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1. Vì x, y, z > 0
\(xy+yz+zx\ge2xyz\)
\(\Leftrightarrow\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge2\)
Suy ra:
\(\dfrac{1}{x}\ge1-\dfrac{1}{y}+1-\dfrac{1}{z}=\dfrac{y-1}{y}+\dfrac{z-1}{z}\ge2\sqrt{\dfrac{\left(y-1\right)\left(z-1\right)}{yz}}\). (1)
Tương tự \(\dfrac{1}{y}\ge2\sqrt{\dfrac{\left(z-1\right)\left(x-1\right)}{zx}}\) (2)
và \(\dfrac{1}{z}\ge2\sqrt{\dfrac{\left(x-1\right)\left(y-1\right)}{xy}}\) (3)
Nhân (1), (2), (3) với nhau theo vế ta được
\(\dfrac{1}{xyz}\ge\dfrac{8\left(x-1\right)\left(y-1\right)\left(z-1\right)}{xyz}\)
\(\Leftrightarrow\left(x-1\right)\left(y-1\right)\left(z-1\right)\le\dfrac{1}{8}\)
Đẳng thức xảy ra \(\Leftrightarrow x=y=z=\dfrac{3}{2}\)
a) \(x>1\Rightarrow x>\sqrt{x}\Rightarrow x-\sqrt{x}-\left|x-\sqrt{x}\right|=x-\sqrt{x}-x+\sqrt{x}=0\)
b) \(M=x-\sqrt{x}=x-\sqrt{x}+\frac{1}{4}-\frac{1}{4}=\left(\sqrt{x}-\frac{1}{2}\right)^2-\frac{1}{4}\ge-\frac{1}{4}\)
Xảy ra đẳng thức khi và chỉ khi \(x=\frac{1}{4}\)
\(\Delta=a^2-4b=a^2-4\left(2008-a\right)=a^2+4a-8032\)
Để pt có nghiệm nguyên \(\Rightarrow\Delta\) là số chính phương
\(\Rightarrow a^2+4a-8032=k^2\)
\(\Leftrightarrow\left(a+2\right)^2-8036=k^2\)
\(\Leftrightarrow\left(a+2-k\right)\left(a+2+k\right)=8036\)
Mặt khác do \(\left(a+2-k\right)+\left(a+2+k\right)=2\left(a+2\right)\) chẵn nên ta chỉ cần xét các cặp ước cùng tính chẵn lẻ của 8036 là \(\left(4018;2\right);\left(2;4018\right);\left(-2;-4018\right);\left(-4018;-2\right)\)
\(\left\{{}\begin{matrix}a+2-k=4018\\a+2+k=2\end{matrix}\right.\) \(\Rightarrow a=2008\Rightarrow x^2+2008x=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-2008\end{matrix}\right.\)
\(\left\{{}\begin{matrix}a+2-k=2\\a+2+k=4018\end{matrix}\right.\) \(\Rightarrow a=2008\) giổng TH trên
\(\left\{{}\begin{matrix}a+2-k=-2\\a+2+k=-4018\end{matrix}\right.\) \(\Rightarrow a=-2012\Rightarrow x^2-2012x+4020=0\Rightarrow\left[{}\begin{matrix}x=2\\x=2010\end{matrix}\right.\)
\(\left\{{}\begin{matrix}a+2-k=-4018\\a+2+k=-2\end{matrix}\right.\) \(\Rightarrow a=-2012\) giống TH trên
Vậy nghiệm nguyên của pt là \(x=\left\{0;2;2008;2010\right\}\)
ĐK: \(x>0\).
a)\(A=\dfrac{x^2+x+1}{x-\sqrt{x}+1}-2\sqrt{x}-1\)
\(A=\dfrac{x^2+x+1-\left(2\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}\)
\(=\dfrac{-2x\sqrt{x}+x^2+3x-2\sqrt{x}-x+\sqrt{x}}{x-\sqrt{x}+1}\)
\(=\dfrac{-2x\sqrt{x}+x^2+2x-\sqrt{x}}{x-\sqrt{x}+1}\)
b)Với x>1 thì A>0 nên |A|=A do đó A-|A|=0.
