2+xy+y^2\right)}{xy\left(x+y\right)}.\dfrac{\left(x-y\right)\left(x+y\right)}{x^2+xy+y^2}\)
\(P=\dfrac{2}{x}+\dfrac{x-y}{xy}=\dfrac{2}{x}+\dfrac{1}{y}-\dfrac{1}{x}=\dfrac{1}{x}+\dfrac{1}{y}\)
b/ \(x^2+y^2+10=2x-6y\Leftrightarrow x^2-2x+1+y^2+6y+9=0\)
\(\Leftrightarrow\left(x-1\right)^2+\left(y+3\right)^2=0\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-3\end{matrix}\right.\)
\(\Rightarrow P=\dfrac{1}{1}-\dfrac{1}{3}=\dfrac{2}{3}\)
Ta có : \(\dfrac{\left(ax+by+cz\right)^2}{x^2+y^2+z^2}=a^2+b^2+c^2\)
\(\Leftrightarrow\left(ax+by+cz\right)^2=\left(a^2+b^2+c^2\right)\left(x^2+y^2+z^2\right)\)
\(\Leftrightarrow a^2x^2+b^2y^2+c^2z^2+2axby+2axcz+2bycz=a^2x^2+b^2x^2+c^2x^2+a^2y^2+b^2y^2+c^2y^2+a^2z^2+b^2z^2+c^2z^2\)
\(\Leftrightarrow2axby+2axvz+2bycz=a^2y^2+b^2x^2+a^2z^2+c^2x^2+b^2z^2+c^2y^2\)
\(\Leftrightarrow a^2y^2+b^2x^2+a^2z^2+c^2x^2+b^2z^2+c^2y^2-2axby-2azcx-2bycz=0\)
\(\Leftrightarrow\left(a^2y^2-2axby+b^2x^2\right)+\left(a^2z^2-2azcx+c^2x^2\right)+\left(b^2z^2-2bycz+c^2y^2\right)=0\)
\(\Leftrightarrow\left(ay-bx\right)^2+\left(az-cx\right)^2+\left(bz-cy\right)^2=0\)
Do \(\left(ay-bx\right)^2\ge0;\left(az-cx\right)^2\ge0;\left(bz-cy\right)^2\ge0\)
\(\Rightarrow\left\{{}\begin{matrix}ay-bx=0\\az-cx=0\\bz-cy=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}ay=bx\\az=cx\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{x}=\dfrac{b}{y}\\\dfrac{c}{z}=\dfrac{a}{x}\end{matrix}\right.\)
\(\Rightarrow\dfrac{a}{x}=\dfrac{b}{y}=\dfrac{c}{z}\left(đpcm\right)\)
:D
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 \(-a+b+1=0\Leftrightarrow x^2+y^2+xy=1\)