Ну если они красные, какими они могут еще быть?
Соsx≥√3\2 -π/6+2πk≤x≤π/6+2πk k∈Z
441.
3х²+9=12х-х²
Решение:
4х²-12х+9=0
D=144-4*4*9=144-144=0
х=12/8=3/2=1,5
Ответ: х=1,5
442.
5х²+1=6х-4х²
Решение:
9х²-6х+1=0
D=36-4*9*1=0
х=6/18=1/3
Ответ: х=1/3
443.
х(х+2)=3
Решение:
х²+2х-3=0
D=4+4*1*3=16
х1=-2+4/2=1
х2=-2-4/2=-3
Ответ: х1=1; х2=-3
444.
х(х+3)=4
Решение:
х²+3х-4=0
D=9+4*1*4=25
х1=-3+5/2=1
х2=-3-5/2=-4
Ответ: х1=1; х2=-4
445.
х(х-5)=-4
Решение:
х²-5х+4=0
D=25-4*1*4=9
х1=5+3/2=4
х2=5-3/2=1
Ответ: х1=4; х2=1
446.
х(х-4)=-3
Решение:
х²-4х+3=0
D=16-4*1*3=4
х1=4+2/2=3
х2=4-2/2=1
Ответ: х1=3; х2=1
447.
х(2х+1)=3х+4
Решение:
2х²+х-3х-4=0
2х²-2х-4=0 |:2
х²-х-2=0
D=1+4*1*2=9
х1=1+3/2=2
х2=1-3/2=-1
Ответ: х1=2; х2=-1
448.
х(2х-3)=4х-3
Решение:
2х²-3х-4х+3=0
2х²-7х+3=0
D=49-4*2*3=25
х1=7+5/4=3
х2=7-5/4=0,5
Ответ: х1=3; х2=0,5
454.
(х-1)(5х+1/2)=0
Решение:
5х²-5х+1/2х-1/2=0
5х²-4,5х-0,5=0 |*2
10х²-9х-1=0
D=81+4*10*1=121
х1=9+11/20=1
х2=9-11/20=-0,1
Ответ: х1=1; х2=-0,1
455.
6(10-х)(3х+4)=0
Решение:
(60-6х)(3х+4)=0
60-6х=0 3х+4=0
-6х=-60 3х=-4
х1=10 х2=-1 целая 1/3
Ответ: х1=10; х2=-1ц 1/3
456.
2(5х-7)(1+х)=0
Решение:
(10х-14)(1+х)=0
10х-14=0 1+х=0
10х=14 х2=-1
х1=1,4
Ответ: х1=1,4; х2=-1
457.
(3х+18)(2-х)=0
Решение:
3х+18=0 2-х=0
3х=-18 х2=2
х1=-6
Ответ: х1=-6; х2=2
458.
(6-х)(5х+40)=0
Решение:
6-х=0 5х+40=0
х1=6 5х=-40
х2=-8
Ответ: х1=6; х2=-8
![1) \lim_{x \to \infty} ( \sqrt{3x+1} - \sqrt{x+5} )= \{\infty-\infty \}= \\ \\ \lim_{x \to \infty} \frac{( \sqrt{3x+1} - \sqrt{x+5})( \sqrt{3x+1} - \sqrt{x+5})}{( \sqrt{3x+1} - \sqrt{x+5})} = \lim_{x \to \infty} \frac{3x+1-x-5}{ \sqrt{3x+1} +\sqrt{x+5}} = \\ \\ =\lim_{x \to \infty} \frac{2x-4}{ \sqrt{3x+1} +\sqrt{x+5}} =\{ \frac{\infty}{\infty} \}= \lim_{x \to \infty} \frac{ \frac{2x}{x} - \frac{4}{x} }{ \sqrt{ \frac{3x+1}{x^2} }+ \sqrt{ \frac{x+5}{x^2} } } =](https://tex.z-dn.net/?f=1%29+%5Clim_%7Bx+%5Cto+%5Cinfty%7D+%28+%5Csqrt%7B3x%2B1%7D+-+%5Csqrt%7Bx%2B5%7D+%29%3D+%5C%7B%5Cinfty-%5Cinfty+%5C%7D%3D++%5C%5C++%5C%5C+%5Clim_%7Bx+%5Cto+%5Cinfty%7D+%5Cfrac%7B%28+%5Csqrt%7B3x%2B1%7D+-+%5Csqrt%7Bx%2B5%7D%29%28+%5Csqrt%7B3x%2B1%7D+-+%5Csqrt%7Bx%2B5%7D%29%7D%7B%28+%5Csqrt%7B3x%2B1%7D+-+%5Csqrt%7Bx%2B5%7D%29%7D+%3D+%5Clim_%7Bx+%5Cto+%5Cinfty%7D++%5Cfrac%7B3x%2B1-x-5%7D%7B+%5Csqrt%7B3x%2B1%7D+%2B%5Csqrt%7Bx%2B5%7D%7D+%3D+%5C%5C+%5C%5C+%3D%5Clim_%7Bx+%5Cto+%5Cinfty%7D++%5Cfrac%7B2x-4%7D%7B+%5Csqrt%7B3x%2B1%7D+%2B%5Csqrt%7Bx%2B5%7D%7D+%3D%5C%7B+%5Cfrac%7B%5Cinfty%7D%7B%5Cinfty%7D++%5C%7D%3D+%5Clim_%7Bx+%5Cto+%5Cinfty%7D++%5Cfrac%7B+%5Cfrac%7B2x%7D%7Bx%7D+-++%5Cfrac%7B4%7D%7Bx%7D+%7D%7B+%5Csqrt%7B+%5Cfrac%7B3x%2B1%7D%7Bx%5E2%7D+%7D%2B+%5Csqrt%7B+%5Cfrac%7Bx%2B5%7D%7Bx%5E2%7D+%7D++%7D+%3D)
