![\frac{a^2b^2}{a^2+2ab+b^2}: \frac{ab}{a+b}= \frac{(ab)^2}{(a+b)^2}* \frac{a+b}{ab}= \frac{ab}{a+b}](https://tex.z-dn.net/?f=+%5Cfrac%7Ba%5E2b%5E2%7D%7Ba%5E2%2B2ab%2Bb%5E2%7D%3A+%5Cfrac%7Bab%7D%7Ba%2Bb%7D%3D+%5Cfrac%7B%28ab%29%5E2%7D%7B%28a%2Bb%29%5E2%7D%2A+%5Cfrac%7Ba%2Bb%7D%7Bab%7D%3D+%5Cfrac%7Bab%7D%7Ba%2Bb%7D+)
<span>если а=4-√3, b=4+√3
</span>
![\frac{(4- \sqrt{3}) *(4+ \sqrt{3}) }{(4- \sqrt{3} )+(4+ \sqrt{3} )} = \frac{16-3}{8} = \frac{13}{8}=1,625](https://tex.z-dn.net/?f=%5Cfrac%7B%284-+%5Csqrt%7B3%7D%29+%2A%284%2B+%5Csqrt%7B3%7D%29+%7D%7B%284-+%5Csqrt%7B3%7D+%29%2B%284%2B+%5Csqrt%7B3%7D+%29%7D+%3D+%5Cfrac%7B16-3%7D%7B8%7D+%3D+%5Cfrac%7B13%7D%7B8%7D%3D1%2C625+)
Х²+у²-8х+2у+17=0
(х²-8х+16)+(у²+2у+1)=0
(х-4)²+(у+1)²=0
(х-4)²>0;(у+1)²>0
х-4=0;х=4
у+1=0;у=-1
ответ х=4;у=-1
<span>(p−q)²=p²-2pq+q²...................</span>
F(x)=x^3-3x^2+5x+3
F'=3x^2-6x+5
F'(-1)=3(-1)^2-6(-1)+5=3+6+5=14