<em><u> X^2-16 : x^2-4x</u></em>
<em><u>------------ -------- = (x+4)/x(x+2)</u></em>
<em><u>x^2+4x+4 : x+2</u></em>
(x-4)(x+4) : x(x-4)
------------ -------- =
(x+2)^2 : (x+2)
<u> (x-4)</u>(x+4) <em> (x+2)</em>
------------ * -------- =<span> </span>
<em>(x+2)</em>^2 x<u>(x-4)</u>
<em><u>(x+4)/x(x+2)</u></em>
По неравенству о среднем арифметическом и геометрическом
![a^4+a^3b+a^3b+ab^3+ab^3+b^4 \geq 6a^2b^2\\ \frac{ a^4+a^3b+a^3b+ab^3+ab^3+b^4}{6} \geq a^2b^2\\ \sqrt[6]{a^{12}*b^{12}} \geq a^2b^2\\ a^2b^2=a^2b^2](https://tex.z-dn.net/?f=++a%5E4%2Ba%5E3b%2Ba%5E3b%2Bab%5E3%2Bab%5E3%2Bb%5E4+%5Cgeq+6a%5E2b%5E2%5C%5C%0A+%5Cfrac%7B+a%5E4%2Ba%5E3b%2Ba%5E3b%2Bab%5E3%2Bab%5E3%2Bb%5E4%7D%7B6%7D+%5Cgeq+a%5E2b%5E2%5C%5C%0A++%5Csqrt%5B6%5D%7Ba%5E%7B12%7D%2Ab%5E%7B12%7D%7D+%5Cgeq+a%5E2b%5E2%5C%5C%0A+a%5E2b%5E2%3Da%5E2b%5E2)
Верно
X^2+3x-(1-2x^2)=3x^2-x+4
x^2+3x-1+2x^2-3x^2+x-4=0
4x-5=0
4x=5
x=1,25
По моему так
cos(x-y)=cosx*cosy+sinx*siny cos(x+y)=cosx*cosy-sinx*sinysin(x-y)=sinx*cosy-cosx*siny sin(x+y)=sinx*cosy+cosx*siny<span>.
Ответ: </span>sinx*cosy+cosx*siny
3tg(x/3+pi/6)=3
tg(x/3+pi/6)=3/3
tg(x/3+pi/6)=1
x/3+pi/6=arctg 1 + pi n,n£Z
x/3=-pi/6 +pi/4+pi n,n£Z
x=-pi/3+pi/2+3pi n,n£Z
x= -2pi/6+3pi/6+3pi n ,n£Z
x=pi/6 + 3pi n ,n £Z
