Sưả câu 2. a²+b²+c²=3abc

Kiểm tra lại đề bài nhé.

Với a = 2; b = 2; c = -1 thỏa mãn đề bài : (a+b+c)^2 = a^2 + b^2 + c^2 

Nhưng không thỏa mãn đẳng thức cần chứng minh.

1, Ta có a^3+b^3+c^3=3abc

-> a^3+b^3+c^3+3a^2b+3ab^2=3abc+3a^2b+3ab^2

-> (a+b)^{3 +} c^3 - 3ab(a+b+c)=0

-> (a+b+c). ((a+b)^2-(a+b).c+c^2)-3ab(a+b+c)=0

-> (a+b+c)(a^2+2ab+b^2-ac-bc+c^2-3ab)=0

Th1: a+b+c=0

->P= a+b/2 . b+c/2 . c+a/2

= (-c)(-a)(-b)/2=-1

TH2 a^2+b^2+c^2-ab-bc-ca=0

->2a^2+2b^2+2c^2-2ab-abc-2ac=0

->(a^2-2ab+b^2)+(a^2-2ac+c^2)+(b^2-2bc+c^2)=0

-> (a-b)^2+(a-c)^2+(b-c)^2=0

Mà (a-b)^2+(a-c)^2+(b-c)^2>= 0

Dấu = xảy ra (=)a-b=0

                         b-c=0

                          a-c=0

-> a=b=c

->P= 1+a/b+1+b/c+1+c/a=2+2+2= 8

bn có thể giải thích phần TH1 ko?

Phần í cx dễ hiểu mà

Đề bài : Cho \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\left(a,b,c\ne0\right)\)và  \(M=\frac{b^2c^2}{a}+\frac{c^2a^2}{b}+\frac{a^2b^2}{c}\)

Chứng minh M=3abc.

Trước tiên, ta chứng minh bài toán phụ : Cho x+y+z=0 . Chứng minh \(x^3+y^3+z^3=3xyz\)

Giải bài toán phụ như sau : Ta có : \(x+y+z=0\Rightarrow z=-\left(x+y\right)\Rightarrow z^3=-\left[x^3+y^3+3xy\left(x+y\right)\right]\)

\(\Rightarrow x^3+y^3+z^3=-3xy\left(x+y\right)=-3xy\left(-z\right)\)

\(\Rightarrow x^3+y^3+z^3=3xyz\)

Áp dụng vào bài đã cho, ta suy ra : \(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\)

Do đó : \(M=\frac{b^2c^2}{a}+\frac{c^2a^2}{b}+\frac{a^2b^2}{c}=\frac{a^2b^2c^2}{a^3}+\frac{a^2b^2c^2}{b^3}+\frac{a^2b^2c^2}{c^3}=a^2b^2c^2\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)=a^2b^2c^2.\frac{3}{abc}=3abc\)Vậy \(M=3abc\)(đpcm)

Cảm ơn bạn nha :*

Không có chi :))

Chúc bạn học tốt ! ^.^

Ta có: \(a^3+b^3+c^3=3abc\Rightarrow\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc=0\Rightarrow\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\Rightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)\cdot c+c^2\right]-3ab\left(a+b+c\right)=0\)\(\Rightarrow\left(a+b+c\right)\left[a^2+b^2+2ab-ac-bc+c^2\right]-3ab\left(a+b+c\right)=0\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)=0\)\(\Rightarrow\left[{}\begin{matrix}a+b+c=0\left(1\right)\\a^2+b^2+c^2-ab-bc-ca=0\left(2\right)\end{matrix}\right.\) 

Từ (1) \(\Rightarrow\left\{{}\begin{matrix}a+b=-c\\a+c=-b\\b+c=-a\end{matrix}\right.\) \(\Rightarrow P=\left(\dfrac{a+b}{b}\right)\cdot\left(\dfrac{b+c}{c}\right)\cdot\left(\dfrac{c+a}{a}\right)=\left(\dfrac{-c}{b}\right)\cdot\left(-\dfrac{a}{c}\right)\cdot\left(-\dfrac{b}{a}\right)=-1\) 

Từ (2) \(\Rightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\Rightarrow a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ca+a^2=0\) \(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

\(\Rightarrow\left\{{}\begin{matrix}a=b\\c=b\\a=c\end{matrix}\right.\)  \(\Rightarrow a=b=c\)  \(\Rightarrow P=\left(\dfrac{a+b}{b}\right)\cdot\left(\dfrac{b+c}{c}\right)\cdot\left(\dfrac{c+a}{a}\right)=\dfrac{2b}{b}\cdot\dfrac{2c}{c}\cdot\dfrac{2a}{a}=8\) 

Vậy...

