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2)Đầu tiên ta cm bđt:\(xy+yz+zx\le\dfrac{\left(x+y+z\right)^2}{3}\)
\(\Leftrightarrow3\left(xy+yz+zx\right)\le x^2+y^2+z^2+2\left(xy+yz+zx\right)\)
\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2zx\ge0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\)(luôn đúng)
\(\Rightarrow xy+yz+zx\le3\)
"="<=>x=y=z=1
Ta có:\(\left\{{}\begin{matrix}a+b+c=0\\a^2+b^2+c^2=2009\end{matrix}\right.\)
\(\Rightarrow ab+bc+ca=\dfrac{-2009}{2}\)
\(\Rightarrow a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)=1009020,25\)
\(\Rightarrow2\left(a^2b^2+b^2c^2+c^2a^2\right)=2018040,5\)
\(\Rightarrow a^4+b^4+c^4=\left(a^2+b^2+c^2\right)^2-2\left(a^2b^2+b^2c^2+c^2a^2\right)=\)2018040,5
Áp dụng BĐT Bunhiacốpxki dạng phân thức có
\(\dfrac{a^2}{a+2b^2}+\dfrac{b^2}{b+2c^2}+\dfrac{c^2}{c+2a^2}\ge\dfrac{\left(a+b+c\right)^2}{a+2b^2+b+2c^2+c+2a^2}=\dfrac{9}{3+2\left(a^2+b^2+c^2\right)}\) (1)
Áp dụng BĐT Bunhiacốpxki có:
\(\left(a.1+b.1+c.1\right)^2\ge\left(1+1+1\right)\left(a^2+b^2+c^2\right)\)
\(\Rightarrow9\ge3\left(a^2+b^2+c^2\right)\Rightarrow3\ge a^2+b^2+c^2\Rightarrow2\left(a^2+b^2+c^2\right)\le6\) (2)
Thay (2) vào (1) có \(\dfrac{a^2}{a+b^2}+\dfrac{b^2}{b+2c^2}+\dfrac{c^2}{c+a^2}\ge\dfrac{9}{3+6}=1\) (đpcm)
Dấu = xảy ra khi a= b=c=1
Cái này mình biết chút... nhưng mà giải trên đây không tiện lắm bạn có chới zalo ko gửi ad qua cho mình để kp rồi mình gửi lời giải qua luôn...
b)Đặt $S=x+y,P=xy$ thì được:
\(\left\{ \begin{align} & S+P=2+3\sqrt{2} \\ & {{S}^{2}}-2P=6 \\ \end{align} \right.\Rightarrow {{S}^{2}}+2S+1=11+6\sqrt{2}={{\left( 3+\sqrt{2} \right)}^{2}}\)
\(\begin{array}{l} \Rightarrow \left\{ \begin{array}{l} S = 2 + \sqrt 2 \\ P = 2\sqrt 2 \end{array} \right. \Rightarrow \left( {x;y} \right) \in \left\{ {\left( {2;\sqrt 2 } \right),\left( {\sqrt 2 ;2} \right)} \right\}\\ \left\{ \begin{array}{l} S = - 4 - \sqrt 2 \\ P = 6 + 4\sqrt 2 \end{array} \right.\left( {VN} \right) \end{array} \)
\( c)\left\{ \begin{array}{l} 2{x^2} + xy + 3{y^2} - 2y - 4 = 0\\ 3{x^2} + 5{y^2} + 4x - 12 = 0 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} 2\left( {2{x^2} + xy + 3{y^2} - 2y - 4} \right) - \left( {3{x^2} + 5{y^2} + 4x - 12} \right) = 0\\ 3{x^2} + 5{y^2} + 4x - 12 = 0 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} {x^2} + 2xy + {y^2} - 4x - 4y + 4 = 0\\ 3{x^2} + 5{y^2} + 4x - 12 = 0 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} {\left( {x + y - 2} \right)^2} = 0\\ 3{x^2} + 5{y^2} + 4x - 12 = 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x + y - 2 = 0\\ 3{x^2} + 5{y^2} + 4x - 12 = 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x = 1\\ y = 1 \end{array} \right. \)
