a 2 x + 3 a + 9 = a 2   ⇔   a x a + 3 = a 2 - 9

\(a^2x+3ax+9=a^2\)

\(a^2x+3ax+9-a^2=0\)

\(ax\left(a+3\right)+\left(3-a\right)\left(a+3\right)=0\)

\(\left(a+3\right)\left(ax+3-a\right)=0\)

\(\left(a+3\right)\left[a\left(x-1\right)+3\right]=0\)

\(\Leftrightarrow\orbr{\begin{cases}a+3=0\\a\left(x-1\right)+3=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}a=-3\left(L\right)\\a=\left\{\pm1;3\right\}\left(N\right);a=-3\left(L\right)\end{cases}}\)

Vậy \(a=\left\{\pm1;3\right\}\)

Áp dụng tính chất các dãy tỉ số bằng nhau ta có:

\(\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}=\dfrac{x+y+z}{a+b+c}=\dfrac{x+y+z}{1}\)

\(x=a\left(x+y+z\right)=x^2=a^2.\left(x+y+z\right)^2\)

\(y=b\left(x+y+z\right)=y^2=b^2\left(x+y+z\right)^2\)

\(z=c\left(x+y+z\right)=z^2=c^2.\left(x+y+z\right)^2\)

\(\Rightarrow x^2+y^2+z^2=a^2\left(x+y+z\right)^2+b^2\left(x+y+z\right)^2+c^2\left(x+y+z\right)^2\)

                         \(=\left(x+y+z\right)^2\left(a^2+b^2+c^2\right)=\left(x+y+z\right)^2\) (do \(a^2+b^2+c^2=1\))

lol

Đáp án A

Gọi số khối của X lần lượt là A₁, A₂, A3

Ta có: 

A 1   +   A 2   +   A 3   = 75 A 2 - A 1   = 1 0 , 79 . A 1   +   0 , 1 .   A 2   +   0 , 11   A 3 1 =   24 , 32

⇒ A 1 = 24 A 2 = 25 A 3 = 26