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\(M=\frac{2004a}{ab+a^2bc+abc}+\frac{b}{bc+b+abc}+\frac{c}{ac+c+1}\)
\(M=\frac{2004a}{ab\left(1+ac+c\right)}+\frac{b}{b\left(c+1+ac\right)}+\frac{c}{ac+c+1}\)
\(M=\frac{2004ac+abc+abc^2}{abc\left(ac+c+1\right)}=\frac{a^2bc^2+abc+abc^2}{abc\left(ac+c+1\right)}=\frac{abc\left(ac+1+c\right)}{abc\left(ac+c+1\right)}=1\)
Lời giải:
Vì $a+b+c=1$ nên:
\(a^2+b^2+abc-1=(a+b)^2-2ab+abc-1\)
\(=(a+b)^2-1+ab(c-2)=(1-c)^2-1+ab(c-2)\)
\(=-c(2-c)+ab(c-2)=c(c-2)+ab(c-2)=(c+ab)(c-2)\)
Do đó:
\(\frac{c+ab}{a^2+b^2+abc-1}=\frac{c+ab}{(c+ab)(c-2)}=\frac{1}{c-2}\)
Hoàn toàn tương tự với các phân thức còn lại, suy ra:
\(\frac{c+ab}{a^2+b^2+abc-1}+\frac{a+bc}{b^2+c^2+abc-1}+\frac{b+ac}{a^2+c^2+abc-1}=\frac{1}{c-2}+\frac{1}{a-2}+\frac{1}{b-2}=\frac{(a-2)(b-2)+(b-2)(c-2)+(c-2)(a-2)}{(a-2)(b-2)(c-2)}\)
\(=\frac{ab+bc+ac-4(a+b+c)+12}{(a-2)(b-2)(c-2)}=\frac{ab+bc+ac+8}{(a-2)(b-2)(c-2)}\)
Ta có đpcm.
\(\Leftrightarrow ab\left(\dfrac{1}{b+c}-\dfrac{1}{a+c}\right)+bc\left(\dfrac{1}{a+c}-\dfrac{1}{a+b}\right)+ca\left(\dfrac{1}{a+b}-\dfrac{1}{b+c}\right)=0\)
\(\Leftrightarrow\dfrac{ab\left(a-b\right)}{\left(b+c\right)\left(a+c\right)}+\dfrac{bc\left(b-c\right)}{\left(a+b\right)\left(a+c\right)}+\dfrac{ca\left(c-a\right)}{\left(a+b\right)\left(b+c\right)}=0\)
\(\Leftrightarrow\dfrac{ab\left(a^2-b^2\right)+bc\left(b^2-c^2\right)+ca\left(c^2-a^2\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=0\)
\(\Leftrightarrow\dfrac{\left(a-b\right)\left(b-c\right)\left(a-c\right)\left(a+b+c\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=b\\b=c\\c=a\end{matrix}\right.\) hay tam giác cân
Các bợn làm nhanh dùm mk nha. Bài kiểm tra sáng mai mình nộp rồi. Ai nhanh nhất mình tick cho nha
sai đề nha bn : là \(\dfrac{2}{\left(a+b\right)^{1002}}\) mới đúng
+ \(\dfrac{x^4}{a}+\dfrac{y^4}{b}=\dfrac{1}{a+b}=\dfrac{x^2+y^2}{a+b}\)
\(\Rightarrow\dfrac{bx^4+ay^4}{ab}=\dfrac{x^2+y^2}{a+b}\)
\(\Rightarrow\left(bx^4+ay^4\right)\left(a+b\right)=ab\left(x^2+y^2\right)\)\(\Rightarrow abx^4+aby^4+a^2y^4+b^2x^4=abx^2+aby^2\)
\(\Rightarrow a^2y^4+b^2x^4=abx^2\left(1-x^2\right)+aby^2\left(1-y^2\right)\)
