2^( 4х - 1 ) + 2^( 4х - 2 ) - 2^( 4х - 3 ) = 160
2^4х•( ( 2^-1 ) + ( 2^( - 2 ) - 2^( - 3 ) ) = 160
2^4х•( ( 1/2 ) + ( 1/4 ) - ( 1/8 )) = 160
1/2 + 1/4 - 1/8 = 4/8 + 2/8 - 1/8 = 5/8
2^4х • ( 5/8 ) = 160
2^4х = 160 : ( 5/8 )
2^4х = 256
2^4х = 2^8
4х = 8
Х = 2
6х2-3х=0
12-3х=0
-3х=0-12
-3х=-12 | : (-3)
х=4
ответ : 4
1.
![(3x^2-19x+20)(2cosx+\sqrt{3})=0](https://tex.z-dn.net/?f=%283x%5E2-19x%2B20%29%282cosx%2B%5Csqrt%7B3%7D%29%3D0)
, ответ:
![x_1=5;x_2=1\frac{1}{3};x_3=\frac{17\pi}{6}](https://tex.z-dn.net/?f=x_1%3D5%3Bx_2%3D1%5Cfrac%7B1%7D%7B3%7D%3Bx_3%3D%5Cfrac%7B17%5Cpi%7D%7B6%7D)
1,1)
![3x^2-19x+20=0\\D=b^2-4ac=(-19)^2-4*3*20=361-240=121\\x=\frac{-bб\sqrt{D}}{2a}=\frac{-(-19)б\sqrt{121}}{2*3}=\frac{19б11}{6}\to\left[\begin{array}{ccc}x_1=\frac{19+11}{6}=5\\x_2=\frac{19-11}{6}=\frac{8}{6}\end{array}\right](https://tex.z-dn.net/?f=3x%5E2-19x%2B20%3D0%5C%5CD%3Db%5E2-4ac%3D%28-19%29%5E2-4%2A3%2A20%3D361-240%3D121%5C%5Cx%3D%5Cfrac%7B-b%D0%B1%5Csqrt%7BD%7D%7D%7B2a%7D%3D%5Cfrac%7B-%28-19%29%D0%B1%5Csqrt%7B121%7D%7D%7B2%2A3%7D%3D%5Cfrac%7B19%D0%B111%7D%7B6%7D%5Cto%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx_1%3D%5Cfrac%7B19%2B11%7D%7B6%7D%3D5%5C%5Cx_2%3D%5Cfrac%7B19-11%7D%7B6%7D%3D%5Cfrac%7B8%7D%7B6%7D%5Cend%7Barray%7D%5Cright)
1,2)
![2cosx+\sqrt{3}=0\\cosx=-\frac{\sqrt{3}}{2}\\x=\frac{5\pi}{6}+2\pi n,n\in Z](https://tex.z-dn.net/?f=2cosx%2B%5Csqrt%7B3%7D%3D0%5C%5Ccosx%3D-%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B2%7D%5C%5Cx%3D%5Cfrac%7B5%5Cpi%7D%7B6%7D%2B2%5Cpi+n%2Cn%5Cin+Z)
на
![[\frac{3\pi}{2};3\pi]](https://tex.z-dn.net/?f=%5B%5Cfrac%7B3%5Cpi%7D%7B2%7D%3B3%5Cpi%5D)
лежит только 1 корень, и он равен
![\frac{5\pi}{6}+2\pi=\frac{17\pi}{6}](https://tex.z-dn.net/?f=%5Cfrac%7B5%5Cpi%7D%7B6%7D%2B2%5Cpi%3D%5Cfrac%7B17%5Cpi%7D%7B6%7D)
2.
![(2-3x-2x^2)(2sinx-\sqrt{3})=0](https://tex.z-dn.net/?f=%282-3x-2x%5E2%29%282sinx-%5Csqrt%7B3%7D%29%3D0)
, ответ:
![x_1=0,5;x_2=-2;x_3=\frac{\pi}{3};x_4=\frac{7\pi}{3}](https://tex.z-dn.net/?f=x_1%3D0%2C5%3Bx_2%3D-2%3Bx_3%3D%5Cfrac%7B%5Cpi%7D%7B3%7D%3Bx_4%3D%5Cfrac%7B7%5Cpi%7D%7B3%7D)
2,1)
![2x^2+3x-2=0\\D=b^2-4ac=3^2-4*2*(-2)=9+16=25\\x=\frac{-bб\sqrt{D}}{2a}=\frac{-3б\sqrt{25}}{2*2}=\frac{-3б5}{4}\to\left[\begin{array}{ccc}x_1=\frac{-3+5}{4}=\frac{2}{4}\\x_2=\frac{-3-5}{4}=-2\end{array}\right](https://tex.z-dn.net/?f=2x%5E2%2B3x-2%3D0%5C%5CD%3Db%5E2-4ac%3D3%5E2-4%2A2%2A%28-2%29%3D9%2B16%3D25%5C%5Cx%3D%5Cfrac%7B-b%D0%B1%5Csqrt%7BD%7D%7D%7B2a%7D%3D%5Cfrac%7B-3%D0%B1%5Csqrt%7B25%7D%7D%7B2%2A2%7D%3D%5Cfrac%7B-3%D0%B15%7D%7B4%7D%5Cto%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx_1%3D%5Cfrac%7B-3%2B5%7D%7B4%7D%3D%5Cfrac%7B2%7D%7B4%7D%5C%5Cx_2%3D%5Cfrac%7B-3-5%7D%7B4%7D%3D-2%5Cend%7Barray%7D%5Cright)
2,2)
![2sinx-\sqrt{3}=0\\sinx=\frac{\sqrt{3}}{2}\\x=\frac{\pi}{3}+2\pi n,n\in Z](https://tex.z-dn.net/?f=2sinx-%5Csqrt%7B3%7D%3D0%5C%5Csinx%3D%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B2%7D%5C%5Cx%3D%5Cfrac%7B%5Cpi%7D%7B3%7D%2B2%5Cpi+n%2Cn%5Cin+Z)
на
![[-\pi;\frac{\pi}{2}]](https://tex.z-dn.net/?f=%5B-%5Cpi%3B%5Cfrac%7B%5Cpi%7D%7B2%7D%5D)
лежат 2 корня, и они равны
![\left[\begin{array}{ccc}x_1=\frac{\pi}{3}\\x_2=\frac{\pi}{3}+2\pi=\frac{7\pi}{3}\end{array}\right](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx_1%3D%5Cfrac%7B%5Cpi%7D%7B3%7D%5C%5Cx_2%3D%5Cfrac%7B%5Cpi%7D%7B3%7D%2B2%5Cpi%3D%5Cfrac%7B7%5Cpi%7D%7B3%7D%5Cend%7Barray%7D%5Cright)
<em>2x^2-10x=0 (Вынесем за скобку 2х)</em>
<em>2x(x-5)=0</em>
<em>x=0</em>
<em>x=5</em>
Ответ: x=0; x=5