Question

Determinați numărul real x pentru care B(1) + b(2)+b(3)+... +b(9) = 9 b(x)
B(x) = (2x+1. X
1. X+1) ​

Answer 1

Răspuns:

x=5

Explicație pas cu pas:

$B(x)=\left[\begin{array}{cc}2x+1&x\\1&x+1\end{array}\right]$

$B(1)+B(2)+...+B(9)=9B(x)$

$\left[\begin{array}{cc}2\cdot1+1&1\\1&1+1\end{array}\right] + \left[\begin{array}{cc}2\cdot2+1&2\\1&2+1\end{array}\right]+...+\left[\begin{array}{cc}2\cdot9+1&9\\1&9+1\end{array}\right]=9\cdot \left[\begin{array}{cc}2\cdot x+1&x\\1&x+1\end{array}\right]$

$\left[\begin{array}{cc}(2\cdot 1 + 1)+(2\cdot 2 + 1)+...+(2\cdot 9 + 1)&1+2+...+9\\1+1+1+1+1+1+1+1+1&(1+1)+(2+1)+...+(9+1)\end{array}\right]=9\cdot \left[\begin{array}{cc}2\cdot x+1&x\\1&x+1\end{array}\right]$

$\left[\begin{array}{cc}2(1+2+..+9)+9&1+2+..+9\\9&(1+2+..+9)+9 \end{array}\right]=9\cdot \left[\begin{array}{cc}2\cdot x+1&x\\1&x+1\end{array}\right]$

$\left[\begin{array}{cc}2\cdot \frac{9\cdot 10}{2}+9&\frac{9\cdot10}{2}\\ 9&\frac{9\cdot10}{2}+9\end{array}\right]=9\cdot \left[\begin{array}{cc}2\cdot x+1&x\\1&x+1\end{array}\right]$

$\left[\begin{array}{cc}9(2\cdot\frac{10}{2}+1)&9\cdot \frac{10}{2}\\ 9\cdot 1&9(\frac{10}{2}+1)\end{array}\right]=9\cdot \left[\begin{array}{cc}2\cdot x+1&x\\1&x+1\end{array}\right]$

$9\cdot \left[\begin{array}{cc}2\cdot 5+1&5\\ 1&5+1\end{array}\right]=9\cdot \left[\begin{array}{cc}2\cdot x+1&x\\1&x+1\end{array}\right]$

$\Rightarrow \boxed{x=5}$