a) a² + b² + c² = ab + bc + ac

\(\Rightarrow\) a² + b² + c² - ab - bc - ac = 0

\(\Rightarrow\) 2(a² + b² + c² - ab - bc - ac) = 0

\(\Rightarrow\) a² + a² + b² + b² + c² + c² - 2ab - 2bc - 2ac = 0

\(\Rightarrow\) (a² - 2ab + b²) + (a² - 2ac + c²) + (b² - 2bc + c²) = 0

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

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

\(\Rightarrow\) a = b = c

b) (a + b + c)² = 3(a² + b² + c²)

a² + b² + c² + 2ab + 2bc + 2ac = 3a² + 3b² + 3c²

\(\Rightarrow\) 2ab + 2ac + 2bc = 2a² + 2b² + 2c²

\(\Rightarrow\) 0 = a² + a² + b² + b² + c² + c² - 2ab - 2bc - 2ac

Hay: a² + a² + b² + b² + c² + c² - 2ab - 2bc - 2ac = 0

\(\Rightarrow\)(a² - 2ab + b²) + (a² - 2ac + c²) + (b² - 2bc + c²) = 0

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

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

\(\Rightarrow\) a = b = c

c) (a + b + c)² = 3(ab + bc + ac)

a² + b² + c² + 2ab + 2ac + 2bc = 3ab + 3bc + 3ac

\(\Rightarrow\) a² + b² + c² = ab + ac + bc2

\(\Rightarrow\) 2(a² + b² + c²) = 2(ab + ac + bc)

\(\Rightarrow\) 2a² + 2b² + 2c² = 2ab + 2bc + 2ac

\(\Rightarrow\) a² + a² + b² + b² + c² + c² - 2ab - 2bc - 2ac = 0

\(\Rightarrow\) (a² - 2ab + b²) + (a² - 2ac + c²) + (b² - 2bc + c²) = 0

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

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

\(\Rightarrow\) a = b = c

CHÚC BN HOK TỐT(nhớ tik mik nha)

a)Cmr : Nếu : a² + b² + c² = ab + bc + ac thì a = b =c

Bài làm

2a² + 2b² + 2c² = 2ab + 2bc + 2ca

=> 2a² + 2b² + 2c² - 2ab - 2bc - 2ac = 0

=> ( a² - 2ab + b²) + ( a² - 2ac + c²⁾ + ( b² - 2bc + c²) =0

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

Vậy :

* ( a - b)² = 0

* ( a - c)² =0

* (b -c)² =0

Suy ra :

* a =b

* a =c

* b = c

Suy ra : a = b =c ( đpcm)

a) \(a^2+b^2+c^2=ab+bc+ac\)

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

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

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

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

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

=> ĐPCM

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

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

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

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

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

\(\Leftrightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}\Leftrightarrow\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}\Leftrightarrow}a=b=c}\)

=> ĐPCM

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

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

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

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

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

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

\(\Leftrightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}\Leftrightarrow\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}\Leftrightarrow}a=b=c}\)

=> ĐPCM

a,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≧0 ; (b-c)^2≧0 ; (c -a)^2 ≧ 0 với mọi a,b,c.
=> (a-b)^2 + (b-c)^2 + (c -a)^2 ≧ 0 (2)
Từ (1) và (2) :
=>a - b = 0; b - c = 0 ; c - a = 0 => a=b=c
Vậy a=b=c.

b,Ta có:(a+b+c)^2=3(a^2+b^2+c^2)

<=>a^2+b^2+c^2+2ab+2ac+2bc=3a^2+3b^2+3c^2

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

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

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

<=>(-a+b)^2+(-b+c)^2+(-a+c)^2=0(1)

ta có:(-a+b)^2≧0, (-b+c)^2≧0, (-a+c)^2≧0(2)với mọi a,b,c.

từ (1)và (2)=>(-a+b)^2=0; (-b+c)^2=0; (-a+c)^2=0

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

<=>a=b=c

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

Chúc bạn học tốt

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

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

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

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

\(\Leftrightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}\Leftrightarrow\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}\Leftrightarrow}a=b=c}\)

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

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

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

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

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

\(\Leftrightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}\Leftrightarrow\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}\Leftrightarrow}a=b=c}\)

(Nhớ k cho mình với nhé!)

`1)(a+b+c)^2=3(a^2+b^2+c^2)`

`<=>a^2+b^2+c^2+2ab+2bc+2ca=3a^2+3b^2+3c^2`

`<=>2ab+2bc+2ca=2a^2+2b^2+2c^2`

`<=>a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ca+a^2=0`

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

Mà `(a-b)^2+(b-c)^2+(c-a)^2>=0`

Vậy dấu "=" xảy ra chỉ có thể là `a=b=c`

`2)(a+b+c)^2=3(ab+bc+ca)`

`<=>a^2+b^2+c^2+2ab+2bc+2ca=3ab+3bc+3ca`

`<=>a^2+b^2+c^2=ab+bc+ca`

`<=>2ab+2bc+2ca=2a^2+2b^2+2c^2`

`<=>a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ca+a^2=0`

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

Mà `(a-b)^2+(b-c)^2+(c-a)^2>=0`

Vậy dấu "=" xảy ra chỉ có thể là `a=b=c`

Vậy nếu `a=b=c` thì ....

1) Ta có: \(3\left(a^2+b^2+c^2\right)=\left(a+b+c\right)^2\)

\(\Leftrightarrow3a^2+3b^2+3c^2-a^2-b^2-c^2-2ab-2bc-2ac=0\)

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

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

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

a) a2+b2+c2 = ab+bc+ca nhân 2 vào cả 2 vế, chuyển tất cả sang vế trái thành 3 HĐT=>đpcm

b) (a+b+c)2 = 3(a2+b2+c2) tách (a+b+c)² thành a²+b²+c²+2ab+2bc+2ac, bỏ ngoặc vế phải, chuyển hết sang vế phaỉ tạo ra 3 HĐT=> dpcm

c) (a+b+c)2 = 3(ab+bc+ca) tách (a+b+c)² thành a²+b²+c²+2ab+2bc+2ac, bỏ ngoặc vế phải, chuyển hết sang vế trái rồi làm như câu a

Hãy nhấn k nếu bạn thấy đây là câu tl đúng :)

a) Ta có: a²+b²+c²=ab+bc+ca

=>2(a²+b²+c²)=2(ab+bc+ca)

<=>2a²+2b²+2c²=2ab+2bc+2ca

<=>2a²+2b²+2c²-2ab-2bc-2ca=0

<=>a²+a²+b²+b²+c²+c²-2ab-2bc=2ca=0

<=>(a^a-2ab+b²)+(b²-2bc+b²)+(a²-2ca+c²)=0

<=>(a-b)²+(b-c)²+(a-c)²=0

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

=>a=b=c

không biết

:) :)

em học lớp 4 nên em không biết

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

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

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

=>a=b=c

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

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

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

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

=>a=b=c