Ta có: \(a^3+b^3+c^3=3abc\)

\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)

\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)

\(\Leftrightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)=0\)

\(\Leftrightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)=0\)

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2=0\end{matrix}\right.\)

Ta có: \(B=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)\)

\(\Leftrightarrow B=\dfrac{a+b}{b}\cdot\dfrac{b+c}{c}\cdot\dfrac{a+c}{a}\)

Trường hợp 1: a+b+c=0

\(\Leftrightarrow\left\{{}\begin{matrix}a+b=-c\\b+c=-a\\c+a=-b\end{matrix}\right.\)

Thay a+b=-c; b+c=-a và c+a=-b vào biểu thức \(B=\dfrac{a+b}{b}\cdot\dfrac{b+c}{c}\cdot\dfrac{a+c}{a}\), ta được:

\(B=\dfrac{-c}{b}\cdot\dfrac{-a}{c}\cdot\dfrac{-b}{a}=\dfrac{-\left(a\cdot b\cdot c\right)}{abc}=-1\)

Trường hợp 2: \(\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}a-b=0\\b-c=0\\a-c=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=b\\b=c\\c=a\end{matrix}\right.\Leftrightarrow a=b=c\)

Ta có: \(B=\dfrac{a+b}{b}\cdot\dfrac{b+c}{c}\cdot\dfrac{a+c}{a}\)

mà a=b=c(cmt)

nên \(B=\dfrac{b+b}{b}\cdot\dfrac{c+c}{c}\cdot\dfrac{a+a}{a}=\dfrac{2b}{b}\cdot\dfrac{2c}{c}\cdot\dfrac{2a}{a}=2\cdot2\cdot2=8\)

Ta có: \(a^3+b^3+c^3=3abc\)

\(\Rightarrow a^3+b^3+c^3-3abc=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}a+b+c=0\\a^2+b^2+c^2-ab-bc-ca=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}a+b+c=0\\2a^2+2b^2+2c^2-2ab-2bc-2ca=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}a+b+c=0\\\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}a+b+c=0\\a=b=c\end{cases}}\)

bạn thay vào M giải tiếp nha

Ta có: \(a^3+b^3+c^3=3abc\)

\(\Leftrightarrow\left(a^3+b^3\right)+c^3-3abc=0\)

\(\Leftrightarrow\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc=0\)

\(\Leftrightarrow\left[\left(a+b\right)^3+c^3\right]-\left[3ab\left(a+b\right)+3abc\right]=0\)

\(\Leftrightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)=0\)

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

Nếu \(a^2+b^2+c^2-ab-bc-ca\)

\(=\frac{1}{2}\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]\ge0\left(\forall a,b,c\right)\)

Dấu "=" xảy ra khi: a = b = c

Khi đó: \(M=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\left(1+1\right)^3=8\)

Nếu \(a+b+c=0\Rightarrow\hept{\begin{cases}a+b=-c\\b+c=-a\\c+a=-b\end{cases}}\)

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

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

\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)

\(\Leftrightarrow\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc=0\)

\(\Leftrightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}a+b+c=0\\a^2+b^2+c^2-ab-bc-ca=0\end{cases}}\)

- \(a+b+c=0\): 

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

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

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

- \(a^2+b^2+c^2-ab-bc-ca=0\)

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

\(\Leftrightarrow a=b=c\).

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

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

a³+b³+c³=3abc

<=>(a+b)³-3ab(a+b)-3abc+c³=0

<=>(a+b+c)[(a+b)²-(a+b)c+c²]-3ab.(a+b+c)=0

<=>(a+b+c)(a²+b²+c²-ab-bc-ac)=0

<=>(a+b+c)(2a²+2b²+2c²-2ab-2bc-2ac)=0

<=>(a+b+c)[(a-b)²+(b-c)²+(c-a)²]=0

<=>a+b+c=0 [(a-b)²+(b-c)²+(c-a)² khác 0]

=>a²+b²-c²=-2ab;b²+c²-a²=-2bc;c²+a²-b²=-2ac

Suy ra : P=\(-\left(\dfrac{1}{2ab}+\dfrac{1}{2bc}+\dfrac{1}{2ac}\right)=-\dfrac{a+b+c}{2abc}=0\)

GT không hợp lí 

Theo định lí cosi 3 số

a^3+b^3+c^3>=3*canbacba(a^3*b^3*c^3)

<=> a^3+b^3+c^3>=3abc

dấu"=" khi a=b=c

trái Gt a,b,c đôi một khác nhau

Bạn sai rồi. Sao ngu vậy. Giải đến thế mà ko làm ra

a³ + b³ + c³ = 3abc 
<=> (a+b)³+c³-3a²b-3ab²-3abc = 0
<=> (a+b+c)[(a+b)²-(a+b)c+c²]-3ab(a+b+c) = 0
<=> (a+b+c)[(a+b)²-(a+b)c+c²-3ab]= 0
<=> (a+b+c)[a²+b²+c²-ab-bc-ac]=0
<=> 2(a+b+c)[a²+b²+c²-ab-bc-ac]=0
<=> (a+b+c)[2a²+2b²+2c²-2ab-2bc-2ac]=0
<=> (a+b+c)[(a-b)²+(b-c)²+(a-c)²]=0
=> a+b+c=0 hoặc (a-b)²+(b-c)²+(a-c)² =0
<=> a+b+c=0 hoặc a=b=c ( Mà a,b,c đổi 1 khác nhau nên TH này loại )
Ta có :  a+b+c=0
Thay -a²=-(-b-c)²=-b²-2bc-c² ; -b²= -a^2-2ac-c² ; -c²= -a^2-2ab-b²  vào B ,Ta được
=>B =  -1/2( 1/ab + 1/ac + 1/bc ) = -1/2 ( (a+b+c)/abc) 
Mà a+b+c = 0 => B=0