\(1,\) Áp dụng BĐT: \(x^2+y^2\ge\dfrac{\left(x+y\right)^2}{2}\text{ và }\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\)
Dấu \("="\Leftrightarrow x=y\)
\(A=\left(a+\dfrac{1}{a}\right)^2+\left(b+\dfrac{1}{b}\right)^2+17\ge\dfrac{1}{2}\left(a+b+\dfrac{1}{a}+\dfrac{1}{b}\right)^2+17\\ A\ge\dfrac{1}{2}\left(1+\dfrac{1}{a}+\dfrac{1}{b}\right)^2+17\ge\dfrac{1}{2}\left(1+\dfrac{4}{a+b}\right)^2+17=\dfrac{25}{2}+17=\dfrac{59}{2}\\ \text{Dấu }"="\Leftrightarrow\left\{{}\begin{matrix}a+\dfrac{1}{a}=b+\dfrac{1}{b}\\a+b=1\end{matrix}\right.\Leftrightarrow a=b=\dfrac{1}{2}\)
\(2,\text{Đặt }A=\dfrac{xy}{z}+\dfrac{yz}{x}+\dfrac{xz}{y}\\ \Leftrightarrow A^2=\dfrac{x^2y^2}{z^2}+\dfrac{y^2z^2}{x^2}+\dfrac{x^2z^2}{y^2}+2\left(\dfrac{xy^2z}{xz}+\dfrac{xyz^2}{xy}+\dfrac{x^2yz}{yz}\right)\\ \Leftrightarrow A^2=\dfrac{x^2y^2}{z^2}+\dfrac{y^2z^2}{x^2}+\dfrac{x^2z^2}{y^2}+2\left(x^2+y^2+z^2\right)\\ \Leftrightarrow A^2=\dfrac{x^2y^2}{z^2}+\dfrac{y^2z^2}{x^2}+\dfrac{x^2z^2}{y^2}+6\)
Áp dụng Cosi: \(\dfrac{x^2y^2}{z^2}+\dfrac{y^2z^2}{x^2}\ge2y^2\)
CMTT: \(\left\{{}\begin{matrix}\dfrac{y^2z^2}{x^2}+\dfrac{x^2z^2}{y^2}\ge2z^2\\\dfrac{x^2y^2}{z^2}+\dfrac{x^2z^2}{y^2}\ge2x^2\end{matrix}\right.\)
Cộng VTV \(\Leftrightarrow A^2\ge2\left(x^2+y^2+z^2\right)+6=12\\ \Leftrightarrow A\ge2\sqrt{3}\)
Dấu \("="\Leftrightarrow x=y=z=1\)
\(ax+by+cz=0\Rightarrow\left(ax+by+cz\right)^2=0\)
\(\Rightarrow a^2x^2+b^2y^2+c^2z^2=-2\left(axby+bycz+axcz\right)\)
Ta co
\(\dfrac{ax^2+by^2+cz^2}{bc\left(y-z\right)^2+ac\left(z-x\right)^2+ab\left(x-y\right)^2}\)
\(=\dfrac{ax^2+by^2+cz^2}{bcy^2-2bcyz+bcz^2+acz^2-2aczx+acx^2+abx^2-2abxy+aby^2}\)
\(=\dfrac{ax^2+by^2+cz^2}{bcy^2+bcz^2+acz^2+acx^2+abx^2+aby^2-2\left(axby+bcyz+axcz\right)}\)
\(=\dfrac{ax^2+by^2+cz^2}{bcy^2+bcz^2+acz^2+acx^2+abx^2+aby^2+a^2x^2+b^2y^2+c^2z^2}\)
\(=\dfrac{ax^2+by^2+cz^2}{\left(acx^2+abx^2+a^2x^2\right)+\left(bcy^2+aby^2+b^2y^2\right)+\left(c^2z^2+acz^2+bcz^2\right)}\)
\(=\dfrac{ax^2+by^2+cz^2}{ax^2\left(a+b+c\right)+by^2\left(a+b+c\right)+cz^2\left(a+b+c\right)}\)
\(=\dfrac{ax^2+by^2+cz^2}{\left(a+b+c\right)\left(ax^2+by^2+cz^2\right)}=\dfrac{1}{a+b+c}\) ( dpcm)
a: Khi m=-1 thì phương trình sẽ là:
x^2-(-3-1)x+2-1-1=0
=>x^2+4x=0
=>x=0 hoặc x=-4
Theo BĐT \(AM-GM\) ta có :
\(xy+yz+zx\le\dfrac{\left(x+y+z\right)^2}{3}=\dfrac{12^2}{3}=48\)
\(x^2+y^2+z^2\ge8\left(x+y+z\right)-\left(16+16+16\right)=48\)
Theo BĐT Cauchy schwarz dưới dạng en-gel ta có :
\(\dfrac{x^3}{y+1}+\dfrac{y^3}{z+1}+\dfrac{z^3}{x+1}=\dfrac{x^4}{xy+z}+\dfrac{y^4}{yz+y}+\dfrac{z^4}{zx+z}\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{xy+yz+zx+x+y+z}=\dfrac{48^2}{48+12}=\dfrac{192}{5}\)
Vậy \(MIN_Q=\dfrac{192}{5}\) . Dấu \("="\Leftrightarrow z=y=z=4\)
Lời giải:
Áp dụng BĐT Cô - si:
\(P=ax^m+\frac{b}{x^n}=\frac{a}{n}x^m+\frac{a}{n}x^m+...+\frac{a}{n}x^m+\frac{b}{mx^n}+...+\frac{b}{mx^n}\)
\(=(m+n)\sqrt[m+n]{(\frac{a}{n})^n.x^{mn}.(\frac{b}{m})^m.\frac{1}{x^{mn}}}\)
\(=(m+n)\sqrt[m+n]{\frac{a^nb^m}{n^n.m^m}}\)