знатоки алгебры сюда!
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Ответы на вопрос
Ответил mappku
0
1)
![ax^2+3ax+6=0;\
x_1^2+x_2^2=1;\
D=9a^2-24a;\
x_1=frac{-3a-sqrt{9a^2-24a}}{2a};\
x_2=frac{-3a+sqrt{9a^2-24a}}{2a};\
x_1^2+x_2^2=1=left(frac{-3a-sqrt{9a^2-24a}}{2a}right)^2+left(frac{-3a+sqrt{9a^2-24a}}{2a}right)^2=\
=frac{9a^2+6asqrt{9a^2-24a}+9a^2-24a+9a^2-6asqrt{9a^2-24}+9a^2-24a}{4a^2}=\
=frac{36a^2-48a}{4a^2}=9-frac{12}{a}=1\
9-1=frac{12}{a};\
8=frac{12}{a};\
frac{a}{12}=frac18;\
a=frac{12}{8}=frac32\
a=1,5;\](https://tex.z-dn.net/?f=ax%5E2%2B3ax%2B6%3D0%3B%5C%0Ax_1%5E2%2Bx_2%5E2%3D1%3B%5C%0AD%3D9a%5E2-24a%3B%5C%0Ax_1%3Dfrac%7B-3a-sqrt%7B9a%5E2-24a%7D%7D%7B2a%7D%3B%5C%0Ax_2%3Dfrac%7B-3a%2Bsqrt%7B9a%5E2-24a%7D%7D%7B2a%7D%3B%5C%0Ax_1%5E2%2Bx_2%5E2%3D1%3Dleft%28frac%7B-3a-sqrt%7B9a%5E2-24a%7D%7D%7B2a%7Dright%29%5E2%2Bleft%28frac%7B-3a%2Bsqrt%7B9a%5E2-24a%7D%7D%7B2a%7Dright%29%5E2%3D%5C%0A%3Dfrac%7B9a%5E2%2B6asqrt%7B9a%5E2-24a%7D%2B9a%5E2-24a%2B9a%5E2-6asqrt%7B9a%5E2-24%7D%2B9a%5E2-24a%7D%7B4a%5E2%7D%3D%5C%0A%3Dfrac%7B36a%5E2-48a%7D%7B4a%5E2%7D%3D9-frac%7B12%7D%7Ba%7D%3D1%5C%0A9-1%3Dfrac%7B12%7D%7Ba%7D%3B%5C%0A8%3Dfrac%7B12%7D%7Ba%7D%3B%5C%0Afrac%7Ba%7D%7B12%7D%3Dfrac18%3B%5C%0Aa%3Dfrac%7B12%7D%7B8%7D%3Dfrac32%5C%0Aa%3D1%2C5%3B%5C%0A "ax^2+3ax+6=0;\
x_1^2+x_2^2=1;\
D=9a^2-24a;\
x_1=frac{-3a-sqrt{9a^2-24a}}{2a};\
x_2=frac{-3a+sqrt{9a^2-24a}}{2a};\
x_1^2+x_2^2=1=left(frac{-3a-sqrt{9a^2-24a}}{2a}right)^2+left(frac{-3a+sqrt{9a^2-24a}}{2a}right)^2=\
=frac{9a^2+6asqrt{9a^2-24a}+9a^2-24a+9a^2-6asqrt{9a^2-24}+9a^2-24a}{4a^2}=\
=frac{36a^2-48a}{4a^2}=9-frac{12}{a}=1\
9-1=frac{12}{a};\
8=frac{12}{a};\
frac{a}{12}=frac18;\
a=frac{12}{8}=frac32\
a=1,5;\")
проверим
![frac32x^2+3cdotfrac32x+6=0;\
frac32x^2+frac92x+6=0;\
D=frac{81}{4}-4cdotfrac32cdot6=frac{81}{4}-36=frac{81-144}{4}=-frac{63}{4};\
x_1=frac{-frac92-isqrt{frac{63}{4}}}{2cdotfrac32}=frac{-9-isqrt{63}}{6};\
x_2=frac{-frac92+isqrt{frac{63}{4}}}{2cdotfrac32}=frac{-9+isqrt{63}}{6};\
1=x_1^2+x_2^2=frac{81+18isqrt{63}-63+81-18isqrt{63}-63}{36}=frac{2cdot81-2cdot63}{36}=\
