Cześć ;-)
Obliczenia do a]
[tex]\dfrac{x-1}{4}+\dfrac{2x-3}{4}=\dfrac{x-1+2x-3}{4}=\dfrac{3x-4}{4}[/tex]
Obliczenia do b]
[tex]\dfrac{2(x-3)}{5}-\dfrac{3(x+2)}{5}=\dfrac{2x-6}{5}-\dfrac{3x+6}{5}=\dfrac{2x-6-(3x+6)}{5}=\\\\\\=\dfrac{2x-6-3x-6}{5}=\dfrac{-x-12}{5}[/tex]
Obliczenia do c]
[tex]c] \ \dfrac{(x-1)(x+1)}{3}+\dfrac{(2x-3)^2}{3}=\dfrac{x^2-1^2}{3}+\dfrac{(2x)^2-2\cdot2x\cdot3+3^2}{3}=\\\\\\=\dfrac{x^2-1}{3}+\dfrac{4x^2-12x+9}{3}=\dfrac{x^2-1+4x^2-12x+9}{3}=\dfrac{5x^2-12x+8}{3}[/tex]
Obliczenia do d]
[tex]\dfrac{(2x-1)(2x+1)}{2}-\dfrac{(3x+2)^2}{2}=\dfrac{(2x)^2-1^1}{2}-\dfrac{(3x)^2+2\cdot3x\cdot2+2^2}{2}=\\\\\\=\dfrac{4x^2-1}{2}-\dfrac{9x^2+12x+4}{2}=\dfrac{4x^2-1-9x^2-12x-4}{2}=\dfrac{-5x^2-12x-5}{2}[/tex]
Wykorzystane wzory skróconego mnożenia
[tex](a+b)^2=a^2+2ab+b^2[/tex]
[tex](a-b)^2=a^2-2ab+b^2[/tex]
[tex](a-b)(a+b)=a^2-b^2[/tex]
Pozdrawiam! ~ JulkaOdMatmy
