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a: ĐKXĐ: \(x,y\in R\)
b: \(A=\dfrac{\dfrac{1}{4}x^2+x^2y+\dfrac{1}{4}y+y^2+x^2y^2+\dfrac{1}{4}+\dfrac{3}{4}y}{x^2y^2+1+x^2-x^2y-y+y^2}\)
\(=\dfrac{\dfrac{1}{4}x^2+x^2y+x^2y^2+y+\dfrac{1}{4}+y^2}{x^2y^2+x^2+1+y^2-x^2y-y}\)
\(=\dfrac{\dfrac{1}{4}\left(x^2+1\right)+y\left(x^2+1\right)+x^2y^2+y^2}{\left(y^2+1\right)\left(x^2+1\right)-y\left(x^2+1\right)}\)
\(=\dfrac{\left(x^2+1\right)\left(y+\dfrac{1}{4}\right)+x^2y^2+y^2}{\left(x^2+1\right)\left(y^2-y+1\right)}\)
\(=\dfrac{\left(x^2+1\right)\left(y+\dfrac{1}{4}\right)+y^2\left(x^2+1\right)}{\left(x^2+1\right)\left(y^2-y+1\right)}=\dfrac{y^2+y+\dfrac{1}{4}}{y^2-y+1}\)
a) \(\dfrac{1}{\left(a-b\right)\left(b-c\right)}+\dfrac{1}{\left(b-c\right)\left(c-a\right)}+\dfrac{1}{\left(c-a\right)\left(a-b\right)}\)
\(=\dfrac{c-a+a-b+b-c}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)
b) \(\dfrac{\left(a^2-\left(b+c\right)^2\right)\left(a+b-c\right)}{\left(a+b+c\right)\left(a^2+c^2-2ac-b^2\right)}\)
\(=\dfrac{\left(a-b-c\right)\left(a+b+c\right)\left(a+b-c\right)}{\left(a+b+c\right)\left(\left(a-c\right)^2-b^2\right)}\)
\(=\dfrac{\left(a-c-b\right)\left(a-c+b\right)}{\left(a-c-b\right)\left(a-c+b\right)}=1\)
c) \(\dfrac{x-1}{x^3}-\dfrac{x+1}{x^3-x^2}+\dfrac{3}{x^3-2x^2+x}\)
\(=\dfrac{x-1}{x^3}-\dfrac{x+1}{x^2\left(x-1\right)}+\dfrac{3}{x\left(x-1\right)^2}\)
\(=\dfrac{\left(x-1\right)^3-x\left(x+1\right)\left(x-1\right)+3x^2}{x^3\left(x-1\right)^2}\)
\(=\dfrac{x^3-3x^2+3x-1-x^3+x+3x^2}{x^3\left(x-1\right)^2}\)
\(=\dfrac{4x-1}{x^3\left(x-1\right)^2}\)
d) \(\left(\dfrac{x^2-y^2}{xy}-\dfrac{1}{x+y}\left(\dfrac{x^2}{y}-\dfrac{y^2}{x}\right)\right):\dfrac{x-y}{x}\)
\(=\left(\dfrac{\left(x-y\right)\left(x+y\right)}{xy}-\dfrac{1}{x+y}.\dfrac{x^3-y^3}{xy}\right):\dfrac{x-y}{x}\)
\(=\left(\dfrac{\left(x-y\right)\left(x+y\right)}{xy}-\dfrac{\left(x-y\right)\left(x^2+xy+y^2\right)}{xy\left(x+y\right)}\right):\dfrac{x-y}{x}\)
\(=\dfrac{\left(x-y\right)\left(x^2+2xy+y^2-x^2-xy-y^2\right)}{xy\left(x+y\right)}.\dfrac{x}{x-y}\)
\(=\dfrac{x}{x+y}\)
\(A=\dfrac{x^2}{\left(x+y\right)\left(1-y\right)}-\dfrac{y^2}{\left(x+y\right)\left(1+x\right)}-\dfrac{x^2y^2}{\left(1+x\right)\left(1-y\right)}\)đkxđ: \(y\ne1;x\ne-1;x\ne-y\)\(=\dfrac{x^2\left(1+x\right)}{\left(x+y\right)\left(1-y\right)\left(1+x\right)}-\dfrac{y^2\left(1-y\right)}{\left(x+y\right)\left(1-y\right)\left(1+x\right)}-\dfrac{x^2y^2\left(x+y\right)}{\left(x+y\right)\left(1-y\right)\left(1+x\right)}\)\(=\dfrac{x^2+x^3-y^2+y^3-x^3y^2-x^2y^3}{\left(x+y\right)\left(1-y\right)\left(1+x\right)}\)
\(=\dfrac{\left(x^3+y^3\right)+\left(x^2-y^2\right)-\left(x^3y^2+x^2y^3\right)}{\left(x+y\right)\left(1-y\right)\left(1+x\right)}\)
\(=\dfrac{\left(x+y\right)\left(x^2-xy+y^2\right)+\left(x-y\right)\left(x+y\right)-x^2y^2\left(x+y\right)}{\left(x+y\right)\left(1-y\right)\left(1+x\right)}\)\(=\dfrac{\left(x+y\right)\left(x^2-xy+y^2+x-y-x^2y^2\right)}{\left(x+y\right)\left(1-y\right)\left(1+x\right)}\)\(=\dfrac{\left(x^2+x\right)-\left(xy+y\right)+\left(y^2-x^2y^2\right)}{\left(1-y\right)\left(x+1\right)}\)
