Question

[tex]\frac{c+3}{c^2}-\frac{1}{c}\\ \frac{2}{x}+\frac{3x-2}{x+1}\\ \frac{(a+x)^3}{4a}\cdot\frac{10a(a-x)}{a^2-x^2}\\ \frac{3a^2}{

[tex]<var>\frac{c+3}{c^2}-\frac{1}{c}\\ \frac{2}{x}+\frac{3x-2}{x+1}\\ \frac{(a+x)^3}{4a}\cdot\frac{10a(a-x)}{a^2-x^2}\\ \frac{3a^2}{b}:\frac{b}{a^3}\\ \frac{2a-b}{a}\cdot\frac{(a}{2a-b}+\frac{a}{b})} </var>[/tex

Answer 1

$\frac{c+3}{c^{2}}-\frac{1}{c}=\frac{c+3-c}{c^{2}}=\frac{(c-c)+3}{c^{2}}=\frac{3}{c^{2}}=3c^{-2}$

 

$\frac{2}{x}+\frac{3x-2}{x+1}=\frac{2(x+1)+x(3x-2)}{x(x+1)}=\frac{2x+2+3x^{2}-2x}{x^{2}+x}= \frac{3x^{2}+(2x-2x)+2}{x^{2}+x}=\frac{3x^{2}+2}{x^{2}+x}$

 

$\frac{(a+x)^{3}}{4a}\cdot\frac{10a(a-x)}{a^{2}-x^{2}}=\frac{(a+x)(a+x)^{2}}{4a}\cdot\frac{10a(a-x)}{(a-x)(a+x)}}=\frac{10a(a-x)(a+x)(a+x)^{2}}{4a(a-x)(a+x)}}=\frac{5(a+x)^{2}}{2}=2,5(a+x)^{2}$

 

$\frac{3a^{2}}{b}:\frac{b}{a^{2}}=\frac{3a^{2}}{b}\cdot\frac{a^{2}}{b}=\frac{3a^{4}}{b^{2}}$

 

$\frac{2a-b}{a}\cdot(\frac{a}{2a-b}+\frac{a}{b})=\frac{2a}{b}=2ab^{-1}$

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1) $\frac{a}{2a-b}+\frac{a}{b}=\frac{ab+a(2a-b)}{b(2a-b)}=\frac{ab+2a^{2}-ab}{b(2a-b)}= \frac{2a^{2}+(ab-ab)}{b(2a-b)}=\frac{2a^{2}}{b(2a-b)}$

 

2) $\frac{2a-b}{a}\cdot\frac{2a^{2}}{b(2a-b)}=\frac{2a}{b}=2ab^{-1}$