F(х) = (х + 1)√х=x^(3/2)+x^(1/2) f'(x)=(3/2)*x^(3/2-1)+(1/2)x^(1/2-1)=(3/2)x^(1/2)+(1/2)x^(-1/2)=(3√x)/2+1/(2√x)
f(х) = ctg5 – sin23. f'(x)=0<span>
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1.log75\3⇵5+5log(4)⇵5=log(25)⇵5+4=2+4=6
2. log(x)⇵3=a
a)a^2-10a+21=0
a(1)=7 a(2)=3
log(x)⇵3=7
x=2187
log(x)⇵3=3
x=9
б)x^2-2=x
x^2-x-2=0
x= -1 , x=2
3. D(f)=2x+1>0
2x> -1
x> -1\2
2x+1>4
x>3\2
Ответ: (3/2;+∞)
1) 7х+5у=19 |•3
4х-3у=5 |•5
21х+15у=57
20х-15у=25
2) 21х+15у=57
+
20х-15у=25
=
41х=82
х=2
3) 7•2+5у=19
у=1
(5*2^32-4*2^30)/4^16=4
1) 5*2^32=21 474 836 480
2) 4*2^30=4 294 907 290
3) 21 474 836 480 - 47 294 907 290=17 179 869 184
4) 4^16= 4 294 907 290
5) 17 179 869 184/ 4 294 907 290=4.