Violympic toán 8

\(P=\dfrac{a}{a+1}+\dfrac{b}{b+1}+\dfrac{c}{c+1}\)

\(P=1-\dfrac{1}{a+1}+1-\dfrac{1}{b+1}+1-\dfrac{1}{c+1}\)

\(P=3-\left(\dfrac{1}{a+1}+\dfrac{1}{b+1}+\dfrac{1}{c+1}\right)\)

Lại có:\(\dfrac{1}{a+1}+\dfrac{1}{b+1}+\dfrac{1}{c+1}\ge\dfrac{9}{a+b+c+3}=\dfrac{9}{4}\)

\(\Rightarrow P\le3-\dfrac{9}{4}=\dfrac{3}{4}\)

Vậy MAXP=3/4<=>a=b=c=1/3

Thay a + b + c =1 vào biểu thức P , ta có :

\(P=\dfrac{a}{2a+b+c}+\dfrac{b}{a+2b+c}+\dfrac{c}{a+b+2c}\)

\(\Rightarrow P-3=\dfrac{-\left(a+b+c\right)}{2a+b+c}+\dfrac{-\left(a+b+c\right)}{a+2b+c}+\dfrac{-\left(a+b+c\right)}{a+b+2c}\)

\(=-\left(a+b+c\right)\left(\dfrac{1}{2a+b+c}+\dfrac{1}{a+2b+c}+\dfrac{1}{a+b+2c}\right)\)

\(\Rightarrow P-3\le-\dfrac{\left(a+b+c\right).9}{4\left(a+b+c\right)}=-\dfrac{9}{4}\)

\(\Rightarrow P\le\dfrac{3}{4}\)

Dấu ''='' xảy ra \(\Leftrightarrow a=b=c=\dfrac{1}{3}\)

Vậy Max P = \(\dfrac{3}{4}\) \(\Leftrightarrow a=b=c=\dfrac{1}{3}\)

\(\Leftrightarrow x^4+x^3+2x^3+2x^2+2x^2+2x+x+1=0\)

\(\Leftrightarrow x^3\left(x+1\right)+2x^2\left(x+1\right)+2x\left(x+1\right)+\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^3+2x^2+2x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^3+x^2+x^2+x+x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left[x^2\left(x+1\right)+x\left(x+1\right)+x+1\right]=0\)

\(\Leftrightarrow\left(x+1\right)^2\left(x^2+x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x^2+x+1=0\left(loai\right)\end{matrix}\right.\)

\(\Leftrightarrow x=-1\)

Ta có : \(x^4+3x^3+4x^2+3x+1=0\)

\(\Leftrightarrow x^4+x^3+2x^3+2x^2+2x^2+2x+x+1=0\)

\(\Leftrightarrow x^3\left(x+1\right)+2x^2\left(x+1\right)+2x\left(x+1\right)+x+1=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^3+2x^2+2x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)^2\left(x^2+x+1\right)=0\)

Vì \(x^2+x+1>0\)

\(\Rightarrow\left(x+1\right)^2=0\)

\(\Rightarrow x=-1\)

Vậy S={-1}

Ta có : \(\dfrac{x^2}{2}+\dfrac{y^2}{3}+\dfrac{z^2}{4}=\dfrac{x^2+y^2+z^2}{5}\)

\(\Leftrightarrow\dfrac{30x^2+20y^2+15z^2}{60}=\dfrac{12\left(x^2+y^2+z^2\right)}{60}\)

\(\Leftrightarrow30x^2-18x^2+20y^2-12y^2+15z^2-12z^2=0\)

\(\Leftrightarrow18x^2+8y^2+3z^2=0\)

Vì \(\left\{{}\begin{matrix}18x^2\ge0\\8y^2\ge0\\3z^2\ge0\end{matrix}\right.\Rightarrow18x^2+8y^2+3z^2\ge0\)

Dấu '' = '' xảy ra \(\Leftrightarrow x=y=z=0\)

Vậy với \(x=y=z=0\) thì \(\dfrac{x^2}{2}+\dfrac{y^2}{3}+\dfrac{z^2}{4}=\dfrac{x^2+y^2+z^2}{5}\)

Ta có : \(3y-x=6\)

\(\Rightarrow x=3y-6\)

Thay \(x=3y-6\) vào biểu thức A , ta có :

\(\Rightarrow A=\dfrac{3y-6}{y-2}+\dfrac{2\left(3y-6\right)-3y}{3y-6-6}\)

\(=\dfrac{3\left(y-2\right)}{y-2}+\dfrac{3y-12}{3y-12}=3+1=4\)

Vậy A = 4 .

\(=x^8+x^2+x+1-x^2\)

\(=x^8+x+1\)

\(=x^8-x^7+x^5-x^4+x^2+x^7-x^6+x^4-x^3+x+x^6-x^5+x^3-x^2+1\)

\(=\left(x^2+x+1\right)\left(x^6-x^5+x^3-x^2+1\right)\)

\(a^4+b^4+c^4=2a^2b^2+2a^2c^2+2b^2c^2\)

\(\Leftrightarrow a^4+b^4+c^4-2a^2b^2-2a^2c^2-2b^2c^2=0\)

\(\Leftrightarrow\left(a^4-2a^2b^2+b^4\right)+\left(b^4-2b^2c^2+c^4\right)+\left(c^4-2c^2a^2+a^4\right)-a^4-b^4-c^4=0\)

\(\Leftrightarrow\left(a^2-b^2\right)^2+\left(c^2-b^2\right)^2+\left(c^2-a^2\right)^2-a^4-b^4-c^4=0\)

\(\Leftrightarrow\left(a-b\right)^2c^2+a^2\left(b+c\right)^2+b^2\left(c+a\right)^2-a^4-b^4-c^4=0\)

\(\Leftrightarrow c^2\left[\left(a-b\right)^2-\left(a+b\right)^2\right]+a^2\left[\left(b+c\right)^2-a^2\right]+b^2\left[\left(c+a\right)^2-b^2\right]=0\)

\(\Leftrightarrow c^2\left[\left(a-b\right)^2-\left(a+b\right)^2\right]+a^2\left[\left(b+c\right)^2-\left(c-b\right)^2\right]+b^2\left[\left(c+a\right)^2-\left(c-a\right)^2\right]=0\)

\(\Leftrightarrow-4abc^2+4a^2bc+4ab^2c=0\)

\(\Leftrightarrow4abc\left(a+b-c\right)=0\)

\(\Leftrightarrow0=0\)(luôn đúng)

=>đpcm

+ a + b + c = 2

+ a,b,c là độ dài 3 cạnh 1 tam giác

\(\Rightarrow a< b+c\)

=> a + a < a + b + c

=> 2a < 2 => a < 1

+ Tương tự ta cm đc : b < 1; c < 1

+ \(\left(1-a\right)\left(1-b\right)\left(1-c\right)>0\)

=> \(1-\left(a+b+c\right)+\left(ab+bc+ca\right)-abc>0\)

\(\Rightarrow2-2\left(a+b+c\right)+2\left(ab+bc+ca\right)-2abc>0\)

\(\Rightarrow2-\left(a+b+c\right)^2+2\left(ab+bc+ca\right)-2abc>0\)

( do a + b + c = 2 )

\(\Rightarrow2-\left(a^2+b^2+c^2\right)-2abc>0\)

\(\Rightarrow a^2+b^2+c^2+2abc< 2\)

Câu hỏi của Nguyễn Anh Minh - Toán lớp 9 | Học trực tuyến