P(x²+1) и P(3x²-2)
252.в)
P(x)=3x²+2x
P(x²+1)=3(x²+1)²+2(x²+1)=3(x⁴+2x²+1)+2x²+2=3x⁴+6x²+3+2x²+1=3x⁴+8x²+4
P(3x²-2)=3(3x²-2)²+2(3x²-2)=3(9x⁴-12x²+4)+6x²-4= 27x⁴-36x²+12+6x²-4= 27x⁴-30x²+8
252.г)
P(x)=4x-5x²
P(x²+1)=4(x²+1)-5(x²+1)²=4x²+4-5(x⁴+2x+1)=4x²+4-5x⁴-10x-5=-5x⁴-6x²-1
P(3x²-2)=4(3x²-2)-5(3x²-2)²=12x²-8-5(9x⁴-12x²+4)=12x²-8-45x⁴+60x²-20= -45x⁴+72x²-28
253.в)
P(x)=2x²+3x
P(x²+1)= 2(x²+1)²+3(x²+1)=2(x⁴+2x²+1)+3x²+3= 2x⁴+4x²+2+3x²+3= 2x⁴+7x²+5
P(3x²-2)=2(3x²-2)²+3(3x²-2)=2(9x⁴-12x²+4)+9x²-6=18x⁴-24x²+8+9x²-6 = 18x⁴-14x² -2
253.г)
P(x)=5x-2x²
P(x²+1)=5(x²+1)-2(x²+1)²=5x²+5-2(x⁴+2x²+1)=5x²+5-2x⁴-4x²-2=-2x⁴+x²+3
P(3x²-2)=5(3x²-2)-2(3x²-2)²=15x²-10-2(9x⁴-12x²+4)= 15x²-10-18x⁴+24x²-8=-18x⁴+39x²-18
256 в) P(x)=4x²+3x-2 y=2x+1
2x=y-1
x=(y-1)/2
R(y)=4((y-1)/2)²+3(y-1)/2-2=(y-1)²+1,5(y-1)-2=y²-2y+1+1,5y-1,5-2=y²-0,5y-2,5
256 г) P(x)=-5x²-2x+6 y=2-3x
-3x=y-2
x=(2-y)/3
![R(y)=-5( \frac{2-y}{3} )^2-2\frac{2-y}{3}+6=- \frac{5}{9} ( y^2-4y+4)+ \frac{2}{3}y -\frac{4}{3}+6= \\ =-\frac{5}{9} y^2+\frac{20}{9}y-\frac{20}{9}+ \frac{2}{3}y -\frac{4}{3}+6=-\frac{5}{9} y^2+\frac{26}{9}y +\frac{22}{9} = \\ -\frac{5}{9} y^2+2\frac{8}{9}y +2\frac{4}{9}](https://tex.z-dn.net/?f=R%28y%29%3D-5%28+%5Cfrac%7B2-y%7D%7B3%7D+%29%5E2-2%5Cfrac%7B2-y%7D%7B3%7D%2B6%3D-+%5Cfrac%7B5%7D%7B9%7D+%28+y%5E2-4y%2B4%29%2B+%5Cfrac%7B2%7D%7B3%7Dy+-%5Cfrac%7B4%7D%7B3%7D%2B6%3D++%5C%5C+%3D-%5Cfrac%7B5%7D%7B9%7D+y%5E2%2B%5Cfrac%7B20%7D%7B9%7Dy-%5Cfrac%7B20%7D%7B9%7D%2B+%5Cfrac%7B2%7D%7B3%7Dy+-%5Cfrac%7B4%7D%7B3%7D%2B6%3D-%5Cfrac%7B5%7D%7B9%7D+y%5E2%2B%5Cfrac%7B26%7D%7B9%7Dy+%2B%5Cfrac%7B22%7D%7B9%7D+%3D++%5C%5C+-%5Cfrac%7B5%7D%7B9%7D+y%5E2%2B2%5Cfrac%7B8%7D%7B9%7Dy+%2B2%5Cfrac%7B4%7D%7B9%7D)
Объяснение:
Используйте распределительный закон:
(a + b) (c + d) = ac + cd + bc + bd
К примеру первый пример:
(a + 2) (b - 3) = ab -3b + 2b - 6 = ab - b - 6
Выражение (ab - b - 6) и есть стандартный вид.
