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2)
\(a^3+b^3+c^3=3abc\)
\(\Rightarrow a^3+b^3+c^3-3abc=0\)
\(\Rightarrow\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)=0\)
\(\Rightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)=0\)
\(\Rightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)=0\)
\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-cb-ac\right)\)
\(\Rightarrow a+b+c=0\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=-c\\b+c=-a\\a+c=-b\end{matrix}\right.\)
\(\Rightarrow N=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)\)
\(\Rightarrow N=\dfrac{a+b}{b}.\dfrac{b+c}{c}.\dfrac{a+c}{a}\)
\(\Rightarrow N=\dfrac{-c}{b}.\dfrac{-a}{c}.\dfrac{-b}{a}\)
\(\Rightarrow N=-1\)
Bài 1:
Thay 2006 = abc vào biểu thức A ,có :
\(\dfrac{a}{ab+a+abc}+\dfrac{b}{bc+b+1}+\dfrac{abc^2}{ac+abc^2+abc}\)
\(=\dfrac{a}{a+ab+abc}+\dfrac{ab}{a\left(1+b+bc\right)}+\dfrac{c.abc}{c\left(a+ab+abc\right)}\)
\(=\dfrac{a}{a+ab+abc}+\dfrac{ab}{a+ab+abc}+\dfrac{abc}{a+ab+abc}\)
\(=\dfrac{a+ab+abc}{a+ab+abc}=1\)
Vậy tại abc = 2006 giá trị biểu thức A là 1
Ta có \(a+b+c\ge3\sqrt[3]{abc}\Leftrightarrow3\ge3\sqrt[3]{abc}\Leftrightarrow\sqrt[3]{abc}\le1\Leftrightarrow abc\le1\)(bđt AM-GM)
Khi đó \(P=2\left(ab+bc+ca\right)-abc\ge2\left(ab+bc+ca\right)-1\)
\(=2\left(\frac{abc}{c}+\frac{abc}{a}+\frac{abc}{b}\right)-1=2\left[abc\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\right]-1\)
\(=2abc\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-1=2.\frac{\left(1+1+1\right)^2}{a+b+c}-1=\frac{2.9}{3}-1=5\)
Đẳng thức xảy ra khi và chỉ khi \(a=b=c=1\)
Vậy GTNN của \(P=5\)đạt được khi \(a=b=c=1\)
p/s : nói chung hướng làm là vậy thôi :v chứ minh làm sai chỗ nào rồi ý
a: \(=\dfrac{x+1}{x+2}\cdot\dfrac{x+3}{x+2}\cdot\dfrac{x+1}{x+3}=\dfrac{\left(x+1\right)^2}{\left(x+2\right)^2}\)
b: \(=\dfrac{x+1}{x+2}:\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x+3\right)^2}\)
\(=\dfrac{x+1}{x+2}\cdot\dfrac{\left(x+3\right)^2}{\left(x+1\right)\left(x+2\right)}=\dfrac{\left(x+3\right)^2}{\left(x+2\right)^2}\)
c: \(=\dfrac{\left(x+3\right)\left(x-1\right)-\left(2x-1\right)\left(x+1\right)-\left(x-3\right)}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x^2+2x-3-2x^2-2x+x+1-x+3}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{-x^2+1}{\left(x-1\right)\left(x+1\right)}=-1\)
\(\left(1+\dfrac{b}{a}\right)\left(1+\dfrac{c}{b}\right)\left(1+\dfrac{a}{c}\right)=8\)
\(\Leftrightarrow\dfrac{a+b}{a}\times\dfrac{b+c}{b}\times\dfrac{a+c}{c}=8\)
\(\Leftrightarrow\left(a+b\right)\left(a+c\right)\left(b+c\right)=8abc\)
~*~*~*~*~
\(\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{a+c}\)
\(=\dfrac{3}{4}+\dfrac{ab}{\left(a+b\right)\left(b+c\right)}+\dfrac{bc}{\left(b+c\right)\left(c+a\right)}+\dfrac{ac}{\left(c+a\right)\left(a+b\right)}\) (1)
\(\Leftrightarrow\dfrac{a}{a+b}-\dfrac{ab}{\left(a+b\right)\left(b+c\right)}+\dfrac{b}{b+c}-\dfrac{bc}{\left(b+c\right)\left(c+a\right)}+\dfrac{c}{c+a}-\dfrac{ac}{\left(c+a\right)\left(a+b\right)}\)
\(=\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{a}{a+b}\left(1-\dfrac{b}{b+c}\right)+\dfrac{b}{b+c}\left(1-\dfrac{c}{c+a}\right)+\dfrac{c}{a+c}\left(1-\dfrac{a}{a+b}\right)\)
\(=\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{a}{a+b}\times\dfrac{c}{b+c}+\dfrac{b}{b+c}\times\dfrac{a}{a+c}+\dfrac{c}{a+c}\times\dfrac{b}{a+b}\)
\(=\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{ac\left(a+c\right)+ab\left(a+b\right)+bc\left(b+c\right)}{\left(a+c\right)\left(b+c\right)\left(a+b\right)}=\dfrac{3}{4}\)
\(\Leftrightarrow ac\left(a+c\right)+ab\left(a+b\right)+bc\left(b+c\right)=\dfrac{3}{4}\times8abc\)
\(\Leftrightarrow ac\left(a+c\right)+ab\left(a+b\right)+bc\left(b+c\right)+2abc=8abc\)
\(\Leftrightarrow\left(a+b\right)\left(a+c\right)\left(b+c\right)=8abc\) luôn đúng
=> (1) đúng
Bạn cũng có thể giải bằng cách đặt \(x=\dfrac{a}{a+b};y=\dfrac{b}{b+c};z=\dfrac{c}{a+c}\).
Cho \(a+b+c=1\) nhé các bạn.
Đặt ab + bc + ca = q; abc = r. Ta có:
\(A=\dfrac{\left(ab+bc+ca\right)+6\left(a+b+c\right)+27}{abc+3\left(ab+bc+ca\right)+9\left(a+b+c\right)+27}-\dfrac{1}{3\left(ab+bc+ca\right)}\)
\(A=\dfrac{q+33}{r+3q+36}-\dfrac{1}{3q}\).
Theo bất đẳng thức Schur: \(a^3+b^3+c^3+3abc\ge a^2b+b^2c+c^2a+ab^2+bc^2+ca^2\)
\(\Leftrightarrow\left(a+b+c\right)^3+9abc\ge4\left(a+b+c\right)\left(ab+bc+ca\right)\)
\(\Leftrightarrow9r\ge4q-1\Leftrightarrow r\ge\dfrac{4q-1}{9}\).
Từ đó \(A\le\dfrac{q+33}{\dfrac{4q-1}{9}+3q+36}-\dfrac{1}{3q}\)
\(\Rightarrow A\leq \frac{27q^2+860q-323}{93q^2+969q}\)
\(\Rightarrow A+\dfrac{1}{10}=\dfrac{\left(3q-1\right)\left(121q+3230\right)}{30q\left(31q+323\right)}\le0\). (Do \(q=ab+bc+ca\le\dfrac{\left(a+b+c\right)^2}{3}=\dfrac{1}{3}\))
\(\Rightarrow A\leq \frac{-1}{10}\). Dấu "=" xảy ra khi và chỉ khi a = b = c = 1.