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oh no bài thứ nhất là dạng chứng minh cs đúng ko ,
ko thể nào là dạng tìm a,b,c đc-.-
Ta có: \(A=a\left(a^2-bc\right)+b\left(b^2-ac\right)+c\left(c^2-ab\right)=0\)
\(\Rightarrow A=a^3+b^3+c^3-3abc=0\) \(\Rightarrow A=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)
\(\Rightarrow A=\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)=0\)
\(\Rightarrow A=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)
Vì \(a+b+c\ne0\Rightarrow a^2+b^2+c^2-ab-ac-bc=0\)
Xét \(M=a^2+b^2+c^2-ab-ac-bc=0\)
\(\Rightarrow2M=2a^2+2b^2+2c^2-2ab-2ac-2bc=0\)
\(\Rightarrow2M=\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
Vì \(\left(a-b\right)^2\ge0;\left(b-c\right)^2\ge0;\left(c-a\right)^2\ge0\forall a,b,c\)
\(\Rightarrow a-b=0;b-c=0;c-a=0\) \(\Rightarrow a=b=c\)
\(\Rightarrow P=\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}=1+1+1=3\)
\(a^2+b^2+c^2=ab+bc+ca\Rightarrow2a^2+2b^2+2c^2=2ab+2bc+2ca\)
\(\Rightarrow\left(2a^2+2b^2+2c^2\right)-\left(2ab+2bc+2ca\right)=0\)
\(\Rightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)
\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\)\(\Rightarrow a-b=b-c=c-a=0\)
\(\Rightarrow P=\left(a-b\right)^{2015}+\left(b-c\right)^{2016}+\left(c-a\right)^{2017}=0\)
b: \(M=\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}=\dfrac{a+b+c}{abc}=0\)
c: \(B=\dfrac{y}{\left(x-y\right)\left(y-z\right)}-\dfrac{z}{\left(x-z\right)\left(y-z\right)}-\dfrac{x}{\left(x-z\right)\left(x-y\right)}\)
\(=\dfrac{y\left(x-z\right)-z\left(x-y\right)-x\left(y-z\right)}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(=\dfrac{xy-yz-xz+zy-xy+xz}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}=0\)
\(\dfrac{bc}{a}+\dfrac{ca}{b}+\dfrac{ab}{c}=a+b+c\)
\(\Leftrightarrow\dfrac{abc}{a^2}+\dfrac{abc}{b^2}+\dfrac{abc}{c^2}=a+b+c\)
\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=\dfrac{a+b+c}{abc}=\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}\)
\(\Leftrightarrow\left(\dfrac{1}{a}-\dfrac{1}{b}\right)^2+\left(\dfrac{1}{b}-\dfrac{1}{c}\right)^2+\left(\dfrac{1}{c}-\dfrac{1}{a}\right)^2=0\)
\(\Leftrightarrow a=b=c\)
Thay vào A r tính thôi
Ta có: \(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)
\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}a+b+c=0\\a^2+b^2+c^2-ab-bc-ca=0\end{matrix}\right.\)
Vì a, b, c là các số dương \(\Rightarrow a=b=c=0\) ( loại )
\(\Rightarrow a^2+b^2+c^2-ab-bc-ca=0\)
\(\Rightarrow a=b=c\) ( tự chứng minh )
\(\Rightarrow M=\left(\dfrac{a}{b}-1\right)+\left(\dfrac{b}{c}-1\right)+\left(\dfrac{c}{a}-1\right)=0\)
Vậy M = 0
\(P=\dfrac{1}{bc\left(b+c\right)+2023}+\dfrac{1}{ca\left(c+a\right)+2023}+\dfrac{1}{ab\left(a+b\right)+2023}\left(abc=2023\right)\)
\(\Leftrightarrow P=\dfrac{1}{bc\left(b+c\right)+abc}+\dfrac{1}{ca\left(c+a\right)+abc}+\dfrac{1}{ab\left(a+b\right)+abc}\)
\(\Leftrightarrow P=\dfrac{1}{bc\left(a+b+c\right)}+\dfrac{1}{ca\left(a+b+c\right)}+\dfrac{1}{ab\left(a+b+c\right)}\)
\(\Leftrightarrow P=\dfrac{1}{\left(a+b+c\right)}\left(\dfrac{1}{bc}+\dfrac{1}{ca}+\dfrac{1}{ab}\right)\)
\(\Leftrightarrow P=\dfrac{1}{\left(a+b+c\right)}\left[\dfrac{a^2bc+b^2ca+c^2ab}{\left(abc\right)^2}\right]\)
\(\Leftrightarrow P=\dfrac{1}{\left(a+b+c\right)}\left[\dfrac{abc\left(a+b+c\right)}{\left(abc\right)^2}\right]\)
\(\Leftrightarrow P=\dfrac{1}{abc}=\dfrac{1}{2023}\)