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\(A=x^3+y^3+z^3-3xyz\)
\(=\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\)
\(=\left(x+y+z\right)\left[\left(x^2+2xy+y^2\right)-\left(xz+yz\right)+z^2\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-xz-yz\right)\)
\(=0\)
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\(A=\left(\dfrac{x}{y}+1\right)\left(\dfrac{y}{z}+1\right)\left(\dfrac{z}{x}+1\right)\)
\(=\dfrac{x+y}{y}\times\dfrac{y+z}{z}\times\dfrac{z+x}{x}\)
\(=\dfrac{-z}{y}\times\dfrac{-x}{z}\times\dfrac{-y}{x}\)
\(=-1\)
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\(A=\dfrac{1}{y^2+z^2-x^2}+\dfrac{1}{x^2+z^2-y^2}+\dfrac{1}{x^2+y^2-z^2}\)
\(=\dfrac{1}{\left(y+z\right)^2-2yz-x^2}+\dfrac{1}{\left(x+z\right)^2-2xz-y^2}+\dfrac{1}{\left(x+y\right)^2-2xy-z^2}\)
\(=\dfrac{1}{\left(-x\right)^2-2yz-x^2}+\dfrac{1}{\left(-y\right)^2-2xz-y^2}+\dfrac{1}{\left(-z\right)^2-2xy-z^2}\)
\(=-\dfrac{1}{2}\left(\dfrac{1}{yz}+\dfrac{1}{xz}+\dfrac{1}{xz}\right)\)
\(=-\dfrac{1}{2}\times\dfrac{x+y+z}{xyz}\)
\(=0\)
Bài 1:
a: \(A=\dfrac{x^4+x^3+x+1}{x^4-x^3+2x^2-x+1}=\dfrac{x^3\left(x+1\right)+\left(x+1\right)}{x^4-x^3+x^2+x^2-x+1}\)
\(=\dfrac{\left(x+1\right)\left(x^3+1\right)}{\left(x^2-x+1\right)\left(x^2+1\right)}=\dfrac{\left(x+1\right)^2}{x^2+1}\)
Để A=0 thì x+1=0
hay x=-1
b: \(B=\dfrac{x^4-5x^2+4}{x^4-10x^2+9}=\dfrac{\left(x^2-1\right)\left(x^2-4\right)}{\left(x^2-1\right)\left(x^2-9\right)}=\dfrac{x^2-4}{x^2-9}\)
Để B=0 thi (x-2)(x+2)=0
=>x=2 hoặc x=-2
a, \(9x^2+y^2+2z^2-18x-6y+4z+20=0\)
\(\Leftrightarrow\left(9x^2-18x+9\right)+\left(y^2-6y+9\right)+\left(2z^2+4z+2\right)=0\)
\(\Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\)
Vì \(\left\{{}\begin{matrix}9\left(x-1\right)^2\ge0\\\left(y-3\right)^2\ge0\\2\left(z+1\right)^2\ge0\end{matrix}\right.\Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\)
Mà \(9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}9\left(x-1\right)^2=0\\\left(y-3\right)^2=0\\2\left(z+1\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\\z=-1\end{matrix}\right.\)
Vậy...
Lời giải:
Đặt $\frac{x}{a}=m; \frac{y}{b}=n; \frac{z}{c}=p$ với $m,n,p>0$.
BĐT cần chứng minh tương đương với:
(m^2a+n^2b+p^2c)(a+b+c)\geq (am+bn+cp)^2$
$\Leftrightarrow m^2(ab+ac)+n^2(ba+bc)+p^2(ca+cb)\geq 2abmn+2amcp+2bncp$
$\Leftrightarrow ab(m^2-2mn+n^2)+bc(n^2-2np+p^2)+ca(m^2-2mp+p^2)\geq 0$
$\Leftrightarrow ab(m-n)^2+bc(n-p)^2+ca(m-p)^2\geq 0$
(luôn đúng với $a,b,c>0$)
Ta có đpcm.
