Rozwiązane

Proszę o pomoc w tych zadaniach:(



Proszę O Pomoc W Tych Zadaniach class=

Odpowiedź :

Krysia

[tex]1)\\\\\frac{x+6}{x+3}=0\\\\mianownik\ musi\ byc\ rozny\ od\ zera\\\\x+3\neq 0\\\\x\neq -3\\\\D=R\setminus \left\{ -3\right\}\\\\x+6=0\\\\x=-6[/tex]

[tex]2)\\\\\frac{6}{x}=12\\\\x\neq 0\\\\D=R\setminus \left\{ 0\right\}\\\\ 12x=6\ \ |:12\\\\x=\frac{6}{12}=\frac{1}{2}[/tex]

[tex]3)\\\\ \frac{3x-2}{x-1} =4 \\\\x-1\neq 0\\\\x\neq 1\\\\D=R\setminus \left\{ 1\right\}\\\\ 4(x-1)=3x-2\\\\4x-4=3x-2\\\\4x-3x=-2+4\\\\x=2[/tex]

[tex]4)\\\\ \frac{3 }{2x-3} =2x-5 \\\\2x-3\neq 0\\\\2x\neq 3\ \ |:2\\\\x\neq \frac{3}{2}\\\\D=R\setminus \left\{ \frac{3}{2}\right\}\\\\(2x-3)(2x-5)=3\\\\4x^2-10x-6x+15-3=0\\\\4x^2-16x+12=0\\\\4(x^2-4x+3)=0\\\\x^2-4x+3=0\\\\\Delta =b^2-4ac=(-4)^2-4*1*3=16-12=4\\\\\sqrt{\Delta } =\sqrt{4}=2[/tex]

[tex]x_{1}=\frac{-b-\sqrt{\Delta }}{2a}=\frac{4-2}{2*1}=\frac{2}{2}=1\\\\ x_{2}=\frac{-b+\sqrt{\Delta }}{2a}=\frac{4+2}{2*1}=\frac{6}{2}=3 \\\\x\in\left\{ 1,3\right\}[/tex]

[tex]5)\\\\\frac{ (x+3)(x-2) }{(x-3)(x-4)}=0 \\\\x-3\neq 0\ \ i\ \ x-4\neq 0\\\\x\neq 3\ \ i\ \ x\neq 4\\\\D=R\setminus \left\{ 3,4\right\}[/tex]

[tex](x+3)(x-2)=0\\\\x+3=0\ \ lub\ \ x-2=0\\\\x=-3\ \ lub\ \ x=2\\\\x\in\left\{ -3,2\right\}[/tex]