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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 ...
Từ \(x\left(\dfrac{1}{y}+\dfrac{1}{z}\right)+y\left(\dfrac{1}{z}+\dfrac{1}{x}\right)+z\left(\dfrac{1}{x}+\dfrac{1}{y}\right)=-2\) ta có:
\(x^2y+y^2z+z^2x+xy^2+yz^2+zx^2+2xyz=0\)
\(\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+y=0\\y+z=0\\z+x=0\end{matrix}\right.\).
Không mất tính tổng quát, giả sử x + y = 0
\(\Leftrightarrow x=-y\)
\(\Leftrightarrow x^3=-y^3\).
Kết hợp với \(x^3+y^3+z^3=1\) ta có \(z^3=1\Leftrightarrow z=1\).
Vậy \(P=\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{-y}+\dfrac{1}{y}+\dfrac{1}{1}=1\).
Do \(x+y+z=0;xy+yz+xz=0\)
\(\Rightarrow\left(x+y+z\right)^2=x^2+y^2+z^2+2\left(xy+yz+xz\right)=0\)
\(\Rightarrow\left(x+y+z\right)^2=x^2+y^2+z^2=0\)\(\Rightarrow x=y=z=0\)
\(\Rightarrow S=\left(x-1\right)^{2011}+\left(y-1\right)^{2012}+\left(z+1\right)^{2013}=\left(-1\right)^{2011}+\left(-1\right)^{2012}+1^{2013}=1\)
a) \(2\left(x^3+y^3\right)-3\left(x^2+y^2\right)\)
\(=2\left[\left(x+y\right)^3-3xy\left(x+y\right)\right]-3\left[\left(x+y\right)^2-2xy\right]\)
\(=2\left(1-3xy\right)-3\left(1-2xy\right)\)
\(=2-6xy-3+6xy=-1\)
\(\Rightarrow\) Giá trị của biểu thức không phụ thuộc vào biến \(x,y\)
b) \(\dfrac{\left(x+5\right)^2+\left(x-5\right)^2}{x^2+25}\)
\(=\dfrac{x^2+10x+25+x^2-10x+25}{x^2+25}\)
\(=\dfrac{2x^2+50}{x^2+25}=\dfrac{2\left(x^2+25\right)}{x^2+25}=2\)
\(\Rightarrow\) Giá trị của biểu thức không phụ thuộc vào biến \(x\)
\(A=\dfrac{x^2}{\left(x+y\right)\left(1-y\right)}-\dfrac{y^2}{\left(x+y\right)\left(1+x\right)}-\dfrac{x^2y^2}{\left(1+x\right)\left(1-y\right)}\)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=|x2+y2+5+2x-4y|-|-(x+y-1)2|+2xy
<=>A=||(x²+2x+1)+(y²-4y+4)| - (x+y-1)² + 2xy
= |(x+1)²+(y-2)²| - (x+y-1)² + 2xy
= (x+1)²+(y-2)²-(x+y-1)²+2xy
Đặt x+1=a và y-2=b
=> A = a² + b² - (a+b)² + 2(a-1).(b+2)
= a² + b² - a² - 2ab - b² - 2ab + 4a - 2b - 2
= 4a - 2b - 2
= 4(x + 1)-2(y-2)-2
= 4x+4-2y-4-2
= 4x-2y-2
Thay x = 2²⁰¹⁹ và y = 16⁵⁰³ = 2²⁰¹² vào A, ta có:
A = 4.2²⁰¹⁹ - 2.2²⁰¹² - 2
= 2²⁰²¹ - 2²⁰¹³ - 2
\(\left(x+y+t\right)^3-x^3-y^3-t^3\\ =\left(x+y\right)^3+3\left(x+y\right)^2t+3\left(x+y\right)t^2+t^3-x^3-y^3-t^3\\ =\left(x+y\right)^3+3\left(x+y\right)^2t+3\left(x+y\right)t^2-\left(x^3+y^3\right)\\ =\left(x+y\right)\left[\left(x+y\right)^2+3\left(x+y\right)t+3t^2\right]-\left(x+y\right)\left(x^2-xy+y^2\right)\\ =\left(x+y\right)\left[x^2+2xy+y^2+3\left(x+y\right)t+3t^2-x^2+xy-y^2\right]\\ =\left(x+y\right)\left[3\left(x+y\right)t+3xy+3t^2\right]\\ =3\left(x+y\right)\left(xt+yt+xy+t^2\right)\\ =3\left(x+y\right)\left[t\left(x+t\right)+y\left(x+t\right)\right]\\ =3\left(x+y\right)\left[\left(x+y\right)t+xy+t^2\right]\\ =3\left(x+y\right)\left(y+t\right)\left(t+x\right)\\ \Rightarrow2011=3\left(x+y\right)\left(y+t\right)\left(t+x\right)\\ \Rightarrow D=\dfrac{3\left(x+y\right)\left(y+t\right)\left(t+x\right)}{\left(x+y\right)\left(y+t\right)\left(t+x\right)}=3\)