=frac{2(81-63)}{36}=frac{2cdot18}{36}=frac{36}{36}=1](https://tex.z-dn.net/?f=frac32x%5E2%2B3cdotfrac32x%2B6%3D0%3B%5C%0Afrac32x%5E2%2Bfrac92x%2B6%3D0%3B%5C%0AD%3Dfrac%7B81%7D%7B4%7D-4cdotfrac32cdot6%3Dfrac%7B81%7D%7B4%7D-36%3Dfrac%7B81-144%7D%7B4%7D%3D-frac%7B63%7D%7B4%7D%3B%5C%0Ax_1%3Dfrac%7B-frac92-isqrt%7Bfrac%7B63%7D%7B4%7D%7D%7D%7B2cdotfrac32%7D%3Dfrac%7B-9-isqrt%7B63%7D%7D%7B6%7D%3B%5C%0Ax_2%3Dfrac%7B-frac92%2Bisqrt%7Bfrac%7B63%7D%7B4%7D%7D%7D%7B2cdotfrac32%7D%3Dfrac%7B-9%2Bisqrt%7B63%7D%7D%7B6%7D%3B%5C%0A1%3Dx_1%5E2%2Bx_2%5E2%3Dfrac%7B81%2B18isqrt%7B63%7D-63%2B81-18isqrt%7B63%7D-63%7D%7B36%7D%3Dfrac%7B2cdot81-2cdot63%7D%7B36%7D%3D%5C%0A%3Dfrac%7B2%2881-63%29%7D%7B36%7D%3Dfrac%7B2cdot18%7D%7B36%7D%3Dfrac%7B36%7D%7B36%7D%3D1 "frac32x^2+3cdotfrac32x+6=0;\
frac32x^2+frac92x+6=0;\
D=frac{81}{4}-4cdotfrac32cdot6=frac{81}{4}-36=frac{81-144}{4}=-frac{63}{4};\
x_1=frac{-frac92-isqrt{frac{63}{4}}}{2cdotfrac32}=frac{-9-isqrt{63}}{6};\
x_2=frac{-frac92+isqrt{frac{63}{4}}}{2cdotfrac32}=frac{-9+isqrt{63}}{6};\
1=x_1^2+x_2^2=frac{81+18isqrt{63}-63+81-18isqrt{63}-63}{36}=frac{2cdot81-2cdot63}{36}=\
=frac{2(81-63)}{36}=frac{2cdot18}{36}=frac{36}{36}=1")
Значит
2)
![4x^2-7x+1=0;\
(1+x_1)(1+x_2)-?\
teorema Vietta:\
1+x_1+x_2+x_1cdot x_2=1+(x_1+x+2)+x_1cdot x_2\
4(x^2-frac74+frac14)=0;\
x_1+x_2=frac74;\
x_1cdot x_2=frac14;\
1+(x_1+x_2)+x_1cdot x_2=1+frac74+frac14=1+frac84=1+2=3;\
(1+x_1)cdot(1+x_2)=3.](https://tex.z-dn.net/?f=4x%5E2-7x%2B1%3D0%3B%5C%0A%281%2Bx_1%29%281%2Bx_2%29-%3F%5C%0Ateorema+Vietta%3A%5C%0A1%2Bx_1%2Bx_2%2Bx_1cdot+x_2%3D1%2B%28x_1%2Bx%2B2%29%2Bx_1cdot+x_2%5C%0A4%28x%5E2-frac74%2Bfrac14%29%3D0%3B%5C%0Ax_1%2Bx_2%3Dfrac74%3B%5C%0Ax_1cdot+x_2%3Dfrac14%3B%5C%0A1%2B%28x_1%2Bx_2%29%2Bx_1cdot+x_2%3D1%2Bfrac74%2Bfrac14%3D1%2Bfrac84%3D1%2B2%3D3%3B%5C%0A%281%2Bx_1%29cdot%281%2Bx_2%29%3D3. "4x^2-7x+1=0;\
(1+x_1)(1+x_2)-?\
teorema Vietta:\
1+x_1+x_2+x_1cdot x_2=1+(x_1+x+2)+x_1cdot x_2\
4(x^2-frac74+frac14)=0;\
x_1+x_2=frac74;\
x_1cdot x_2=frac14;\
1+(x_1+x_2)+x_1cdot x_2=1+frac74+frac14=1+frac84=1+2=3;\
(1+x_1)cdot(1+x_2)=3.")
проверим
Значит
2)
Ответил mmb1
0
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Ответил Аноним
0
я опубликовала еще задачки)) посмотрите))
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