a,cho \(\left(x+y+t\right)^{^3}-x^3-y^3-t^3=2011\) .tính giá trị của D=\(\dfrac{2011}{\left(x+y\right)\left(y+t\right)\left(t+x\right)}\)
b,cho ba số dương a,b,c thoả mãn \(a+b+c=18099\).tìm GTNN của H=\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)
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\(\left(x+y+t\right)^3-x^3-y^3-t^3\\ =\left(x+y\right)^3+3\left(x+y\right)^2t+3\left(x+y\right)t^2+t^3-x^3-y^3-t^3\\ =\left(x+y\right)^3+3\left(x+y\right)^2t+3\left(x+y\right)t^2-\left(x^3+y^3\right)\\ =\left(x+y\right)\left[\left(x+y\right)^2+3\left(x+y\right)t+3t^2\right]-\left(x+y\right)\left(x^2-xy+y^2\right)\\ =\left(x+y\right)\left[x^2+2xy+y^2+3\left(x+y\right)t+3t^2-x^2+xy-y^2\right]\\ =\left(x+y\right)\left[3\left(x+y\right)t+3xy+3t^2\right]\\ =3\left(x+y\right)\left(xt+yt+xy+t^2\right)\\ =3\left(x+y\right)\left[t\left(x+t\right)+y\left(x+t\right)\right]\\ =3\left(x+y\right)\left[\left(x+y\right)t+xy+t^2\right]\\ =3\left(x+y\right)\left(y+t\right)\left(t+x\right)\\ \Rightarrow2011=3\left(x+y\right)\left(y+t\right)\left(t+x\right)\\ \Rightarrow D=\dfrac{3\left(x+y\right)\left(y+t\right)\left(t+x\right)}{\left(x+y\right)\left(y+t\right)\left(t+x\right)}=3\)
a: ĐKXĐ: \(x,y\in R\)
b: \(A=\dfrac{\dfrac{1}{4}x^2+x^2y+\dfrac{1}{4}y+y^2+x^2y^2+\dfrac{1}{4}+\dfrac{3}{4}y}{x^2y^2+1+x^2-x^2y-y+y^2}\)
\(=\dfrac{\dfrac{1}{4}x^2+x^2y+x^2y^2+y+\dfrac{1}{4}+y^2}{x^2y^2+x^2+1+y^2-x^2y-y}\)
\(=\dfrac{\dfrac{1}{4}\left(x^2+1\right)+y\left(x^2+1\right)+x^2y^2+y^2}{\left(y^2+1\right)\left(x^2+1\right)-y\left(x^2+1\right)}\)
\(=\dfrac{\left(x^2+1\right)\left(y+\dfrac{1}{4}\right)+x^2y^2+y^2}{\left(x^2+1\right)\left(y^2-y+1\right)}\)
\(=\dfrac{\left(x^2+1\right)\left(y+\dfrac{1}{4}\right)+y^2\left(x^2+1\right)}{\left(x^2+1\right)\left(y^2-y+1\right)}=\dfrac{y^2+y+\dfrac{1}{4}}{y^2-y+1}\)
\(A=\left|-x-2011\right|+\left|x+2012\right|\ge\left|-x-2011+x+2012\right|=1\)
\(\Rightarrow A_{min}=1\) khi \(\left\{{}\begin{matrix}x+2011\le0\\x+2012\ge0\end{matrix}\right.\) \(\Rightarrow-2012\le x\le-2011\)
Bài 2:
\(x-y-z=0\Rightarrow\left\{{}\begin{matrix}y-x=-z\\x-z=y\\y+z=x\end{matrix}\right.\)
\(B=\left(\frac{x-z}{x}\right)\left(\frac{y-x}{y}\right)\left(\frac{y+z}{z}\right)=\frac{y.\left(-z\right).x}{xyz}=-1\)
Bài 3:
Gọi chiều dài 3 cạnh tương ứng là \(a,b,c\)
\(\Rightarrow4a=12b=cx\Rightarrow\left\{{}\begin{matrix}a=\frac{cx}{4}\\b=\frac{cx}{12}\end{matrix}\right.\)
Mặt khác theo BĐT tam giác ta có: \(a-b< c< a+b\)
\(\Rightarrow\frac{cx}{4}-\frac{cx}{12}< c< \frac{cx}{4}+\frac{cx}{12}\Rightarrow\frac{x}{4}-\frac{x}{12}< 1< \frac{x}{4}+\frac{x}{12}\)
\(\Rightarrow\frac{x}{6}< 1< \frac{x}{3}\) \(\Rightarrow3< x< 6\) \(\Rightarrow\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)
