Приращение функции, называется следующее:
![\Delta y=f(x+\Delta x)-f(x)](https://tex.z-dn.net/?f=%5CDelta+y%3Df%28x%2B%5CDelta+x%29-f%28x%29)
В нашем случае:
![\Delta y=2(x+\Delta x)^2-2x^2](https://tex.z-dn.net/?f=%5CDelta+y%3D2%28x%2B%5CDelta+x%29%5E2-2x%5E2)
![\Delta y=2x^2+4x\Delta x+2\Delta x^2-2x^2](https://tex.z-dn.net/?f=%5CDelta+y%3D2x%5E2%2B4x%5CDelta+x%2B2%5CDelta+x%5E2-2x%5E2)
![\Delta y=4x\Delta x+2\Delta x^2=\Delta x(4x+2\Delta x)](https://tex.z-dn.net/?f=%5CDelta+y%3D4x%5CDelta+x%2B2%5CDelta+x%5E2%3D%5CDelta+x%284x%2B2%5CDelta+x%29)
Давайте проверим, через производную функции:
![\lim_{\Delta x \to 0} \frac{\Delta x(4x+2\Delta x)}{\Delta x}=4x+2\Delta x=4x](https://tex.z-dn.net/?f=+%5Clim_%7B%5CDelta+x+%5Cto+0%7D++%5Cfrac%7B%5CDelta+x%284x%2B2%5CDelta+x%29%7D%7B%5CDelta+x%7D%3D4x%2B2%5CDelta+x%3D4x++)
Действительно.
А значит, приращение функции равно:
![\Delta y=2\Delta x(2x+\Delta x)=4x\Delta x+2\Delta x^2](https://tex.z-dn.net/?f=%5CDelta+y%3D2%5CDelta+x%282x%2B%5CDelta+x%29%3D4x%5CDelta+x%2B2%5CDelta+x%5E2)
Формулы
![\dfrac D4=\Big(\dfrac b2\Big)^2-ac;~~~~x_{1,2}=\dfrac{-\frac b2\pm \sqrt {\frac D4}}a](https://tex.z-dn.net/?f=%5Cdfrac+D4%3D%5CBig%28%5Cdfrac+b2%5CBig%29%5E2-ac%3B~~~~x_%7B1%2C2%7D%3D%5Cdfrac%7B-%5Cfrac+b2%5Cpm+%5Csqrt+%7B%5Cfrac+D4%7D%7Da)
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1. 3x² - 4x + 1 = 0
![\dfrac D4=2^2-3=1;~~~~x_{1,2}=\dfrac{2\pm 1}3\\\\ \boxed{x_1=1;~~x_2=\dfrac 13}](https://tex.z-dn.net/?f=%5Cdfrac+D4%3D2%5E2-3%3D1%3B~~~~x_%7B1%2C2%7D%3D%5Cdfrac%7B2%5Cpm+1%7D3%5C%5C%5C%5C+%5Cboxed%7Bx_1%3D1%3B~~x_2%3D%5Cdfrac+13%7D)
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2. 5x² + 14x - 3 = 0
![\dfrac D4=7^2-5\cdot(-3)=64=8^2;~~~~x_{1,2}=\dfrac{-7\pm 8}5\\\\ \boxed{x_1=\dfrac 15;~~x_2=-3}](https://tex.z-dn.net/?f=%5Cdfrac+D4%3D7%5E2-5%5Ccdot%28-3%29%3D64%3D8%5E2%3B~~~~x_%7B1%2C2%7D%3D%5Cdfrac%7B-7%5Cpm+8%7D5%5C%5C%5C%5C+%5Cboxed%7Bx_1%3D%5Cdfrac+15%3B~~x_2%3D-3%7D)
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3. 7x² + 18x - 9 = 0
![\dfrac D4=9^2-7\cdot(-9)=144=12^2;~~~~x_{1,2}=\dfrac{-9\pm 12}7\\\\ \boxed{x_1=\dfrac 37;~~x_2=-3}](https://tex.z-dn.net/?f=%5Cdfrac+D4%3D9%5E2-7%5Ccdot%28-9%29%3D144%3D12%5E2%3B~~~~x_%7B1%2C2%7D%3D%5Cdfrac%7B-9%5Cpm+12%7D7%5C%5C%5C%5C+%5Cboxed%7Bx_1%3D%5Cdfrac+37%3B~~x_2%3D-3%7D)
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4. 12x² - 16x - 3 = 0
![\dfrac D4=8^2-12\cdot(-3)=100=10^2;~~~~x_{1,2}=\dfrac{8\pm 10}{12}\\\\ \boxed{x_1=1\dfrac 12;~~x_2=-\dfrac 16}](https://tex.z-dn.net/?f=%5Cdfrac+D4%3D8%5E2-12%5Ccdot%28-3%29%3D100%3D10%5E2%3B~~~~x_%7B1%2C2%7D%3D%5Cdfrac%7B8%5Cpm+10%7D%7B12%7D%5C%5C%5C%5C+%5Cboxed%7Bx_1%3D1%5Cdfrac+12%3B~~x_2%3D-%5Cdfrac+16%7D)
Ответ:
Объяснение:cos2x-3cosx+2=0;
2cos²x-1-3cosx+2=0;
2cos²x-3cosx+1=0;( кв. ур-ие отн-но cosx) D=9-8=1
cosx=1/2⇔x=±π/3+2πn,n∈z
или cosx=1-- x=2πn,n∈z
2*(x-1/2)*3(x-1/3)*4(x-1/4)=0
Сокращаем, приравниваем каждую скобку к нулю. Получаем ответы:
х1=1/2
x2=1/3
x3=1/4
-40a^6b^5
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