Proszę o uzupełnienie temat 2.... Question from @XPolishManx

January 2019 1 11 Report

Proszę o uzupełnienie temat 2.

ghe 5.
a)
$8x-\frac{5x-4}{7}=2\ / \cdot 7$

$56x-(5x-4)=14$

$56x-5x+4=14$

$56x-5x=14-4$

$51x=10\ /:51$

$x= \frac{10}{51}$

b)
$(4x-7)(25x^2+16)=0$

$4x-7=0 \ \ \ lub \ \ \ 25x^2+16=0$

$4x=7\ /:4 \ \ \ lub \ \ \ 25x^2=-16- brak \ rozwiazan$

$x=1 \frac{3}{4}$

c)
$x^2+8-6x=0$

$x^2-6x+8=0$

$\Delta=(-6)^2-4 \cdot 1 \cdot 8=36-32=4$

$\sqrt{\Delta}= \sqrt{4} =2$

$x_1= \frac{6-2}{2} = \frac{4}{2}=2$

$x_2= \frac{6+2}{2} = \frac{8}{2}=4$

d)
$9x^2-16=0$

$(3x-4)(3x+4)=0$

$3x-4=0 \ \ \ lub \ \ \ 3x+4=0$

$3x=4\ /:3 \ \ \ lub \ \ \ 3x=-4\ /:3$

$x=1 \frac{1}{3} \ \ \ lub \ \ \ x=-1 \frac{1}{3}$

e)
$-15x-x^2=0$

$-x(15+x)=0$

$-x=0\ /:(-1) \ \ \ lub \ \ \ 15+x=0$

$x=0 \ \ \ lub \ \ \ x=-15$

f)
$\begin{cases}5x+2y=9\ / \cdot 5\\ 2x-5y=-8\ / \cdot 2\end{cases}$

$\begin{cases}25x+10y=45\\ 4x-10y=-16\end{cases}$
+_________
$29x=29\ /:29$

$x=1$

$\begin{cases}x=1\\ 2x-5y=-8\end{cases}$

$\begin{cases}x=1\\ 2 \cdot 1-5y=-8\end{cases}$

$\begin{cases}x=1\\ 2-5y=-8\end{cases}$

$\begin{cases}x=1\\ -5y=-8-2\end{cases}$

$\begin{cases}x=1\\ -5y=-10\ /:(-2)\end{cases}$

$\begin{cases}x=1\\ y=-2\end{cases}$
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6.
$7x \le -6\ /:7$

$x \le -\frac{6}{7}$

$x\in \left(- \infty ;- \frac{6}{7}\right\rangle$
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7.
załącznik 1
$A=\left(7;11\right\rangle$

$B=\left\langle 3;9\right\rangle$

$A \cup B=\left\langle 3;11\right\rangle$

$A \cap B=\left(7;9 \right\rangle$

$A \setminus B=\left(9;11\right\rangle$

$B \setminus A=\left\langle 3;7\right\rangle$
==================
8.
załącznik 2

1 votes Thanks 1