Bài 1
\(M=\dfrac{2x+y+z-15}{x}+\dfrac{x+2y+z-15}{y}+\dfrac{x+y+2z-15}{z}\)
\(M=\dfrac{x+12-15}{x}+\dfrac{y+12-15}{y}+\dfrac{z+12-15}{z}\)
\(M=\dfrac{x-3}{x}+\dfrac{y-3}{y}+\dfrac{z-3}{z}\)
\(M=1-\dfrac{3}{x}+1-\dfrac{3}{y}+1-\dfrac{3}{z}\)
\(M=3-\left(\dfrac{3}{x}+\dfrac{3}{y}+\dfrac{3}{z}\right)\)
\(M=3-3\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)
Áp dụng bất đẳng thức Cauchy - Schwarz dạng phân thức
\(\Rightarrow\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{\left(1+1+1\right)^2}{x+y+z}=\dfrac{9}{x+y+z}=\dfrac{3}{4}\)
\(\Rightarrow3\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\ge\dfrac{9}{4}\)
\(\Rightarrow3-3\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\le\dfrac{3}{4}\)
\(\Leftrightarrow M\le\dfrac{3}{4}\)
Vậy \(M_{max}=\dfrac{3}{4}\)
Dấu " = " xảy ra khi \(x=y=z=4\)
Bài 2
\(P=\dfrac{\left(a+b+c\right)^2}{30\left(a^2+b^2+c^2\right)}+\dfrac{a^3+b^3+c^3}{4abc}-\dfrac{131\left(a^2+b^2+c^2\right)}{60\left(ab+bc+ca\right)}\)
Xét \(\dfrac{a^3+b^3+c^3}{4abc}\)
\(=\dfrac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)+3abc}{4abc}\)
\(=\dfrac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)}{4abc}+\dfrac{3}{4}\)
\(=\dfrac{1}{4}\left(\dfrac{1}{bc}+\dfrac{1}{ca}+\dfrac{1}{ab}\right)\left(a^2+b^2+c^2-ab-bc-ca\right)+\dfrac{3}{4}\)
Áp dụng bất đẳng thức Cauchy - Schwarz dạng phân thức
\(\Rightarrow\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\ge\dfrac{\left(1+1+1\right)^2}{ab+bc+ca}=\dfrac{9}{ab+bc+ca}\)
\(\Rightarrow\dfrac{1}{4}\left(\dfrac{1}{bc}+\dfrac{1}{ca}+\dfrac{1}{ab}\right)\left(a^2+b^2+c^2-ab-bc-ca\right)+\dfrac{3}{4}\ge\dfrac{9\left(a^2+b^2+c^2-ab-bc-ca\right)}{4\left(ab+bc+ca\right)}+\dfrac{3}{4}\)
\(\Rightarrow\dfrac{1}{4}\left(\dfrac{1}{bc}+\dfrac{1}{ca}+\dfrac{1}{ab}\right)\left(a^2+b^2+c^2-ab-bc-ca\right)+\dfrac{3}{4}\ge\dfrac{9\left(a^2+b^2+c^2\right)-9\left(ab+bc+ca\right)}{4\left(ab+bc+ca\right)}+\dfrac{3}{4}\)
\(\Rightarrow\dfrac{1}{4}\left(\dfrac{1}{bc}+\dfrac{1}{ca}+\dfrac{1}{ab}\right)\left(a^2+b^2+c^2-ab-bc-ca\right)+\dfrac{3}{4}\ge\dfrac{9\left(a^2+b^2+c^2\right)}{4\left(ab+bc+ca\right)}-\dfrac{9}{4}+\dfrac{3}{4}\)
\(\Rightarrow\dfrac{1}{4}\left(\dfrac{1}{bc}+\dfrac{1}{ca}+\dfrac{1}{ab}\right)\left(a^2+b^2+c^2-ab-bc-ca\right)+\dfrac{3}{4}\ge\dfrac{9\left(a^2+b^2+c^2\right)}{4\left(ab+bc+ca\right)}-\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{a^3+b^3+c^3}{4abc}\ge\dfrac{9\left(a^2+b^2+c^2\right)}{4\left(ab+bc+ca\right)}-\dfrac{3}{2}\)
\(\Rightarrow\dfrac{a^3+b^3+c^3}{4abc}-\dfrac{131\left(a^2+b^2+c^2\right)}{60\left(ab+bc+ca\right)}\ge\dfrac{9\left(a^2+b^2+c^2\right)}{4\left(ab+bc+ca\right)}-\dfrac{131\left(a^2+b^2+c^2\right)}{60\left(ab+bc+ca\right)}-\dfrac{3}{2}\)
