\(\Leftrightarrow\left(a-1\right)^2+\left(b-2\right)^2+\left(c-5\right)^2+12=0\)

Khi \(a=1;b=2;c=5\)

Good luck :3

a² + b² + c² + 42=2a +8b +10c

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

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

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

\(\Rightarrow\hept{\begin{cases}a-1=0\\b-4=0\\c-5=0\end{cases}}\Rightarrow\hept{\begin{cases}a=1\\b=4\\c=5\end{cases}}\)

Khi đó \(a+b+c=1+4+5=10\)

cho x<0 thỏa mãn  \(\frac{1}{x^2+9x+20}\)+\(\frac{1}{x^2+11x+30}\)+\(\frac{1}{x^2+13x+42}\)=\(\frac{1}{18}\)  tìm  x=?

mn giải giúp mk với

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

2) \(A=\frac{x^3-27}{x-3}+5x\)

      \(=\frac{\left(x-3\right).\left(x^2+3x+9\right)}{x-3}+5x\)

       \(=x^2+3x+9+5x=x^2+8x+9\)

        \(=\left(x+\text{4}\right)^2-7\ge-7\)

Vậy \(A_{min}=-7\)

4) Số đỉnh của đa giác có tổng các góc trong bằng \(1080^o\)là 8

P/s cn mấy cái kia kh bk =))

số cặp bằng = 0 nha

Không ghi lại đề

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

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

Suy ra: \(\left\{{}\begin{matrix}a=1\\b=4\\c=5\end{matrix}\right.\)

Vậy: \(a+b+c=1+4+5=10\)

Đáp án D

Bài toán trở thành: Tìm M nằm trên đường tròn giao tuyến của mặt cầu  (S) và mặt phẳng (P) sao cho KM lớn nhất

Ta thấy rằng : 3a-7b+5c=30

1.

\(a+b+c=0\)

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

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

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

Ta có:

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

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

\(=\dfrac{5\left(a^2+b^2+c^2\right)+4\left(ab+bc+ca\right)}{5\left(a^2+b^2+c^2\right)-4\left(ab+bc+ca\right)}\)

\(=\dfrac{-10\left(ab+bc+ca\right)+4\left(ab+bc+ca\right)}{-10\left(ab+bc+ca\right)-4\left(ab+bc+ca\right)}\)

\(=\dfrac{-6}{-14}=\dfrac{3}{7}\)

b.

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

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

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

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

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

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

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

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

\(\Rightarrow\dfrac{ab+2bc+3ca}{3a^2+4b^2+5c^2}=\dfrac{a^2+2a^2+3a^2}{3a^2+4a^2+5a^2}=\dfrac{6}{12}=\dfrac{1}{2}\)