Câu a : Ta có : \(x^3+x^2z+y^2z-xyz+y^3=0\)

\(\Leftrightarrow\left(x^3+y^3\right)+\left(x^2z+y^2z-xyz\right)=0\)

\(\Leftrightarrow\left(x+y\right)\left(x^2-xy+y^2\right)+z\left(x^2-xy+y^2\right)=0\)

\(\Leftrightarrow\left(x^2-xy+y^2\right)\left(x+y+z\right)=0\)

\(\Leftrightarrow x+y+z=0\) ( đpcm )

Câu b : \(VT=\left(a+b+c\right)^3-a^3-b^3-c^3\)

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

\(=3\left(a+b\right)\left(b+c\right)\left(c+a\right)=VP\)

Câu c : Ta có : \(a^3+b^3+c^3=3abc\)

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

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

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

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

\(\Leftrightarrow a+b+c=0\) ( đúng )

1. \(a^3+b^3+c^3-3abc\)

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

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

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

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

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

2. \(a+b+c=0\)

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

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

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

3.Còn có a + b + c = 0 nữa mà bn.

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

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

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

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

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

\(\Rightarrow2a^2+2b^2+2c^2=2ab+2bc+2ac\)

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

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

\(\Rightarrow a=b=c\)

làm đúng mà ko hiểu

e)

\(\dfrac{a^2+b^2+c^2}{3}\ge\left(\dfrac{a+b+c}{3}\right)^2\)

\(\Leftrightarrow3\left(a^2+b^2+c^2\right)\ge a^2+b^2+c^2+2\left(ab+bc+ca\right)\)

\(\Leftrightarrow2\left(a^2+b^2+c^2\right)\ge2\left(ab+bc+ac\right)\)

\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2ac-2bc\ge0\)

\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(a^2-2ac+c^2\right)+\left(b^2-2bc+c^2\right)\ge0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2\ge0\) ( luôn đúng)

=> ĐPCM

BPT?

Tách VP ra

đặt biến phụ như sau:

x=a+b-c

y=a-b+c

z=b+c-a

=> x+y=(a+b-c)+(a-b+c)=2a

=> \(a=\frac{\left(x+y\right)}{2}\)

y+z=(a-b+c)+(b+c-a)=2c

=>\(c=\frac{\left(y+z\right)}{2}\)

z+x=(b+c-a)+(a+b-c)=2b

=> \(b=\frac{\left(z+x\right)}{2}\)

thay vào phương trình ban đầu

=> \(\left(\frac{x+y}{2}\right)^3+\left(\frac{z+x}{2}\right)^3+\left(\frac{y+z}{2}\right)^3=x^3+y^3+z^3\)

nhân đều cả hai vế với 8 để khử VT

=> \(\left(x+y\right)^3+\left(z+x\right)^3+\left(y+z\right)^3=8\left(x^3+y^3+z^3\right)\)

=\(\left(x^3+3x^2y+3xy^2+y^3+z^3+3z^2x+3zx^2+x^3+y^3+3y^2z+3yz^2+z^2\right)=8\left(x^3+y^3+z^3\right)\)

=\(\left(2x^3+y^3+z^3\right)+3\left(x^2y+xy^2+z^2x+zx^2+y^2z+yz^2\right)=8\left(x^3+y^3+z^3\right)\)

\(3\left(x^2y+xy^2+z^2x+zx^2+y^2z+yz^2\right)=6\left(x^3+y^3+z^3\right)\)

chia cả hai vế cho 3

=> \(x^2y+xy^2+z^2x+zx^2+y^2z+yz^2=2\left(x^3+y^3+z^3\right)\)

=> \(2x^3+2y^3+2z^3-\left(x^2y+xy^2+z^2x+zx^2+y^2z+yz^2\right)=0\)

\(\left(x^3-x^2y-xy^2+y^3\right)+\left(y^3-y^2z-yz^2+z^3\right)+\left(z^3-z^2x-zx^2+x^3\right)=0\)

=> \(\left(x-y\right)\left(x^2-y^2\right)+\left(y-z\right)\left(y^2-z^2\right)+\left(z-x\right)\left(z-x\right)^2\)

=> \(\left(x-y\right)^2\left(x+y\right)+\left(y-z\right)^2\left(y+z\right)+\left(z-x\right)^2\left(z+x\right)=0\)

thay ngược lại các giá trị và

x-y=(a+b-c)-(a-b+c)=2b-2c=2(b-c)

làm tương tự ta có:

y-z=2(a-b)

z-x=2(c-a)

thay vòa phương trình vừa suy ra trước đó:

\(2\left(b-c\right)^2\cdot2a+2\left(a-b\right)^2\cdot2c+2\left(c-a\right)^2\cdot2b=0\)

\(8a\left(b-c\right)^2+8c\left(a-b\right)^2+8b\left(c-a\right)^2=0\)

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

theo giả thiết a,b,c>0

=> \(a\left(b-c\right)^2\ge0\)

\(c\left(a-b\right)^2\ge0\)

\(b\left(c-a\right)^2\ge0\)

=> b-c=0;a-b=0;c-a=0

=>a=b=c(đpcm)

a/

\(a^2+b^2+c^2+29ab+bc+ca=3\left(ab+bc+ca\right)\)

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

\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)

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

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

b/ \(a^3+b^3+c^3=\left(a+b\right)^3+c^3-3ab\left(a+b\right)\)

\(=\left(a+b+c\right)\left(\left(a+b\right)^2-c\left(a+b\right)+c^2\right)-3ab\left(a+b\right)\)

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

c/ Không, vì \(a=b=c\ne\) thì \(a^3+b^3+c^3=3a^3=3abc\) vẫn đúng

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

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

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

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

\(\Leftrightarrow\frac{1}{2}\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]=0\)

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

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

chữ đẹp đấy

Ta có

\(\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\ge a^2+b^2+c^2\)

\(\dfrac{a^4}{ab}+\dfrac{b^4}{bc}+\dfrac{c^4}{ca}\ge a^2+b^2+c^2\)

Áp dụng bất đẳng thức Svacxo ta có

\(\dfrac{a^4}{ab}+\dfrac{b^4}{bc}+\dfrac{c^4}{ca}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{ab+bc+ca}\) (1)

Chứng minh bất đẳng thức sau:

\(\dfrac{\left(a^2+b^2+c^2\right)^2}{ab+bc+ca}\ge\left(a^2+b^2+c^2\right)\) (2)

Rút gọn 2 bên ta được

\(\left(a^2+b^2+c^2\right)\ge ab+bc+ca\)

\(2\left(a^2+b^2+c^2\right)\ge2ab+2bc+2ca\)

\(a^2-2ab+b^2+a^2-2ac+c^2+b^2-2bc+c^2\ge0\)

\(\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2\ge0\) (luôn đúng)

Từ(1) và (2) suy ra đpcm

Lời giải hay quá bạn

Lời giải:

$a+b-c=0\Rightarro a+b=c$. Kết hợp sử dụng đẳng thức quen thuộc \(a^3+b^3=(a+b)^3-3ab(a+b)\) ta có:

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

\(=c^3-3ab.c-c^3+3abc=0\) (đpcm)

xin chào

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

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

=> a=b=c

b, \(0=\left(a+b+c\right)^3=a^3+b^3+c^3+6abc+3a^2b+3ab^2+3b^2c+3bc^2+3c^2a+3ca^2\)

\(=a^3+b^3+c^3+6abc+3ab\left(a+b\right)+3bc\left(b+c\right)+3ac\left(a+c\right)\)

\(=a^3+b^3+c^3+6abc-3abc-3abc-3abc\)

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