$\frac{1}{x+2}$+$\frac{1}{(x+2)(4x+7)}$=
$\frac{2x²-x}{x-1}$+$\frac{x+1}{1-x}$+ $\frac{2-x²}{x-1}$=
$\frac{3x+5}{x²-5x}$+ $\frac{25-x}{25-5x}$=
$\frac{5}{2x²y}$+$\frac{3}{5xy²}$+$\frac{x}{y³}$=
$\frac{1}{x+2}$+$\frac{1}{(x+2)(4x+7)}$=
$\frac{2x²-x}{x-1}$+$\frac{x+1}{1-x}$+ $\frac{2-x²}{x-1}$=
$\frac{3x+5}{x²-5x}$+ $\frac{25-x}{25-5x}$=
$\frac{5}{2x²y}$+$\frac{3}{5xy²}$+$\frac{x}{y³}$=
`1/(x+2)+1/((x+2)(4x+7))=(1+4x+7)/((x+2)(4x+7))=(4(x+2))/((x+2)(4x+7))=4/(4x+7)`
`(2x^2-x)/(x-1)+(x+1)/(1-x)+(2-x^2)/(x-1)=(2x^2-x-x-1+2-x^2)/(x-1)=(x^2-2x+1)/(x-1)=(x-1)^2/(x-1)=x-1`
`(3x+5)/(x^2-5x)+(25-x)/(25-5x)=(3x+5)/(x(x-5))+(x-25)/(5(x-5))=(15x+25+x^2-25x)/(5x(x-5))=((x^2-10x+25)/(5x(x-5))=(x-5)^2/(5x(x-5))=(x-5)/(5x)`
`5/(2x^2y ) +3/(5xy^2)+x/(y^3)=(5.5y^2+3.2xy+10x.x^2)/(10x^2y^3)=(25y^2+6xy+10x^3)/(10x^2y^3)`
`1)` `1/{x+2}+1/{(x+2)(4x+7)}`
`\qquad (x\ne -2; x\ne -7/4)`
`={4x+7}/{(x+2)(4x+7)}+1/{(x+2)(4x+7)}`
`={4x+7+1}/{(x+2)(4x+7)}`
`={4x+8}/{(x+2)(4x+7)}`
$\\$
`2)` `{2x^2-x}/{x-1}+{x+1}/{1-x}+{2-x^2}/{x-1}` `(x\ne 1)`
`={2x^2-x}/{x-1}-{x+1}/{x-1}+{2-x^2}/{x-1}`
`={2x^2-x-x-1+2-x^2}/{x-1}`
`={x^2-2x+1}/{x-1}={(x-1)^2}/{x-1}`
`=x-1`
$\\$
`3)` `{3x+5}/{x^2-5x}+{25-x}/{25-5x}` `(x\ne 0;x\ne 5)`
`={3x+5}/{x(x-5)}+{x-25}/{5(x-5)}`
`={5(3x+5)}/{5x(x-5)}+{x(x-25)}/{5x(x-5)}`
`={15x+25+x^2-25x}/{5x(x-5)}`
`={x^2-10x+25}/{5x(x-5)}`
`={(x-5)^2}/{5x(x-5)}`
`={x-5}/{5x}`
$\\$
`4)` `5/{2x^2y}+3/{5xy^2}+x/{y^3}` `(x\ne 0;y\ne 0)`
`={5.5.y^2}/{10x^2y^3}+{3.2.xy}/{10x^2y^3}+{x.10x^2}/{10x^2y^3}`
`={25y^2+6xy+10x^3}/{10x^2y^3}`