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\(a\left(a^2-bc\right)+b\left(b^2-ca\right)+c\left(c^2-ab\right)=0\)
\(\Rightarrow a^3-abc+b^3-abc+c^3-abc=0\)
\(\Rightarrow a^3+b^3+c^3-3abc=0\)
\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)=0\)
Mà \(a+b+c\ne0\Rightarrow a^2+b^2+c^2-ab-ac-bc=0\)
\(\Rightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac=0\)
\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\Rightarrow\hept{\begin{cases}a-b=0\\b-c=0\\a-c=0\end{cases}\Rightarrow}a=b=c\)
Vậy \(P=\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}=1+1+1=3\)
Em tham khảo cách làm tại link: Câu hỏi của Cao Chi Hieu - Toán lớp 9 - Học toán với OnlineMath
a) \(x^3+y^3+z^3-3xyz\)
\(=x^3+3x^2y+3xy^2+y^3+z^3-3x^2y-3xy^3-3xyz\)
\(=\left(x+y\right)^3+z^3-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2\right)-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2-3xy\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-xz-yz\right)\)
\(\Rightarrow\frac{a+b+c}{a}+\frac{a+b+c}{b}+\frac{a+b+c}{c}-\frac{a+b+c}{a+b+c}=0\)
\(\Rightarrow\left(a+b+c\right).\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{a+b+c}\right)=0\)
xét: \(\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{a+b+c}=0\left(\text{vì a+b+c khác 0}\right)\)
\(\text{ta có: }\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{a+b+c}=0\)
\(\Rightarrow\frac{ab+bc+ac}{abc}-\frac{1}{a+b+c}=0\)
\(\Rightarrow\frac{\left(ab+bc+ac\right).\left(a+b+c\right)-abc}{abc.\left(a+b+c\right)}=0\)
\(\Rightarrow\left(ab+bc+ac\right).\left(a+b+c\right)-abc=0\)
\(\Rightarrow\left(b+a\right).\left(c+a\right).\left(c+b\right)=0\)
\(\Rightarrow\hept{\begin{cases}b=-a\\a=-c\\c=-b\end{cases}}\)
\(M=\left(-b^{101}+b^{101}\right).\left(-c^{2017}+c^{2017}\right).\left(b^{2019}+-b^{2019}\right)=0\)
p/s: dài nhỉ =)
Từng ý nhé !!!
\(P=\frac{a^2}{bc}+\frac{b^2}{ac}+\frac{c^2}{ab}=\frac{a^3}{abc}+\frac{b^3}{abc}+\frac{c^3}{abc}=\frac{1}{abc}\left(a^3+b^3+c^3\right)\)
\(\frac{1}{abc}.3abc=3\)
\(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left[\frac{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}{2}\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}a+b+c=0\\a=b=c\end{cases}}\)
Xét \(a+b+c=0\) ta có :\(\hept{\begin{cases}a+b=-c\\a+c=-b\\b+c=-a\end{cases}}\)
\(Q=\frac{a^2}{\left(a-b\right)\left(a+b\right)-c^2}+\frac{b^2}{\left(b+c\right)\left(b-c\right)-a^2}+\frac{c^2}{\left(c+a\right)\left(c-a\right)-b^2}\)
\(=\frac{a^2}{-ac+bc-c^2}+\frac{b^2}{-ab+ac-a^2}+\frac{c^2}{-bc+ab-b^2}\)
\(=\frac{a^2}{-c\left(a+c\right)+bc}+\frac{b^2}{-a\left(a+b\right)+ac}+\frac{c^2}{-b\left(c+b\right)+ab}\)
\(=\frac{a^2}{bc+bc}+\frac{b^2}{ac+ac}+\frac{c^2}{ab+ab}\)
\(=\frac{a^2}{2bc}+\frac{b^2}{2ac}+\frac{c^2}{2ab}=\frac{1}{2abc}\left(a^3+b^3+c^3\right)=\frac{1}{2abc}.3abc=\frac{3}{2}\)
Xét \(a=b=c\) ta có :
\(Q=\frac{a^2}{a^2-a^2-a^2}+\frac{b^2}{b^2-b^2-b^2}+\frac{c^2}{c^2-c^2-c^2}=-1-1-1=-3\)
Ai biết cách làm thì nhanh tay giải giùm mình nhé!!!!!!!!!!!!
mk đang cần gấp....<3<3<3<3<3<3
Chú ý rằng, với đa thức \(a^3+b^3+c^3-3abc\) thì ta có thể phân tích đa thức trên thành một nhân tử bằng cách dùng hằng đẳng thức, khi đó:
\(a^3+b^3+c^3-3abc=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)
\(=\left[\left(a+b\right)^3+c^3\right]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-ab+c^2-3ab\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)\)
\(a^3+b^3+c^3-3abc=\frac{1}{2}\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]\)
Nhận xét: Nếu \(a^3+b^3+c^3=3abc\) thì \(a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow\) \(\frac{1}{2}\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]=0\)
\(\Leftrightarrow\) \(^{a+b+c=0}_{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}\) \(\Leftrightarrow\) \(^{a+b+c=0}_{a=b=c}\)
\(------------------\)
Vì \(abc=16\) (theo giả thiết) nên \(a,\) \(b,\) \(c\ne0\) và \(3abc=48\) \(\left(1\right)\)
Ta có: \(a^3+b^3+c^3=48\) \(\left(2\right)\)
Do đó, từ \(\left(1\right)\) và \(\left(2\right)\) suy ra \(a^3+b^3+c^3=3abc\) \(\left(=48\right)\)
\(\Leftrightarrow\) \(a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow\) \(\frac{1}{2}\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]=0\) \(\left(\text{*}\right)\) (theo nhận xét trên)
Mà \(a+b+c\ne0\) nên từ \(\left(\text{*}\right)\) suy ra \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\), tức \(a=b=c\) \(\left(\text{**}\right)\)
Mặt khác, ta cũng có \(abc=16\) và do \(\left(\text{**}\right)\) nên \(a^3=16\)
Khi đó, biểu thức \(P\) sẽ trở thành:
\(P=\frac{\left(a+b\right)}{ab}.\frac{\left(b+c\right)}{bc}.\frac{\left(c+a\right)}{ca}=\frac{2a}{a^2}.\frac{2a}{a^2}.\frac{2a}{a^2}=\frac{8a^3}{a^6}=\frac{8}{a^3}=\frac{8}{16}=\frac{1}{2}\) (do \(a\ne0\))