\(\Rightarrow\dfrac{a^3+b^3+c^3}{4abc}-\dfrac{131\left(a^2+b^2+c^2\right)}{60\left(ab+bc+ca\right)}\ge\dfrac{a^2+b^2+c^2}{15\left(ab+bc+ca\right)}-\dfrac{3}{2}\) (1)
Xét \(\dfrac{\left(a+b+c\right)^2}{30\left(a^2+b^2+c^2\right)}\)
\(=\dfrac{a^2+b^2+c^2+2\left(ab+bc+ca\right)}{30\left(a^2+b^2+c^2\right)}\)
\(=\dfrac{1}{30}+\dfrac{ab+bc+ca}{15\left(a^2+b^2+c^2\right)}\) (2)
Cộng (1) và (2) theo từng vế
\(P\ge\dfrac{a^2+b^2+c^2}{15\left(ab+bc+ca\right)}+\dfrac{ab+bc+ca}{15\left(a^2+b^2+c^2\right)}-\dfrac{22}{15}\)
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\dfrac{a^2+b^2+c^2}{15\left(ab+bc+ca\right)}+\dfrac{ab+bc+ca}{15\left(a^2+b^2+c^2\right)}\ge2\sqrt{\dfrac{\left(a^2+b^2+c^2\right)\left(ab+bc+ca\right)}{225\left(ab+bc+ca\right)\left(a^2+b^2+c^2\right)}}\)
\(\Rightarrow\dfrac{a^2+b^2+c^2}{15\left(ab+bc+ca\right)}+\dfrac{ab+bc+ca}{15\left(a^2+b^2+c^2\right)}\ge2\sqrt{\dfrac{1}{225}}\)
\(\Rightarrow\dfrac{a^2+b^2+c^2}{15\left(ab+bc+ca\right)}+\dfrac{ab+bc+ca}{15\left(a^2+b^2+c^2\right)}\ge\dfrac{2}{15}\)
\(P\ge\dfrac{a^2+b^2+c^2}{15\left(ab+bc+ca\right)}+\dfrac{ab+bc+ca}{15\left(a^2+b^2+c^2\right)}-\dfrac{22}{15}\ge\dfrac{2}{15}-\dfrac{22}{15}=-\dfrac{4}{3}\)
\(\Leftrightarrow P\ge-\dfrac{4}{3}\)
Vậy \(P_{min}=\dfrac{-4}{3}\)
Dấu " = " xảy ra khi \(a=b=c=1\)
Bài 1
\(M=\dfrac{2x+y+z-15}{x}+\dfrac{x+2y+z-15}{y}+\dfrac{x+y+2z-15}{z}\)
a. Đề bài em ghi sai thì phải
Vì:
\(x+y=2\left(\sqrt{x-3}+\sqrt{y-3}\right)\)
\(\Leftrightarrow\left(x-3-2\sqrt{x-3}+1\right)+\left(y-3-2\sqrt{y-3}+1\right)+4=0\)
\(\Leftrightarrow\left(\sqrt{x-3}-1\right)^2+\left(\sqrt{y-3}-1\right)^2+4=0\) (vô lý)
b.
Xét hàm \(f\left(x\right)=x^3+ax^2+bx+c\)
Hàm đã cho là hàm đa thức nên liên tục trên mọi khoảng trên R
Hàm bậc 3 nên có tối đa 3 nghiệm
\(f\left(-2\right)=-8+4a-2b+c>0\)
\(f\left(2\right)=8+4a+2b+c< 0\)
\(\Rightarrow f\left(-2\right).f\left(2\right)< 0\Rightarrow f\left(x\right)\) luôn có ít nhất 1 nghiệm thuộc (-2;2)
\(\lim\limits_{x\rightarrow+\infty}f\left(x\right)=x^3\left(1+\dfrac{a}{x}+\dfrac{b}{x^2}+\dfrac{c}{x^3}\right)=+\infty.\left(1+0+0+0\right)=+\infty\)
\(\Rightarrow\) Luôn tồn tại 1 số thực dương n đủ lớn sao cho \(f\left(n\right)>0\)
\(\Rightarrow f\left(2\right).f\left(n\right)< 0\Rightarrow f\left(x\right)\) luôn có ít nhất 1 nghiệm thuộc \(\left(2;n\right)\) hay \(\left(2;+\infty\right)\)
Tương tự \(\lim\limits_{x\rightarrow-\infty}f\left(x\right)=-\infty\Rightarrow f\left(-2\right).f\left(m\right)< 0\Rightarrow f\left(x\right)\) luôn có ít nhất 1 nghiệm thuộc \(\left(-\infty;-2\right)\)
\(\Rightarrow f\left(x\right)\) có đúng 3 nghiệm pb \(\Rightarrow\) hàm cắt Ox tại 3 điểm pb
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 ...