ta có : \(a^2+b^2+c^2+42=2a+8b+10c\)

\(\Leftrightarrow\) \(a^2+b^2+c^2+42-2a-8b-10c=0\)

\(\Leftrightarrow\) \(\left(a^2-2a+1\right)+\left(b^2-8b+16\right)+\left(c^2-10c+25\right)=0\)

\(\Leftrightarrow\) \(\left(a-1\right)^2+\left(b-4\right)^2+\left(c-5\right)^2=0\)

mà \(\left\{{}\begin{matrix}\left(a-1\right)^2\ge0\forall a\\\left(b-4\right)^2\ge0\forall b\\\left(c-5\right)^2\ge0\forall c\end{matrix}\right.\)

\(\Rightarrow\) \(\left(a-1\right)^2+\left(b-4\right)^2+\left(c-5\right)^2=0\)

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}\left(a-1\right)^2=0\\\left(b-4\right)^2=0\\\left(c-5\right)^2=0\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}a-1=0\\b-4=0\\c-5=0\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}a=1\\b=4\\c=5\end{matrix}\right.\)

khi đó \(a+b+c=1+4+5=10\)

làm ơn trả lời hộ mk với ah mai mk phải nộp bài r

a² +b² +c² +42 = 2a+8b+610c

a² -2a+1 + b²-8b+16 +c² -10c + 24 =0

(a-1)² +(b-4)²+(c-5)²=0

suy ra a= 1 ;b= 4; c= 5

vậy a+b+c = 10

\(a^2+2a+b^2+4b+4c^2-4c+6=0\)

\(\Leftrightarrow\left(a^2+2a+1\right)+\left(b^2+4b+4\right)+\left(4c^2-4c+1\right)=0\)

\(\Leftrightarrow\left(a+1\right)^2+\left(b+2\right)^2+\left(2c-1\right)^2=0\)

Mà \(\begin{cases}\left(a+1\right)^2\ge0\\\left(b+2\right)^2\ge0\\\left(2c-1\right)^2\ge0\end{cases}\)

\(\Rightarrow\left(a+1\right)^2+\left(b+2\right)^2+\left(2c-1\right)^2\ge0\)

\(\Rightarrow\begin{cases}a+1=0\\b+2=0\\2c-1=0\end{cases}\)\(\Rightarrow\begin{cases}a=-1\\b=-2\\c=\frac{1}{2}\end{cases}\)

Câu hỏi của Phạm Thị Thùy Linh - Toán lớp 8 - Học toán với OnlineMath