\(=\dfrac{x\left(x+1\right)-y\left(x+1\right)-y^2\left(x-1\right)\left(x+1\right)}{\left(1-y\right)\left(x+1\right)}\) \(=\dfrac{\left(x+1\right)\left(x-y-y^2x+y^2\right)}{\left(1-y\right)\left(x+1\right)}\)
\(=\dfrac{-\left(y-y^2\right)+\left(x-y^2x\right)}{1-y}\)
\(=\dfrac{-y\left(1-y\right)+x\left(1-y\right)\left(1+y\right)}{1-y}\)
\(=\dfrac{\left(1-y\right)\left(x+xy-y\right)}{1-y}=x+xy-y\)
\(\dfrac{x^2}{\left(x+y\right)\left(1-y\right)}-\dfrac{y^2}{\left(x+y\right)\left(1+x\right)}-\dfrac{x^2y^2}{\left(1+x\right)\left(1-y\right)}\) MTC : (x+y)(1-y)(1+x)
A=
\(\dfrac{x^2\times\left(1+x\right)}{\left(x+y\right)\left(1-y\right)\left(1+x\right)}-\dfrac{y^2\times\left(1-y\right)}{\left(x+y\right)\left(1-y\right)\left(1+x\right)}-\dfrac{x^2y^2\times\left(x+y\right)}{\left(x+y\right)\left(1-y\right)\left(1+x\right)}\)
A= \(\dfrac{x^2+x^3}{\left(x+y\right)\left(1-y\right)\left(1+x\right)}-\dfrac{y^2}{\left(x+y\right)\left(1-y\right)\left(1+x\right)}-\dfrac{x^3y^2+x^2y^3}{\left(x+y\right)\left(1-y\right)\left(1+x\right)}\)
\(\dfrac{x^2+x^3-y^2-x^3y^2-x^2y^3}{\left(x+y\right)\left(1-y\right)\left(1+x\right)}\)
a/ Ta có :
\(\left(x+y+t\right)-x^3-y^3-z^3=2011\)
\(\Leftrightarrow3\left(x+y\right)\left(y+t\right)\left(t+x\right)=2011\)
\(\Leftrightarrow\left(x+y\right)\left(y+t\right)\left(t+x\right)=\dfrac{2011}{3}\)
Thay vào D ta được :
\(D=\dfrac{2011}{\dfrac{2011}{3}}=3\)
Vậy.....
b/ Ta có :
\(H=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)
\(\Leftrightarrow10899H=10899\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
\(\Leftrightarrow10899H=\dfrac{a+b+c}{a}+\dfrac{a+b+c}{b}+\dfrac{a+b+c}{c}\)
\(\Leftrightarrow10899H=1+\dfrac{a}{b}+\dfrac{b}{a}+\dfrac{a}{c}+\dfrac{c}{a}+1+\dfrac{b}{c}+\dfrac{c}{b}+1\)
\(\Leftrightarrow10899H=3+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{a}{c}+\dfrac{c}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)\)
Áp dụng BĐT Cô - si cho các số dương ta có ;
\(+,\dfrac{a}{b}+\dfrac{b}{a}\ge2\sqrt{\dfrac{a}{b}.\dfrac{b}{a}}=2\)
+, \(\dfrac{b}{c}+\dfrac{c}{b}\ge2\sqrt{\dfrac{b}{c}.\dfrac{c}{b}}=2\)
+, \(\dfrac{c}{a}+\dfrac{a}{c}\ge2\sqrt{\dfrac{b}{c}.\dfrac{c}{b}}=2\)
Cộng vế với vế của các BĐT ta có :
\(\dfrac{a}{b}+\dfrac{b}{a}+\dfrac{a}{c}+\dfrac{c}{a}+\dfrac{b}{c}+\dfrac{c}{b}\ge6\)
\(\Leftrightarrow10899H\ge9\)
\(\Leftrightarrow H\ge\dfrac{1}{2011}\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=6033\)
Vậy..
b ) Do a ; b ; c dương \(\Rightarrow\dfrac{1}{a};\dfrac{1}{b};\dfrac{1}{c}\) dương
Áp dụng BĐT Cô - si cho 3 số dương , ta có :
\(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge3\sqrt[3]{abc}.3\sqrt[3]{\dfrac{1}{abc}}=9\)
\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{9}{a+b+c}\)
Theo GT : \(a+b+c=18099\)
\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{9}{18099}=\dfrac{1}{2011}\)
\(\Rightarrow H\ge\dfrac{1}{2011}\)
Dấu " = " xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}a+b+c=18099\\a=b=c\end{matrix}\right.\)
\(\Leftrightarrow a=b=c=6033\)
Vậy ...