a)Trừ theo vế của \(pt\left(2\right)\) cho \(pt\left(1\right)\):
\(\left(5x+3y\right)-\left(3x+2y\right)=-4-1\)
\(\Leftrightarrow2x+y=-5\). Khi đó
\(3x+2y=1\Leftrightarrow2\left(2x+y\right)-x=1\)
\(\Leftrightarrow2\cdot\left(-5\right)-x=1\)\(\Leftrightarrow x=-11\)
\(\Rightarrow3x+2y=1\Rightarrow y=\dfrac{1-3x}{2}=\dfrac{1-3\cdot\left(-11\right)}{2}=17\)
Vậy nghiệm hpt \(\left(x;y\right)=\left(-11;17\right)\)
b)\(2x^2+2\sqrt{3}x-3=0\)
\(\Delta=\left(2\sqrt{3}\right)^2-\left(4\cdot2\cdot\left(-3\right)\right)=36\)
\(\Rightarrow x_{1,2}=\dfrac{-2\sqrt{3}\pm\sqrt{36}}{4}\)
c)\(9x^4+8x^2-1=0\)
\(\Leftrightarrow9x^4-x^2+9x^2-1=0\)
\(\Leftrightarrow x^2\left(9x^2-1\right)+\left(9x^2-1\right)=0\)
\(\Leftrightarrow\left(9x^2-1\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(3x+1\right)\left(x^2+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}3x-1=0\\3x+1=0\\x^2+1=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\pm\dfrac{1}{3}\\x^2+1>0\left(loai\right)\end{matrix}\right.\)
\(P=\dfrac{a}{a^2+bc}+\dfrac{b}{b^2+ca}+\dfrac{c}{c^2+ab}\)
\(\le\dfrac{a}{2a\sqrt{bc}}+\dfrac{b}{2b\sqrt{ca}}+\dfrac{c}{2c\sqrt{ab}}\)
\(=\dfrac{a\sqrt{bc}}{2abc}+\dfrac{b\sqrt{ca}}{2abc}+\dfrac{c\sqrt{ab}}{2abc}\)
\(\le\dfrac{2a^2+b^2+c^2}{8abc}+\dfrac{2b^2+a^2+c^2}{8abc}+\dfrac{2c^2+b^2+a^2}{8abc}\)
\(=\dfrac{4\left(a^2+b^2+c^2\right)}{8abc}=\dfrac{1}{2}\)
Có: \(a+b+c=0\)
\(\Leftrightarrow\left(a+b+c\right)^2=0\)
\(\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ca=0\)
\(\Leftrightarrow2\left(ab+bc+ca\right)=-1\) (do \(a^2+b^2+c^2=1\) )
\(\Leftrightarrow ab+bc+ca=-\dfrac{1}{2}\)
\(\Leftrightarrow\left(ab+bc+ca\right)^2=\dfrac{1}{4}\)
\(\Leftrightarrow\left(ab\right)^2+\left(bc\right)^2+\left(ca\right)^2+2ab.bc+2bc.ca+2ca.ab=\dfrac{1}{4}\)
\(\Leftrightarrow\left(ab\right)^2+\left(bc\right)^2+\left(ca\right)^2+2abc\left(a+b+c\right)=\dfrac{1}{4}\)
\(\Leftrightarrow \left(ab\right)^2+\left(bc\right)^2+\left(ca\right)^2=\dfrac{1}{4}\) (do \(a+b+c=0\))
Lại có: \(M=a^4+b^4+c^4\)
\(=\left(a^2+b^2+c^2\right)^2-2\left(a^2b^2 +b^2c^2+c^2a^2\right)\)
\(=1-2\left[\left(ab\right)^2+\left(bc\right)^2+\left(ca\right)^2\right]\) (do \(a^2+b^2+c^2=1\))
\(=1-2.\dfrac{1}{4}\)(do \(\left(ab\right)^2+\left(bc\right)^2+\left(ca\right)^2=\dfrac{1}{4}\))
\(=1-\dfrac{1}{2}=\dfrac{1}{2}\)
Vậy \(M=\dfrac{1}{2}\)