![= \lim_{x \to +\infty} \frac{2- \frac{4}{x} }{ \sqrt{ \frac{3}{x}+ \frac{1}{x^2} }+ \sqrt{ \frac{1}{x}+ \frac{5}{x^2} } } = \frac{2- \frac{4}{\infty} }{ \sqrt{ \frac{3}{\infty}+ \frac{1}{\infty^2}} + \sqrt{ \frac{1}{\infty}+ \frac{5}{\infty^2} } } =\\ \\ = \frac{2-0}{ \sqrt{0+0}+ \sqrt{0+0} } = \frac{2}{0} =\infty](https://tex.z-dn.net/?f=%3D++%5Clim_%7Bx+%5Cto+%2B%5Cinfty%7D++%5Cfrac%7B2-++%5Cfrac%7B4%7D%7Bx%7D+%7D%7B+%5Csqrt%7B+%5Cfrac%7B3%7D%7Bx%7D%2B+%5Cfrac%7B1%7D%7Bx%5E2%7D+%7D%2B+%5Csqrt%7B+%5Cfrac%7B1%7D%7Bx%7D%2B+%5Cfrac%7B5%7D%7Bx%5E2%7D++%7D++%7D+%3D++%5Cfrac%7B2-++%5Cfrac%7B4%7D%7B%5Cinfty%7D+%7D%7B+%5Csqrt%7B+%5Cfrac%7B3%7D%7B%5Cinfty%7D%2B+%5Cfrac%7B1%7D%7B%5Cinfty%5E2%7D%7D+%2B+%5Csqrt%7B+%5Cfrac%7B1%7D%7B%5Cinfty%7D%2B+%5Cfrac%7B5%7D%7B%5Cinfty%5E2%7D++%7D++%7D+%3D%5C%5C+%5C%5C+%3D+%5Cfrac%7B2-0%7D%7B+%5Csqrt%7B0%2B0%7D%2B++%5Csqrt%7B0%2B0%7D+++%7D+%3D+%5Cfrac%7B2%7D%7B0%7D+%3D%5Cinfty)
![2) \ \lim_{x \to 0}( \frac{2x+1}{x+1} )^{ \frac{1}{x}} =\{1^\infty \}=\lim_{x \to 0}( \frac{x+x+1}{x+1} )^{ \frac{1}{x}}=\lim_{x \to 0}( 1+ \frac{x}{x+1} )^{ \frac{1}{x}}= \\ \\ =\lim_{x \to 0}( 1+ \frac{x}{x+1} )^{ \frac{x+1}{x}* \frac{x}{x+1} * \frac{1}{x}}= \lim_{x \to 0} (e^{\frac{x}{x+1} * \frac{1}{x}}) =\lim_{x \to 0} (e^{\frac{1}{x+1} }) = \\ \\ = e^{ \lim_{x \to0} ( \frac{1}{x+1})}=e^{ \frac{1}{0+1} } =e^1=e](https://tex.z-dn.net/?f=2%29+%5C++%5Clim_%7Bx+%5Cto+0%7D%28+%5Cfrac%7B2x%2B1%7D%7Bx%2B1%7D+%29%5E%7B+%5Cfrac%7B1%7D%7Bx%7D%7D+%3D%5C%7B1%5E%5Cinfty+%5C%7D%3D%5Clim_%7Bx+%5Cto+0%7D%28+%5Cfrac%7Bx%2Bx%2B1%7D%7Bx%2B1%7D+%29%5E%7B+%5Cfrac%7B1%7D%7Bx%7D%7D%3D%5Clim_%7Bx+%5Cto+0%7D%28++1%2B+%5Cfrac%7Bx%7D%7Bx%2B1%7D+%29%5E%7B+%5Cfrac%7B1%7D%7Bx%7D%7D%3D+%5C%5C+%5C%5C+%3D%5Clim_%7Bx+%5Cto+0%7D%28++1%2B+%5Cfrac%7Bx%7D%7Bx%2B1%7D+%29%5E%7B++%5Cfrac%7Bx%2B1%7D%7Bx%7D%2A+%5Cfrac%7Bx%7D%7Bx%2B1%7D+%2A+%5Cfrac%7B1%7D%7Bx%7D%7D%3D+%5Clim_%7Bx+%5Cto+0%7D+%28e%5E%7B%5Cfrac%7Bx%7D%7Bx%2B1%7D+%2A+%5Cfrac%7B1%7D%7Bx%7D%7D%29+%3D%5Clim_%7Bx+%5Cto+0%7D+%28e%5E%7B%5Cfrac%7B1%7D%7Bx%2B1%7D+%7D%29+%3D+%5C%5C+%5C%5C+%3D+e%5E%7B+%5Clim_%7Bx+%5Cto0%7D+%28+%5Cfrac%7B1%7D%7Bx%2B1%7D%29%7D%3De%5E%7B+%5Cfrac%7B1%7D%7B0%2B1%7D+%7D+%3De%5E1%3De)
