![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :
\(\left(a-b-c\right)^2=a^2+b^2+c^2-2ab-2bc-2ac\)
mà theo đề bài \(a^2+b^2+c^2-ab-bc-ac=0\)
\(\Rightarrow\left(a-b-c\right)^2=-ab-bc-ac=0\)
\(\Rightarrow\left(a-b-c\right)^2=-\left(ab+bc+ac\right)=0\)
mà \(-\left(ab+bc+ac\right)\le0\)
\(\Rightarrow a=b=c=0\)
\(\Rightarrow dpcm\)
![](https://rs.olm.vn/images/avt/0.png?1311)
chứng minh rằng
nếu a2 + b2 + c2 = ab +ac + bc thì a = b= c
Giải
Ta có: a^2 + b^2 + c^2 = ab + bc + ca
<=> 2.a^2 + 2.b^2 + 2.c^2 = 2.ab + 2.bc + 2.ca
<=> ( a^2 - 2ab + b^2 ) + ( b^2 - 2bc +c^2 ) + ( c^2 - 2ac + a^2 ) =0
<=> (a-b)^2 + (b-c)^2 + (c -a)^2 =0 (1)
Vì (a-b)^2 ; (b-c)^2 ; (c -a)^2 ≧ 0 với mọi a,b,c.
=> (a-b)^2 + (b-c)^2 + (c -a)^2 ≧ 0 (2)
Từ (1) và (2) khẳng định dấu "=" khi:
a - b = 0; b - c = 0 ; c - a = 0 => a=b=c
Vậy a=b=c.
\(a^2+b^2+c^2=ab+ac+bc\Leftrightarrow a^2+b^2+c^2-ab-ac-bc=0\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2ac-2bc=0\Leftrightarrow\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2=0\)
\(\left\{{}\begin{matrix}a-b=0\\a-c=0\\b-c=0\end{matrix}\right.\)\(\Leftrightarrow a=b=c\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a )
`VP= (a+b)^3-3ab(a+b)`
`=a^3+3a^2b+3ab^2+b^3-3a^2b-3ab^2`
`=a^3+b^3 =VT (đpcm)`
b)
b) Ta có
`VT=a3+b3+c3−3abc`
`=(a+b)3−3ab(a+b)+c3−3abc`
`=[(a+b)3+c3]−3ab(a+b+c)`
`=(a+b+c)[(a+b)2+c2−c(a+b)]−3ab(a+b+c)`
`=(a+b+c)(a2+b2+2ab+c2−ac−bc−3ab)`
`=(a+b+c)(a2+b2+c2−ab−bc−ca)=VP`
a) Ta có:
`VP= (a+b)^3-3ab(a+b)`
`=a^3 + b^3+3ab ( a + b )- 3ab ( a + b )`
`=a^3 + b^3=VT(dpcm)`
b) Ta có
`VT=a^3+b^3+c^3−3abc`
`=(a+b)^3−3ab(a+b)+c^3−3abc`
`=[(a+b)^3+c^3]−3ab(a+b+c)`
`=(a+b+c)[(a+b)^2+c^2−c(a+b)]−3ab(a+b+c)`
`=(a+b+c)(a^2+b^2+2ab+c^2−ac−bc−3ab)`
`=(a+b+c)(a^2+b^2+c^2−ab−bc−ca)=VP`
![](https://rs.olm.vn/images/avt/0.png?1311)
c: Ta có: \(a\left(a+2b\right)^3-b\left(2a+b\right)^3\)
\(=a^4+6a^3b+12a^2b^2+8ab^3-8a^3b-12a^2b^2-6ab^3-b^4\)
\(=a^4-2a^3b+2ab^3-b^4\)
\(=\left(a-b\right)\left(a+b\right)\left(a^2+b^2\right)-2ab\left(a^2-b^2\right)\)
\(=\left(a-b\right)^3\cdot\left(a+b\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Vì a,b,c là 3 cạnh tam giác nên \(a+b>c\Leftrightarrow ac+bc>c^2\)
CMTT: \(ab+bc>b^2;ab+ac>a^2\)
Cộng vế theo vế \(\Leftrightarrow a^2+b^2+c^2< ab+bc+ca+ab+bc+ca\)
\(\Leftrightarrow a^2+b^2+c^2< 2ab+2bc+2ca\\ \Leftrightarrow a^2+b^2+c^2-2ab-2bc-2ca< 0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(a+b+c\right)^2=a^2+b^2+c^2\Leftrightarrow ab+bc+ca=0\)
\(\Rightarrow a^3b^3+b^3c^3+c^3a^3=3a^2b^2c^2\)
Ta có:
\(\dfrac{bc}{a^2}+\dfrac{ac}{b^2}+\dfrac{ab}{c^2}=\dfrac{a^3b^3+b^3c^3+c^3a^3}{a^2b^2c^2}=\dfrac{3a^2b^2c^2}{a^2b^2c^2}=3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a^2+b^2+c^2-ab-ac-bc=0\\\Leftrightarrow 2a^2+2b^2+2c^2-2ab-2ac-2bc=0\\\Leftrightarrow (a^2-2ab+b^2)+(b^2-2bc+c^2)+(a^2-2ac+c^2)=0\\\Leftrightarrow (a-b)^2+(b-c)^2+(a-c)^2=0\)
Ta thấy: \(\left(a-b\right)^2\ge0\forall a;b\)
\(\left(b-c\right)^2\ge0\forall b;c\)
\(\left(a-c\right)^2\ge0\forall a;c\)
\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\ge0\forall a;b;c\)
Mặt khác: \(\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)
nên: \(\left\{{}\begin{matrix}a-b=0\\b-c=0\\a-c=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=b\\b=c\\a=c\end{matrix}\right.\)
\(\Leftrightarrow a=b=c\left(dpcm\right)\)
#\(Toru\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có
$$a^2+b^2+c^2-ab-bc-ca=0,$$
hay $$\dfrac{1}{2}\left[(a-b)^2+(b-c)^2 +(c-a)^2\right[ = 0.$$
Mà vế trái luôn không âm \(\forall a,b,c \in \mathbb{R}\), đẳng thức xảy ra khi $a=b=c.$
Vậy ta có điều cần chứng minh.
Ta có: \(a^2+b^2+c^2=ab+bc+ca\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Leftrightarrow a=b=c\)
a) => 2a^2 + 2b^2 = 2ab + 2ba
=> 2a^2 + 2b^2 - 2ab - 2ba = 0
=> (a-b)^2 + (a-b)^2 = 0
=> 2(a-b)^2 = 0
=> a-b = 0
=> a = b
b) Nhân hai vế với 2 và làm tương tự câu a)
=> (a-b)^2 + (b-c)^2 + (a-c)^2 = 0
=> a = b = c