Question

Układy równań zadanie daje NAJ pilne!!!

Answer 1

Zadanie 7.1.

Odpowiedź:

FP

Szczegółowe wyjaśnienie:

[tex]\left\{\begin{array}{l}ax-3y=1\\-8x+by=-2\end{array}\right.[/tex]

Zdanie 1.

[tex]\left\{\begin{array}{l}2x-3y=1\ |*4\\-8x+6y=-2\end{array}\right.\\\\\left\{\begin{array}{l}8x-12y=4\\-8x+6y=-2\end{array}\right|+\\\\\left\{\begin{array}{l}-6y=2\ |:(-6)\\-8x+6y=-2\end{array}\right.\\\\\left\{\begin{array}{l}y=-\frac{1}{3}\\-8x+6*(-\frac{1}{3})=-2\end{array}\right.\\\\\left\{\begin{array}{l}y=-\frac{1}{3}\\-8x-2=-2\end{array}\right.\\\\\left\{\begin{array}{l}y=-\frac{1}{3}\\-8x=0\end{array}\right.\\\\\left\{\begin{array}{l}y=-\frac{1}{3}\\x=0\end{array}\right.[/tex]

Układ ma rozwiązanie, więc jest oznaczony.

Odp: FAŁSZ

Zdanie 2.

[tex]\left\{\begin{array}{l}-8x-3y=1|*(-1)\\-8x-3y=-2\end{array}\right.\\\\\left\{\begin{array}{l}8x+3y=-1\\-8x-3y=-2\end{array}\right|+\\\\0=-3[/tex]

Układ jest sprzeczny.

Odp: PRAWDA

Zadanie 7.2.

Odpowiedź:

[tex]\left\{\begin{array}{l}y=0\\x=\frac{1}{4}\end{array}\right.[/tex]

Szczegółowe wyjaśnienie:

[tex]a=\sqrt[3]{-32}*\sqrt[3]{-2}=\sqrt[3]{-32*(-2)}=\sqrt[3]{64}=4\\b=|-4-5|=|-9|=9\\\\\left\{\begin{array}{l}4x-3y=1\ |*2\\-8x+9y=-2\end{array}\right.\\\\\left\{\begin{array}{l}8x-6y=2\\-8x+9y=-2\end{array}\right|+\\\\\left\{\begin{array}{l}3y=0\ |:3\\-8x+9y=-2\end{array}\right.\\\\\left\{\begin{array}{l}y=0\\-8x+9*0=-2\end{array}\right.\\\\\left\{\begin{array}{l}y=0\\-8x=-2\ |:(-8)\end{array}\right.\\\\\left\{\begin{array}{l}y=0\\x=\frac{1}{4}\end{array}\right.[/tex]