![4) \lim_{x \to \frac{ \pi }{2} } (sinx)^{tgx}=\{1^\infty \}= \lim_{x \to \frac{ \pi }{2} }{(1+sinx-1)^{ \frac{1}{sinx -1} *(sinx-1)*tgx} }\\ \\ = \lim_{x \to \frac{ \pi }{2} }(e^{(sinx-1)tgx})=e^{ \lim_{x \to \frac{ \pi }{2} }(sinx-1)tgx](https://tex.z-dn.net/?f=4%29++%5Clim_%7Bx+%5Cto++%5Cfrac%7B+%5Cpi+%7D%7B2%7D+%7D+%28sinx%29%5E%7Btgx%7D%3D%5C%7B1%5E%5Cinfty+%5C%7D%3D+%5Clim_%7Bx+%5Cto++%5Cfrac%7B+%5Cpi+%7D%7B2%7D+%7D%7B%281%2Bsinx-1%29%5E%7B+%5Cfrac%7B1%7D%7Bsinx++-1%7D+%2A%28sinx-1%29%2Atgx%7D+%7D%5C%5C++%5C%5C+%3D+%5Clim_%7Bx+%5Cto++%5Cfrac%7B+%5Cpi+%7D%7B2%7D+%7D%28e%5E%7B%28sinx-1%29tgx%7D%29%3De%5E%7B+%5Clim_%7Bx+%5Cto++%5Cfrac%7B+%5Cpi+%7D%7B2%7D+%7D%28sinx-1%29tgx)
Найдем отдельно предел показателя, произведя замену х→π/2, на
х-π/2 →0:
![\lim_{x \to \frac{ \pi }{2} }(sinx-1)tgx= \left[\begin{array}{c}x- \frac{ \pi }{2} =t \\x= \frac{ \pi }{2}+t&t \to 0 \\ \end{array}\right] = \\ \\ \lim_{t \to0} (sin(\frac{ \pi }{2}+t)-1)tg(\frac{ \pi }{2}+t)= \lim_{t \to 0} (cost-1)*(-ctgt) = \\ \\ =\lim_{t \to 0}- (1-cost)*(-ctgt) =\lim_{t \to 0} (1-cost)* \frac{1}{tgt} =](https://tex.z-dn.net/?f=%5Clim_%7Bx+%5Cto+%5Cfrac%7B+%5Cpi+%7D%7B2%7D+%7D%28sinx-1%29tgx%3D+++%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx-+%5Cfrac%7B+%5Cpi+%7D%7B2%7D+%3Dt+%5C%5Cx%3D+%5Cfrac%7B+%5Cpi+%7D%7B2%7D%2Bt%26t+%5Cto+0+%5C%5C++%5Cend%7Barray%7D%5Cright%5D+%3D+%5C%5C++%5C%5C+++%5Clim_%7Bt+%5Cto0%7D+%28sin%28%5Cfrac%7B+%5Cpi+%7D%7B2%7D%2Bt%29-1%29tg%28%5Cfrac%7B+%5Cpi+%7D%7B2%7D%2Bt%29%3D+%5Clim_%7Bt+%5Cto+0%7D+%28cost-1%29%2A%28-ctgt%29+%3D++%5C%5C++%5C%5C++%3D%5Clim_%7Bt+%5Cto+0%7D-+%281-cost%29%2A%28-ctgt%29+%3D%5Clim_%7Bt+%5Cto+0%7D+%281-cost%29%2A+%5Cfrac%7B1%7D%7Btgt%7D+%3D)
Далее пользуемся таблицей эквивалентности: заменяем
1-cost на t²/2
tgt на t
![= \lim_{t \to 0} \frac{t^2}{2} * \frac{1}{t} =\lim_{t \to 0} \frac{t}{2} = \frac{0}{2} =0](https://tex.z-dn.net/?f=%3D+%5Clim_%7Bt+%5Cto+0%7D++%5Cfrac%7Bt%5E2%7D%7B2%7D+%2A+%5Cfrac%7B1%7D%7Bt%7D+%3D%5Clim_%7Bt+%5Cto+0%7D++%5Cfrac%7Bt%7D%7B2%7D++%3D+%5Cfrac%7B0%7D%7B2%7D+%3D0)
![e^ \lim_{x \to \frac{ \pi }{2} } (sinx-1)tgx}=e^0=1](https://tex.z-dn.net/?f=e%5E+%5Clim_%7Bx+%5Cto++%5Cfrac%7B+%5Cpi+%7D%7B2%7D+%7D+%28sinx-1%29tgx%7D%3De%5E0%3D1)
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