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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:
\(\frac{ab}{c^2}+\frac{bc}{a^2}+\frac{ac}{b^2}=\frac{abc}{c^3}+\frac{abc}{a^3}+\frac{abc}{b^3}\)
\(=abc\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)\)
\(=abc\cdot3\cdot\frac{1}{abc}=3\)
a² + b² + c² + d² + e² ≥ a(b + c + d + e)
Ta có: a² + b² + c² + d² + e²
= (a²/4 + b²) + (a²/4 + c²) + (a²/4 + d²) + (a²/4 + e²)
Lại có: (a/2 - b)² ≥ 0 <=> a²/4 - ab + b² ≥ 0 <=> a²/4 + b² ≥ ab
Tương tự ta có:
. a²/4 + c² ≥ ac
. a²/4 + d² ≥ ad
. a²/4 + e² ≥ ae
--> (a²/4 + b²) + (a²/4 + c²) + (a²/4 + d²) + (a²/4 + e²) ≥ ab + ac + ad + ae
<=> a² + b² + c² + d² + e² ≥ a(b + c + d + e) --> đ.p.c.m
Dấu " = " xảy ra <=> a/2 = b = c = d = e
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)
\(\Leftrightarrow\left(\frac{1}{a}+\frac{1}{b}\right)^3=-\frac{1}{c^3}\)
\(\Leftrightarrow\frac{1}{a^3}+\frac{3}{a^2b}+\frac{3}{ab^2}+\frac{1}{b^3}=-\frac{1}{c^3}\)
\(\Leftrightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=-\frac{3}{a^2b}-\frac{3}{ab^2}=-\frac{3}{ab}\left(\frac{1}{a}+\frac{1}{b}\right)\)
\(\Leftrightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\)
\(\Rightarrow abc\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)=\frac{3}{abc}.abc\)
\(\Rightarrow\frac{bc}{a^2}+\frac{ca}{b^2}+\frac{ac}{b^2}=3\)
\(a^3+b^3+c^3-3abc=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)
\(=\left[\left(a+b\right)^3+c^3\right]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+2ab-bc-ca+c^2\right)-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
\(\Rightarrow a^3+b^3+c^3=3abc\Leftrightarrow a+b+c=0\) vì \(a\ne b\ne c\)
\(\Rightarrow P=\frac{ab^2}{\left(a+b\right)^2-c^2-2ab}+...\)
\(=\frac{ab^2}{\left(a+b+c\right)\left(a+b-c\right)-2ab}+..\)
\(=\frac{ab^2}{-2ab}+\frac{bc^2}{-2bc}+\frac{ca^2}{-2ca}=-\left(\frac{a+b+c}{2}\right)=0\)
Ta có:(a+b+c)2=a2+b2c2+2ab+2bc+2ac
=>ab+bc+ac=0=>ab+ac+bc/abc=0
=>1/a+1/b+1/c=0
=>1/a3+1/b3+1/c3=3/abc
=>bc/a2+ac/a2+ab/c2=abc(1/a3+1/b3+1/c3)=3
theo giả thiết $\left(a+b+c\right)^2=a^2+b^2+c^2$ suy ra ab+ac+bc=0
do đó \(\frac{ab+ac+bc}{abc}=0\) hay \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)
có \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\) suy ra \(\frac{1}{a}+\frac{1}{b}=-\frac{1}{c}\)
\(\frac{1}{a^3}+\frac{1}{b^3}=-\frac{1}{c^3}\)
\(\left(\frac{1}{a}+\frac{1}{b}\right)^3=\left(-\frac{1}{c}\right)^3\)
\(\frac{1}{a^3}+\frac{1}{b^3}+3\frac{1}{ab}\left(\frac{1}{a}+\frac{1}{b}\right)=-\frac{1}{c^3}\) \(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=-3\frac{1}{ab}\left(\frac{1}{a}+\frac{1}{b}\right)\)
\(=-3\frac{1}{ab}.\left(\frac{-1}{c}\right)\)
\(=\frac{3}{abc}\)
$\frac{bc}{a^2}+\frac{ac}{b^2}+\frac{ab}{c^2}$
\(=\frac{abc}{a^3}+\frac{abc}{b^3}+\frac{abc}{c^3}\)
\(=abc\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)=abc.\frac{3}{abc}=3\)
\(a\left(a^2-bc\right)+b\left(b^2-ac\right)+c\left(c^2-ab\right)=0\)
\(a^3-abc+b^3-abc+c^3-abc=0\)
\(a^3+b^3+c^3-3abc=0\)
\(\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)=0\)
\(\left(a+b+c\right)\left(a^2+b^2+c^2+2ab-bc-ca\right)-3ab\left(a+b+c\right)=0\)
\(\left(a+b+c\right)\left(a^2+b^2+c^2+2ab-bc-ca-3ab\right)=0\)
\(\left(a+b+c\right)\left(a^2+b^2+c^2-bc-ca-ab\right)=0\)
Mà \(a+b+c\ne0\)
\(\Rightarrow a^2+b^2+c^2-bc-ca-ab=0\)
\(a^2+b^2+c^2=ab+bc+ca\)
\(2\left(a^2+b^2+c^2\right)=2\left(ab+bc+ca\right)\)
\(2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)
\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
mình làm hơi tắt.
Đến đây bạn tự làm nốt nhé~
\(a+b+c=0\Leftrightarrow\frac{a+b+c}{abc}=0\Leftrightarrow\frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}=0\)
\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}\right)=\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\)
ta thấy từ a+b+c=0 \(\Leftrightarrow a^3+b^3+c^3=3abc\)(được cm nhiều trg sách cx như trên mạng)
\(\frac{a^2}{bc}+\frac{b^2}{ac}+\frac{c^2}{ab}=\frac{a^3+b^3+c^3}{abc}=\frac{3abc}{abc}=3\)
suy ra đpcm
Ta có : \(a+b+c=0\)
Lập phương 2 vế lên ta có :
\(\left(a+b+c\right)^3=0^3\)
\(\Leftrightarrow a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)
mà \(a+b+c=0\Rightarrow\hept{\begin{cases}a+b=-c\\b+c=-a\\a+c=-b\end{cases}}\)
\(\Leftrightarrow a^3+b^3+c^3+3\left(-a\right)\left(-b\right)\left(-c\right)=0\)
\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow a^3+b^3+c^3=3abc\)
Ta lại có:
\(\frac{a^2}{bc}+\frac{b^2}{ca}+\frac{c^2}{ab}-3=0\)
\(\Rightarrow\frac{a^3}{abc}+\frac{b^3}{abc}+\frac{c^3}{abc}-3=0\)
\(\Leftrightarrow\frac{a^3+b^3+c^3}{abc}-3=0\)
Theo chứng minh trên có : \(a^3+b^3+c^3=3abc\)
\(\Rightarrow\frac{3abc}{abc}-3=0\)
\(\Leftrightarrow3-3=0\)( đúng )
Vậy với \(a+b+c=0\left(a\ne0;b\ne0;c\ne0\right)\)thì \(\frac{a^2}{bc}+\frac{b^2}{ca}+\frac{c^2}{ab}-3=0\)