Ответ:
Объяснение:
23) f'(x) = 2x*e^(-x) + x^2*(-e^(-x)) = e^(-x)*(2x - x^2) = 0
x1 = 0; x2 = 2
24) f'(x) = 1/2 - (-1/2*sin(x/2)) = 1/2 + 1/2*sin(x/2) = 1/2*(1 + sin(x/2)) = 0
sin(x/2) = -1
x/2 = -П/2 + 2П*k
x = -П + 4П*k
25) ![f'(x)=\frac{1}{2\sqrt{x+4} }-\frac{2}{x+7} = \frac{x+7-4\sqrt{x+4} }{2(x+7)\sqrt{x+4} } =0](https://tex.z-dn.net/?f=f%27%28x%29%3D%5Cfrac%7B1%7D%7B2%5Csqrt%7Bx%2B4%7D%20%7D-%5Cfrac%7B2%7D%7Bx%2B7%7D%20%3D%20%5Cfrac%7Bx%2B7-4%5Csqrt%7Bx%2B4%7D%20%7D%7B2%28x%2B7%29%5Csqrt%7Bx%2B4%7D%20%7D%20%3D0)
Область определения: x >= -4; x ≠ -7
x + 7 - 4√(x+4) = 0
x + 7 = 4√(x+4)
(x+7)^2 = 16(x+4)
x^2 + 14x + 49 = 16x + 64
x^2 - 2x - 15 = 0
(x - 5)(x + 3) = 0
x1 = -3; x2 = 5
26) ![f'(x) = \frac{2}{2\sqrt{x+2} } -\frac{1}{x-4} =\frac{x-4-\sqrt{x+2} }{(x-4)\sqrt{x+2} } =0](https://tex.z-dn.net/?f=f%27%28x%29%20%3D%20%5Cfrac%7B2%7D%7B2%5Csqrt%7Bx%2B2%7D%20%7D%20-%5Cfrac%7B1%7D%7Bx-4%7D%20%3D%5Cfrac%7Bx-4-%5Csqrt%7Bx%2B2%7D%20%7D%7B%28x-4%29%5Csqrt%7Bx%2B2%7D%20%7D%20%3D0)
Область определения: x >= -2; x ≠ 4
x - 4 - √(x+2) = 0
x - 4 = √(x+2)
(x - 4)^2 = x + 2
x^2 - 8x + 16 = x + 2
x^2 - 9x + 14 = 0
(x - 2)(x - 7) = 0
x1 = 2; x2 = 7
6(a-b)(a²+ab+b²)/3(a-b)(a+b)=2(a²+ab+b²)/(a+b)=
=2(42,25+16,25+6,25)/(6,5+2,5)=2*64,75/9=129,5/9=14 35/90=14 7/18
Sin² π/8 + cos² 3π/8 + sin² 5π/8 + cos² 7π/8 = sin² π/8 + cos²(4π/8 - π/8) + sin²(4π/8 + π/8) + cos²(8π/8 - π/8) = sin²π/8 + cos²(π/2 - π/8) + sin²(π/2 + π/8) + cos²(π - π/8)=sin² π/8 + sin² π/8 + cos² π/8 + cos² π/8=1+1=2
tg 435 + tg 375= tg (360+75) + tg (360+15)=tg 75 + tg 15=tg (90-15) + tg 15=
ctg 15+ tg15=(1+tg² 15) : tg 15=1/cos² 15 : (sin 15/cos 15)=1 : sin 15*cos 15=2: sin 30=2: 1/2=4
Решение во вложении. Приятной учёбы
Числа х, (х+1), (х+2), (х+3).
За умовою задачi
(x+2)(x+3)-x(x+1)=34
x^2+3x+2x+6-x^2-x=34
4x=34
x=34/4
x=8,5