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

Khi \(a=1;b=2;c=5\)

Good luck :3

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 \(a+b+c=1+4+5=10\)

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: \(a+b+c=1+4+5=10\)

Ta có BĐT \(a+b+c\le\sqrt{3\left(a^2+b^2+c^2\right)}=3\)

Nên BĐT cần chứng minh là 

\(\frac{a^2}{a+b^2}+\frac{b^2}{b+c^2}+\frac{c^2}{c+a^2}\ge\frac{3}{2}\)

Đặt \(\hept{\begin{cases}a^2=x\\b^2=y\\c^2=z\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x+y+z=3\\x,y,z>0\end{cases}}\)

Áp dụng BĐT AM-GM and Cauchy-Schwarz ta có:

\(Σ\frac{a^2}{a+b^2}=Σ\frac{x}{\sqrt{x}+y}=Σ\frac{x}{\sqrt{\frac{x\left(x+y+z\right)}{3}+y}}\)

\(=Σ\frac{6x}{2\sqrt{3x\left(x+y+z\right)}+6y}\geΣ\frac{6x}{3x+x+y+z+6y}=Σ\frac{6x}{4x+7y+z}\)

\(=Σ\frac{6x^2}{4x^2+7xy+xz}\ge\frac{6\left(x+y+z\right)^2}{Σ\left(4x^2+7xy+xz\right)}=\frac{3}{2}\)

-Nguồn : Xem câu hỏi

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.

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

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

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

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

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

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

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

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

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