Bải 3a
\(\dfrac{-a+b+c}{2a}+\dfrac{-b+c+a}{2b}+\dfrac{-c+a+b}{2c}\ge\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{-a}{2a}+\dfrac{b+c}{2a}+\dfrac{-b}{2b}+\dfrac{c+a}{2b}+\dfrac{-c}{2c}+\dfrac{a+b}{2c}\ge\dfrac{3}{2}\)
\(\Leftrightarrow-\dfrac{3}{2}+\dfrac{b+c}{2a}+\dfrac{c+a}{2b}+\dfrac{a+b}{2c}\ge\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{b+c}{2a}+\dfrac{c+a}{2b}+\dfrac{a+b}{2c}\ge3\)
\(\Leftrightarrow\dfrac{b+c}{a}+\dfrac{c+a}{b}+\dfrac{a+b}{c}\ge6\)
\(\Leftrightarrow\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)+\left(\dfrac{c}{a}+\dfrac{a}{c}\right)\ge6\)
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{b}+\dfrac{b}{a}\ge2\sqrt{\dfrac{ab}{ba}}=2\\\dfrac{b}{c}+\dfrac{c}{b}\ge2\sqrt{\dfrac{bc}{cb}}=2\\\dfrac{c}{a}+\dfrac{a}{c}\ge2\sqrt{\dfrac{ca}{ac}}=2\end{matrix}\right.\)
\(\Rightarrow\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)+\left(\dfrac{c}{a}+\dfrac{a}{c}\right)\ge2+2+2=6\)
\(\Leftrightarrow\dfrac{-a+b+c}{2a}+\dfrac{-b+c+a}{2b}+\dfrac{-c+a+b}{2c}\ge\dfrac{3}{2}\) ( đpcm )
Dấu " = " xảy ra khi \(a=b=c\)
Bài 3b)
\(P=\dfrac{x}{y+z}+\dfrac{y}{x+z}+\dfrac{z}{x+y}\)
\(P=\dfrac{x^2}{xy+xz}+\dfrac{y^2}{xy+yz}+\dfrac{z^2}{xz+yz}\)
Áp dụng bất đẳng thức Cauchy - Schwarz dạng phân thức
\(\Rightarrow\dfrac{x^2}{xy+xz}+\dfrac{y^2}{xy+yz}+\dfrac{z^2}{xz+yz}\ge\dfrac{\left(x+y+z\right)^2}{2\left(xy+yz+xz\right)}\)( 1 )
Theo hệ quả của bất đẳng thức Cauchy
\(\Rightarrow\left(x+y+z\right)^2\ge3\left(xy+yz+xz\right)\)
\(\Rightarrow\dfrac{\left(x+y+z\right)^2}{2\left(xy+yz+xz\right)}\ge\dfrac{3\left(xy+yz+xz\right)}{2\left(xy+yz+xz\right)}=\dfrac{3}{2}\) ( 2 )
Từ ( 1 ) và ( 2 )
\(\Rightarrow\)\(\dfrac{x^2}{xy+xz}+\dfrac{y^2}{xy+yz}+\dfrac{z^2}{xz+yz}\ge\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{x}{y+z}+\dfrac{y}{x+z}+\dfrac{z}{x+y}\ge\dfrac{3}{2}\)
\(\Leftrightarrow P\ge\dfrac{3}{2}\)
Vậy \(P_{min}=\dfrac{3}{2}\)
Dấu " = " xảy ra khi \(a=b=c\)
Bài 2:
a) \(\dfrac{x-17}{33}+\dfrac{x-21}{29}+\dfrac{x}{25}=4\)
\(\Rightarrow\left(\dfrac{x-17}{33}-1\right)+\left(\dfrac{x-21}{29}-1\right)+\left(\dfrac{x}{25}-2\right)=0\)
\(\Rightarrow\dfrac{x-50}{33}+\dfrac{x-50}{29}+\dfrac{x-50}{25}=0\)
\(\Rightarrow\left(x-50\right)\left(\dfrac{1}{33}+\dfrac{1}{29}+\dfrac{1}{25}\right)=0\)
Mà \(\dfrac{1}{33}+\dfrac{1}{29}+\dfrac{1}{25}\ne0\)
\(\Rightarrow x-50=0\)
\(\Rightarrow x=50\)
Vậy x = 50
Cách khác:
Áp dụng BĐT AM-GM:
\(\frac{a}{b^2}+\frac{1}{a}\geq 2\sqrt{\frac{1}{b^2}}=\frac{2}{b}\)
\(\frac{b}{c^2}+\frac{1}{b}\geq 2\sqrt{\frac{1}{c^2}}=\frac{2}{c}\)
\(\frac{c}{a^2}+\frac{1}{c}\geq 2\sqrt{\frac{1}{a^2}}=\frac{2}{a}\)
Cộng theo vế và rút gọn:
\(\frac{a}{b^2}+\frac{b}{c^2}+\frac{c}{a^2}\geq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\) (đpcm)
Đúng rồi bạn nhé.