![(2^{12}+5^3)\div21\\ 2^{12}=1024\cdot4=4096;\\ 5^3=125;\\ 4096+125=4221=21\cdot201](https://tex.z-dn.net/?f=%282%5E%7B12%7D%2B5%5E3%29%5Cdiv21%5C%5C%0A2%5E%7B12%7D%3D1024%5Ccdot4%3D4096%3B%5C%5C%0A5%5E3%3D125%3B%5C%5C%0A4096%2B125%3D4221%3D21%5Ccdot201)
видно, что оно справедливо, теперь докажем это
![2^{12}+5^3=\left(2^4\right)^3+5^3=16^3+5^3=\\ \left|\begin{array}{c}a^3+b^3=(a+b)(a^2-ab+b^2)\\a^3-a^2b+ab^2+a^2b-ab^2+b^3=a^3+b^3\end{array}\right|\\ =(16+5)(16^2-16\cdot5+5^2)=21\cdot(256-80+25)=21\cdot201](https://tex.z-dn.net/?f=2%5E%7B12%7D%2B5%5E3%3D%5Cleft%282%5E4%5Cright%29%5E3%2B5%5E3%3D16%5E3%2B5%5E3%3D%5C%5C%0A++%5Cleft%7C%5Cbegin%7Barray%7D%7Bc%7Da%5E3%2Bb%5E3%3D%28a%2Bb%29%28a%5E2-ab%2Bb%5E2%29%5C%5Ca%5E3-a%5E2b%2Bab%5E2%2Ba%5E2b-ab%5E2%2Bb%5E3%3Da%5E3%2Bb%5E3%5Cend%7Barray%7D%5Cright%7C%5C%5C%0A%3D%2816%2B5%29%2816%5E2-16%5Ccdot5%2B5%5E2%29%3D21%5Ccdot%28256-80%2B25%29%3D21%5Ccdot201)
поскольку один из множителей делится на 21, то тогда и само всё выражение делится на него
Доказано, то-есть
![2^{12}+ 5^{3}](https://tex.z-dn.net/?f=2%5E%7B12%7D%2B+5%5E%7B3%7D++)
делится на 21 нацело!
<span>lg²x+lgx²=3
</span><span>lg²x + 2lgx - 3 = 0
lgx = z
z</span>² + 2z - 3 = 0
z₁ = - 3
z₂ = 1
<span>1) lgx = - 3
x = 10^(-3)
x</span>₁ = 0,001
2) lgx = 1
x₂ = 10
Ответ: b) <span>0,001;10</span>
(3x² -7x+2)/(2-6x) =3(x-1/3)(x-2)/(-6(x-1/3)) = (2-x)/2 . * * * x≠1/3 * * *
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(x+40)/(x³-16x) :((x-4)/(3x² +11x-4) - 16/(16-x²)) =
(x+40)/x(x²-16) :( (x-4)/3(x+4)(x-1/3) +16/(x² -16))=
(x+40)/x(x²-16) :( (x-4)/(x+4)(3x-1) +16/(x² -16))=
(x+40)/x(x²-16) :( ((x-4)² +16(3x-1)) /(x²-16)(3x-1))=
(x+40)/x(x²-16) :( (x²-8x+16 +48x-16) / (x²-16)(3x-1)) =
x+40)/x(x²-16) :( (x² +40x) / (x²-16)(3x-1))
(x+40)/x(x²-16) *(x²-16)(3x-1)/(x²+40) = (3x-1)/x * * * x≠<span>±4 * * *</span>