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$x^{2}-6x$ $+9=0$

$∠$ $F$ From the sum $ot$ $x^{A}\left(2\right)-8$ with $3x-12$ subtract the sum $ofx-9$ with $3x-x^{A}\left(2\right)$ Multinlication takes place $\left(2\times A$ $\left(2\right)+5x+21\right)\left(x-$ $5\right)=$ #Please provide solution by using math formula.

Using the \emph{removal of first derivative} method, the differential equation \( \frac{d^{2}y} $\left(d\times n$ $\left(2\right)\right)+P|ffac\left(dy\right)\left(dx\right)+Qy=F$ $dx\right)+Qy=RN\right)$ is transformed as \). For, the differential equation \frac{d^{2}y} $\left(d^{n}\left(2\right)y\right)$ $dx$ $\left(2\right)+2x$ $\left(0C\left(dy\right)\left(dx\right)+\left(x$ $2+1\right)y=\times n3+3x\right)$ the value of $\left(11\right)$

$x^{2}-6x+9$

$1$ $1.$ $lim _{x→3}\dfrac {x^{2}+3x+2} {x^{2}-6x+9}$

$1.$ The locus of the point which is at a distance $5$ units from $\left(-3,4\right)$ $1s$ $1\right)x^{2}+y^{2}+6x-8y=0$ $2\right)$ $x^{2}+y^{2}-6x+8y=0$ $3\right)$ $x^{2}+y^{2}+6x-8y+25=0$ $4\right)$ $x^{2}+y^{2}$ $-6x+8y+50=0$