Question

rezolvare completa,daca nu se vede tot macar din ce se vede​

Answer 1

Problemele sunt pentru clasa a 7-a, exceptând a doua problemă.

$\it \left.\begin{aligned} \it x+y=s\\ \\ \it x-y=d\end{aligned}\right\}\ \Rightarrow x+y+x-y=s+d\ \Rightarrow 2x=s+d\ \Rightarrow x=\dfrac{s+d}{2}$

Answer 2

Explicație pas cu pas:

$a = 2( \sqrt{3} + 3 \sqrt{2}) - 4(5 \sqrt{2} + 3 \sqrt{3}) \\ = 2 \sqrt{3} + 6 \sqrt{2} - 20 \sqrt{2} - 12 \sqrt{3} \\ = - 14 \sqrt{2} - 10 \sqrt{3}$

$b = 7 \sqrt{2}(2 + 3 \sqrt{2}) - 8 \sqrt{3}(3 \sqrt{3} - 2) \\ = 14 \sqrt{2} + 42 - 72 + 16 \sqrt{3} \\ = 14 \sqrt{2} + 16 \sqrt{3} - 30$

$m_{a} = \frac{a + b}{2} = \frac{- 14 \sqrt{2} - 10 \sqrt{3} + 14 \sqrt{2} + 16 \sqrt{3} - 30}{2} \\ = \frac{6 \sqrt{3} - 30}{2} = 3\sqrt{3} - 15$

.

$2x - 3 = 3x - 2$

$2( - 1) - 3 = 3( - 1) - 2 \\ - 2 - 3 = - 3 - 2 \\ - 5 = - 5 \: adevarat$

$\frac{x + 2}{3} + 0.(6) = \frac{2 - x}{5} - \frac{x - 3}{10}$

$\frac{ - 1 + 2}{3} + \frac{6}{9} = \frac{2 - ( - 1)}{5} - \frac{ - 1 - 3}{10} \\ \frac{1}{3} + \frac{2}{3} = \frac{3}{5} - \frac{ - 4}{10} \iff \frac{3}{3} = \frac{3}{5} + \frac{2}{5} \\ \iff 1 = 1 \: adevarat$

$\frac{x + 2}{3} + \frac{2}{3} = \frac{2 - x}{5} - \frac{x - 3}{10} \\ \frac{x + 2 + 2}{3} = \frac{2(2 - x) - (x - 3)}{10} \\ \frac{x + 4}{3} = \frac{4 - 2x - x + 3}{10} \\ \frac{x + 4}{3} = \frac{7 - 3x}{10} \\ 10(x + 4) = 3(7 - 3x)$

=> cele două ecuații nu sunt echivalente