Question

Bài 1: Cho a² + b² + c² = ab + bc + ca. Chứng minh a = b = c

Bài 2: Cho ( a/ b + c) + ( b/ a + c) + ( c/ a + b) = 1. Chứng minh rằng: ( a²/ b + c) + ( b²/ a + c) + ( c²/ a + b) = 0

Answer 1

1) Ta có a² + b² + c² = ab + bc + ca

=> 2a² + 2b² + 2c² = 2ab + 2bc + 2ca

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

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

=> (a - b)² + (b - c)² + (a - c)² = 0

=> \(\hept{\begin{cases}a-b=0\\b-c=0\\a-c=0\end{cases}}\Rightarrow\hept{\begin{cases}a=b\\b=c\\a=c\end{cases}}\Rightarrow a=b=c\left(\text{đpcm}\right)\)

Answer 2

a^2 + b^2 + c^2 = ab + bc + ca

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

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

<=> a-b = 0 và b-c=0 và c-a=0

<=> a=b=c

a^2/b+c + b^2/a+c + c^2=a+b

= a(a/b+c) + b(b/a+c) + c(c/a+b)

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

= a(a+b+c/b+c) - a + b(a+b+c/a+c) - b + c(a+b+c/a+b) - c

= (a+b+c)(a/b+c + b/a+c + c/a+b) - (A+b+c)

mà a/b+c + b/a+c + c/a+b = 1

= a+b+c - (a+b+c)

= 0

Answer 3

Bài 1: 

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

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

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

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

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

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

Answer 4

Bài 1:

Ta có:\(a^2+b^2+c^2=ab+bc+ca\)

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

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

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

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

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