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ta có \(\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}+a+b+c=a+b+c\)
\(\Leftrightarrow\left(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\right)\left(a+b+c\right)=a+b+c\)
\(\Leftrightarrow\left(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\right)=\frac{\left(a+b+c\right)}{a+b+c}\)
\(\Leftrightarrow\left(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\right)=1\)
Bài 1.
Từ \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Leftrightarrow \frac{ab+bc+ac}{abc}=0\Rightarrow ab+bc+ac=0\)
\(\Rightarrow ab+bc=-ac\)
Khi đó:
\(D=\frac{bc}{a^2}+\frac{ac}{b^2}+\frac{ab}{c^2}=\frac{(ab)^3+(bc)^3+(ca)^3}{a^2b^2c^2}=\frac{(ab+bc)^3-3ab.bc(ab+bc)+(ac)^3}{a^2b^2c^2}\)
\(=\frac{(-ac)^3-3ab.bc(-ac)+(ac)^3}{a^2b^2c^2}=\frac{3a^2b^2c^2}{a^2b^2c^2}=3\)
Bài 2:
\(a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Rightarrow a+b+c=ab+bc+ac=0\)
\(\Rightarrow a^2+b^2+c^2=\frac{(a+b+c)^2-2(ab+bc+ac)}{2}=0\)
\(\Rightarrow a=b=c=0\)
Vô lý do theo đề bài $a,b,c\neq 0$
Bạn xem lại đề.
\(a^3+b^3+c^3=3abc\)
<=> \(a^3+b^3+c^3-3abc=0\)
<=> \(\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
<=> \(\orbr{\begin{cases}a+b+c=0\\a^2+b^2+c^2-ab-bc-ca=0\end{cases}}\)
<=> \(\orbr{\begin{cases}a+b+c=0\\a=b=c\end{cases}}\)
đến đây ez tự làm nốt nhé, ko ra ib mk
Xét \(a^3+b^3+c^3=1\)
\(\Rightarrow\left(a+b\right)^3-3ab\left(a+b\right)+c^3=1\)
\(\Rightarrow\left(a+b\right)^3+c^3-3ab\left(a+b\right)=1\)
\(\Rightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b\right)=1\)
\(\Rightarrow\left(a+b+c\right)^3-3\left(a+b\right)c\left(a+b+c\right)-3ab\left(a+b\right)=1\)
\(\Rightarrow\left(a+b+c\right)^3-3\left(a+b\right)\left[ac+cb+c^2-ab\right]=1\)
Tìm a+b+c rồi thay vô P(a+b+c) ở dưới
\(a\left(\frac{1}{b}+\frac{1}{c}\right)+b\left(\frac{1}{a}+\frac{1}{c}\right)+c\left(\frac{1}{a}+\frac{1}{b}\right)\)
\(=a\left(P-\frac{1}{a}\right)+b\left(P-\frac{1}{b}\right)+c\left(P-\frac{1}{c}\right)\)
\(=P\left(a+b+c\right)-3\)
\(M=\frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}\)
⇔\(M+3=\frac{b+c}{a}+1+\frac{c+a}{B}+1+\frac{a+b}{c}+1\)
⇔\(M+3=\frac{a+b+c}{a}+\frac{a+b+c}{b}+\frac{a+b+c}{c}\)
⇔\(M+3=abc\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
mà \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)
⇔M+3=abc.0=0
⇔M=-3
quy đồng lên ta có bc/abc+ac/abc+ab/abc=0
bc+ac+ab/abc=0
suy ra bc+ac+ab=0
quy đồng M ta có (b+c)bc/abc+(c+a)ac/abc+(a+b)ab/abc
=(b^2c+bc^2+ac^2+a^2c+a^2b+ab^2)/abc
=(b^2c+ab^2+abc+bc^2+ac^2+abc+a^2c+a^2b+abc-3abc)/abc
=(b(bc+ab+ac)+c(bc+ac+ab)+a(ac+ab+bc)-3abc)/abc
=-3abc/abc=-3
de ma abc=3 dua ti thoi kho day