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Câu 1:
- Chứng minh a3+b3+c3=3abc thì a+b+c=0
\(a^3+b^3+c^3=3abc\Rightarrow a^3+b^3+c^3-3abc=0\)
\(\Rightarrow\left(a+b\right)^3-3a^2b-3ab^2+c^3-3abc=0\)
\(\Rightarrow\left[\left(a+b\right)^3+c^3\right]-3abc\left(a+b+c\right)=0\)
\(\Rightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)=0\)
\(\Rightarrow0=0\) Đúng (Đpcm)
- Chứng minh a3+b3+c3=3abc thì a=b=c
Áp dụng Bđt Cô si 3 số ta có:
\(a^3+b^3+c^3\ge3\sqrt[3]{a^3b^3c^3}=3abc\)
Dấu = khi a=b=c (Đpcm)
Câu 2
Từ \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Rightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=3\cdot\frac{1}{abc}\)
Ta có:
\(\frac{ab}{c^2}+\frac{bc}{a^2}+\frac{ac}{b^2}=\frac{abc}{c^3}+\frac{abc}{a^3}+\frac{abc}{b^3}\)
\(=abc\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)\)
\(=abc\cdot3\cdot\frac{1}{abc}=3\)
khowaaaaaaa
tui mới học lớp 6 mà sao trả lời được mà hỏi
tìm đứa khác giúp đi cưng ok
\(\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
Hoặc \(a+b+c=0\)
Hoặc \(\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
TH1 : \(a+b+c=0\Rightarrow a=-\left(b+c\right);b=-\left(a+c\right);c=-\left(a+b\right)\)
\(\Rightarrow\)\(A=\left[1-\frac{\left(b+c\right)}{b}\right]\left[1-\frac{\left(a+c\right)}{c}\right]\left[1-\frac{\left(a+b\right)}{a}\right]\)
\(\Rightarrow\)\(A=\left(1-1-\frac{c}{b}\right)\left(1-1-\frac{a}{c}\right)\left(1-1-\frac{b}{a}\right)\)
\(\Rightarrow\)\(A=\left(\frac{-c}{b}\right)\left(\frac{-a}{c}\right)\left(\frac{-b}{a}\right)=-1\)
TH2 : \(\left(a^2+b^2+c^2-ab-bc-ac\right)=0\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Rightarrow\)\(a-b=b-c=c-a=0\)hay \(a=b=c=0\)
\(\Rightarrow\)\(A=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)
1.CMR:
a) 3.\(\left(x^2+y^2+z^2\right)-\left(x-y\right)^2\) \(-\left(y-z\right)^2-\left(z-x\right)^2=\left(x+y+z\right)^2\)
Thay a3+b3=(a+b)3-3ab(a+b) vào giả thiết ta có:
(a+b)3-3ab(a+b)+c3-3abc=0
<=> [(a+b)+c].\(\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]\)-3ab(a+b+c)=0
<=> (a+b+c) (a2+b2+c2-ab-bc+c2-3ab)=0
<=> (a+b+c)(a2+b2+c2-ab-bc-ca)=0
\(\Leftrightarrow\orbr{\begin{cases}a+b+c=0\\a^2+b^2+c^2-ab-bc-ca=0\end{cases}}\)
- Nếu a+b+c=0
\(\Rightarrow A=\frac{b+a}{b}\cdot\frac{c+b}{c}\cdot\frac{a+c}{a}=\frac{-c}{b}\cdot\frac{-a}{c}\cdot\frac{-b}{a}\Rightarrow A=-1\)
- Nếu \(a^2+b^2+c^2-ab-bc-ca=0\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
<=> a=b=c
Khi đó \(A=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)
giúp vsssssssssss
https://vn.answers.yahoo.com/question/index?qid=20110907041853AA9iaBQ