<span>12x*2-(3x-4)(4x+1)=19
24х - (12х^2 + 3x - 16x - 4) = 19
24x - 12x^2 - 3x + 16x + 4x = 19
41x - 12x^2 = 19
123 - 108 = 19
х = 3
</span>
Продолжение:
x(x-2)/(x-2)(x+2)=0
x/(x+2=0
x=0
<span>a=(2;3) b=(1;8) c=(4;2)</span>
<span>1) a+b=(2+1;3+8)=(3;11)</span>
<span> Ia+bI=sqrt(3^2+11^2)=sqrt130</span>
<span>2) 2a-b=(2*2-1;2*3-8)=(3;-2)</span>
<span> I2a-bI=sqrt(3^2+(-2)^2)=sqrt13</span>
<span>3) a+2b-3c=(2+2-12;3+16-6)=(-8;13)</span>
<span> Ia+2b-3cI=sqrt(64+169)=sqrt233</span>
<span>4) 4a=(8;12)</span>
<span> I4aI=sqrt(64+144)=sqrt208</span>
<span>5) 3a-8b=(-2;-55)</span>
<span> I3a-8bI=sqrt(4+3025)=sqrt3029</span>
Ответ:
<em><u>0</u></em><em><u>.</u></em><em><u>4</u></em><em><u> </u></em><em><u>и</u></em><em><u> </u></em><em><u>1</u></em><em><u>1</u></em>
Объяснение:
![(5x - 2)(11 - x) = 0 \\ 5x - 2 = 0 \\ 5x = 2 \\ x = 0.4 \\ \\ 11 - x = 0 \\ x = 11](https://tex.z-dn.net/?f=%285x%20-%202%29%2811%20-%20x%29%20%3D%200%20%5C%5C%205x%20-%202%20%3D%200%20%5C%5C%205x%20%3D%202%20%5C%5C%20x%20%3D%200.4%20%5C%5C%20%20%5C%5C%2011%20-%20x%20%3D%200%20%5C%5C%20x%20%3D%2011)
![x=\sqrt{2005}-1\\ ](https://tex.z-dn.net/?f=x%3D%5Csqrt%7B2005%7D-1%5C%5C%0A)
если один корень такой , то второй будет сопряженным
![x_{2}=-\sqrt{2005}-1\\ (x-\sqrt{2005}+1)(x+\sqrt{2005}+1)=\\ x^2+x\sqrt{2005}+x-x\sqrt{2005}-2005-\sqrt{2005}+x+\sqrt{2005}+1=x^2+2x-2004](https://tex.z-dn.net/?f=x_%7B2%7D%3D-%5Csqrt%7B2005%7D-1%5C%5C%0A%28x-%5Csqrt%7B2005%7D%2B1%29%28x%2B%5Csqrt%7B2005%7D%2B1%29%3D%5C%5C%0Ax%5E2%2Bx%5Csqrt%7B2005%7D%2Bx-x%5Csqrt%7B2005%7D-2005-%5Csqrt%7B2005%7D%2Bx%2B%5Csqrt%7B2005%7D%2B1%3Dx%5E2%2B2x-2004)
Ответ да и он равен
![x^2+2x-2004](https://tex.z-dn.net/?f=x%5E2%2B2x-2004)