Question

Przedstaw w postaci jak najprostszej sumy algebraicznej: (zdj)

Answer 1

[tex]a)\ \ (x+5)^2=x^2+2x\cdot5+5^2=x^2+10x+25\\\\\\b)\ \ (3a-\frac{b}{2})^2=(3a)^2-2\cdot3a\cdot\frac{b}{2}+(\frac{b}{2})^2=9a^2-\not6^3a\cdot\frac{b}{\not2_{1}}+\frac{b^2}{4}=9a^2-3ab+\frac{b^2}{4}\\\\\\c)\ \ (-a+\sqrt{2}b)^2=(-a)^2+2\cdot(-a)\cdot\sqrt{2}b+(\sqrt{2}b)^2=a^2-2\sqrt{2}ab+2b^2\\\\\\d)\ \ (-2y-\sqrt{x})^2=(-2y)^2-2\cdot(-2y)\cdot\sqrt{x}+(\sqrt{x})^2=4y^2+4y\sqrt{x}+x[/tex]

[tex]e)\ \ (-\frac{1}{2}a-\sqrt{3}b)^2=(-\frac{1}{2}a)^2-\not2\cdot(-\frac{1}{\not2}a)\cdot\sqrt{3}b+(\sqrt{3}b)^2=\frac{1}{4}a^2+\sqrt{3}ab+3b^2\\\\\\f)\ \ (-\frac{2}{3a}+\frac{2a}{3})(\frac{2}{3a}+\frac{2a}{3})=(\frac{2a}{3}-\frac{2}{3a})(\frac{2a}{3}+\frac{2}{3a})=(\frac{2a}{3})^2-(\frac{2}{3a})^2=\frac{4a^2}{9}-\frac{4}{9a^2}[/tex]

[tex]g)\ \ 5(x-1)(x+2)-5(x+1)(x-2)=5(x^2+2x-x-2)-5(x^2-2x+x-2)=\\\\=5(x^2+x-2)-5(x^2-x-2)=5x^2+5x-10-5x^2+5x+10=5x+5x=10x\\\\\\h)\ \ 4(x-1)(x+1)-(2(x+1))^2=4(x^2-1)-(2x+2)^2=\\\\=4x^2-4-(4x^2+8x+4)=4x^2-4-4x^2-8x-4=-8-8x=-8x-8[/tex]

Zastosowane wzory skróconego mnożenia

[tex](a+b)^2=a^2-2ab+b^2\\\\(a-b)^2=a^2-2ab+b^2\\\\(a-b)(a+b)=a^2-b^2[/tex]