Question

Câu 1: CMR : Nếu \(a^3+b^3+c^3=3abc\) thì \(a+b+c=0\) hoặc \(a=b=c\)

Câu 2: Cho \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\) . Tính \(\frac{bc}{a^2}+\frac{ca}{b^2}+\frac{ab}{c^2}\)

Câu 3 : Cho \(a^3+b^3+c^3=3abc\left(a.b.c\ne0\right)\). Tính\(A=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)\)

Answer 1

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)

 

 

 

Answer 2

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:

\(\frac{ab}{c^2}+\frac{bc}{a^2}+\frac{ac}{b^2}=\frac{abc}{c^3}+\frac{abc}{a^3}+\frac{abc}{b^3}\)

\(=abc\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)\)

\(=abc\cdot3\cdot\frac{1}{abc}=3\)

Answer 3

\(a^3+b^3+c^3=3abc\)

\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}a+b+c=0\\a^2+b^2+c^2-ab-bc-ac=0\end{array}\right.\)

Xét \(a+b+c=0\)\(\Rightarrow\begin{cases}a=-\left(b+c\right)\\b=-\left(a+c\right)\\c=-\left(a+b\right)\end{cases}\)

\(\Rightarrow A=\left(1-\frac{b+c}{b}\right)\left(1-\frac{a+c}{c}\right)\left(1-\frac{a+b}{a}\right)\)

\(=\left(1-1-\frac{c}{b}\right)\left(1-1-\frac{a}{c}\right)\left(1-1-\frac{b}{a}\right)\)

\(=\left(-\frac{c}{b}\right)\left(-\frac{a}{c}\right)\left(-\frac{b}{a}\right)=-1\)

Xét \(a^2+b^2+c^2-ab-bc-ca=0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

\(\Leftrightarrow a-b=b-c=c-a=0\Leftrightarrow a=b=c\)

\(\Leftrightarrow A=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)

 

Answer 4

Câu 1 :

nếu bn chưa học COsi thì

a³+b³+c³-abc=(a+b+c)(a²+b²+c²-ab-bc-ca)=0

<=>a+b+c=0 hoặc a²+b²+c²-ab-bc-ca=0

<=>a+b+c=0 hoặc (a - b)² +(b - c)²+ (c - a)²= 0

<=>a+b+c=0 hoặc a=b=c