\(\Rightarrow a^2y^4+b^2x^4=abx^2y^2+abx^2y^2\)
\(\Rightarrow\left(ay^2\right)^2+\left(bx^2\right)^2-2abx^2y^2=0\)
\(\Rightarrow\left(ay^2-bx^2\right)^2=0\)
\(\Rightarrow ay^2-bx^2=0\Rightarrow ay^2=bx^2\)
\(\Rightarrow\dfrac{x^2}{a}=\dfrac{y^2}{b}=\dfrac{x^2+y^2}{a+b}=\dfrac{1}{a+b}\) ( tính chất dãy tỉ số bằng nhau )
\(\Rightarrow\dfrac{x^{2004}}{a^{1002}}=\dfrac{y^{2002}}{b^{1002}}=\dfrac{1}{\left(a+b\right)^{1002}}\)
\(\Rightarrow\dfrac{x^{2004}}{a^{1002}}+\dfrac{y^{2004}}{b^{1002}}=\dfrac{2}{\left(a+b\right)^{1002}}\) ( đpcm )
\(\dfrac{x^2-2x+2004}{x^2}=\dfrac{\dfrac{2003}{2004}x^2+\dfrac{1}{2004}x^2-2x+2004}{x^2}=\dfrac{2003}{2004}+\dfrac{\dfrac{1}{2004}\left(x^2-2.2004+2004^2\right)}{x^2}=\dfrac{2003}{2004}+\dfrac{\dfrac{1}{2004}\left(x-2004\right)^2}{x^2}\ge\dfrac{2003}{2004}\)
\("="\Leftrightarrow x=2004\)
a: \(A=\dfrac{\left(2004+1\right)\left(2004^2-2004+1\right)}{2004^2-2003}=2005\)
b: \(B=\dfrac{\left(2005-1\right)\left(2005^2+2005+1\right)}{2005^2+2006}=2004\)
Áp dụng bất đẳng thức AM - GM ta ccó :
\(\frac{a}{bc}+\frac{b}{ac}\ge2\sqrt{\frac{a}{bc}.\frac{b}{ac}}=2\sqrt{\frac{1}{c^2}}=\frac{2}{c}\)(1)
\(\frac{b}{ac}+\frac{c}{ab}\ge2\sqrt{\frac{b}{ac}.\frac{c}{ab}}=2\sqrt{\frac{1}{a^2}}=\frac{2}{a}\)(2)
\(\frac{a}{bc}+\frac{c}{ab}\ge2\sqrt{\frac{a}{bc}.\frac{c}{ab}}=2\sqrt{\frac{1}{b^2}}=\frac{2}{b}\)(3)
Cộng vế với vế của (1);(2);(3) lại ta được :
\(\frac{2a}{bc}+\frac{2b}{ac}+\frac{2c}{ab}\ge\frac{2}{a}+\frac{2}{b}+\frac{2}{c}\)
\(\Leftrightarrow2\left(\frac{a}{bc}+\frac{b}{ac}+\frac{c}{ab}\right)\ge2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
\(\Rightarrow\frac{a}{bc}+\frac{b}{ac}+\frac{c}{ab}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)(đpcm)
\(\dfrac{2004a}{ab+2004a+2004}+\dfrac{b}{bc+b+2004}+\dfrac{c}{ac+c+1}\)\(=\dfrac{a^2bc}{ab+a^2bc+abc}+\dfrac{b}{bc+b+abc}+\dfrac{c}{ac+c+1}\)\(=\dfrac{a^2bc}{ab\left(1+ac+c\right)}+\dfrac{b}{b\left(c+1+ac\right)}+\dfrac{c}{ac+c+1}\)\(=\dfrac{ac}{ac+c+1}+\dfrac{1}{ac+c+1}+\dfrac{c}{ac+c+1}\)
\(=\dfrac{ac+c+1}{ac+c+1}=1\)
=> đpcm
mk ko hiểu từ bước 1 luôn bn lm rõ ra đc ko
2004 biến đổi tk nào mà mất đi