Odpowiedź :
Cześć ;-)
1]
[tex]-7(x-2)-x=-5(x-2)\\\\-7x+14-x=-5x+10\\\\-7x-x+5x=10-14\\\\-3x=-4 \ \ /:(-3)\\\\x=\frac{4}{3}=1\frac{1}{3}[/tex]
2]
[tex]4(3x-2)+5=2(6x-4)+x\\\\12x-8+5=12x-8+x\\\\12x-12x-x=-8+8-5\\\\-x=-5 \ \ /:(-1)\\\\x=5[/tex]
3]
[tex]5x^2-(3+x)(6+5x)=0\\\\5x^2-(18+15x+6x+5x^2)=0\\\\5x^2-18-15x-6x-5x^2=0\\\\5x^2-15x-6x-5x^2=18\\\\-21x=18 \ \ /:(-21)\\\\x=-\frac{18}{21}=-\frac{6}{7}[/tex]
4]
[tex]x-9x+\frac{1}{2}=\frac{1}{5}\\\\x-9x=\frac{1}{5}-\frac{1}{2}\\\\-8x=\frac{2}{10}-\frac{5}{10}\\\\-8x=-\frac{3}{10} \ \ /\cdot(-\frac{1}{8})\\\\x=\frac{3}{80}[/tex]
5]
Pierwszy możliwy zapis
[tex]\frac{x}{3}-\frac{x}{2}=3-\frac{x}{4} \ \ /\cdot12\\\\\frac{12x}{3}-\frac{12x}{2}=36-\frac{12x}{4}\\\\4x-6x=36-3x\\\\4x-6x+3x=36\\\\x=36[/tex]
Drugi możliwy zapis
[tex]\frac{x}{3}-\frac{x}{2}=\frac{3-x}{4} \ \ /\cdot12\\\\\frac{12x}{3}-\frac{12x}{2}=\frac{\not12^3(3-x)}{\not4_1}\\\\4x-6x=3(3-x)\\\\4x-6x=9-3x\\\\4x-6x+3x=9\\\\x=9[/tex]
Pozdrawiam! ~ JulkaOdMatmy
