Выражение: 2/2-x-0.5=4/x*(2-x)
ответ: 4.5-x-8/x=0
Решаем по действиям:
1) 2/2=1
2.0|2_ _
2_ |1
0
2) 1-0.5=0.5
-1.0
_0_._5_
0.5
3) 4*(2-x)=8-4*x
4*(2-x)=4*2-4*x
3.1) 4*2=8
X4
_2_
8
4) (8-4*x)/x=8/x-4*x/x
5) x/x=1
6) 0.5-x-(8/x-4)=0.5-x-8/x+4
7) 0.5+4=4.5
+0.5
_4_._0_
4.5
Решаем по шагам:
1) 1-x-0.5-4/x*(2-x)=0
1.1) 2/2=1
2.0|2_ _
2_ |1
0
2) 0.5-x-4/x*(2-x)=0
2.1) 1-0.5=0.5
-1.0
_0_._5_
0.5
3) 0.5-x-(8-4*x)/x=0
3.1) 4*(2-x)=8-4*x
4*(2-x)=4*2-4*x
3.1.1) 4*2=8
X4
_2_
8
4) 0.5-x-(8/x-4*x/x)=0
4.1) (8-4*x)/x=8/x-4*x/x
5) 0.5-x-(8/x-4)=0
5.1) x/x=1
6) 0.5-x-8/x+4=0
6.1) 0.5-x-(8/x-4)=0.5-x-8/x+4
7) 4.5-x-8/x=0
7.1) 0.5+4=4.5
+0.5
_4_._0_
4.5
Пусть y = uv, тогда y' = u'v + uv':
Решим левый интеграл:
cosx = \frac{1-t^2}{1+t^2} => dx = \frac{2}{1+t^2}dt\\ \int \frac{2(1+t^2)}{(1+t^2)(1-t^2)} dt = \int \frac{2}{(1-t)(1+t)}dt = \int ( \frac{1}{1-t} + \frac{1}{1+t})dt = ln(1-t)+ln( 1+t) = ln|1-t^2| = ln|1-tg^2\frac{x}{2}| \\" class="latex-formula" id="TexFormula2" src="https://tex.z-dn.net/?f=%5Cint%20%5Cfrac%7Bdx%7D%7Bcosx%7D%3B%5C%5C%20tg%5Cfrac%7Bx%7D%7B2%7D%3Dt%20%3D%3E%20cosx%20%3D%20%5Cfrac%7B1-t%5E2%7D%7B1%2Bt%5E2%7D%20%3D%3E%20dx%20%3D%20%5Cfrac%7B2%7D%7B1%2Bt%5E2%7Ddt%5C%5C%20%20%5Cint%20%5Cfrac%7B2%281%2Bt%5E2%29%7D%7B%281%2Bt%5E2%29%281-t%5E2%29%7D%20dt%20%3D%20%5Cint%20%5Cfrac%7B2%7D%7B%281-t%29%281%2Bt%29%7Ddt%20%3D%20%5Cint%20%28%20%5Cfrac%7B1%7D%7B1-t%7D%20%2B%20%5Cfrac%7B1%7D%7B1%2Bt%7D%29dt%20%3D%20ln%281-t%29%2Bln%28%201%2Bt%29%20%3D%20ln%7C1-t%5E2%7C%20%3D%20ln%7C1-tg%5E2%5Cfrac%7Bx%7D%7B2%7D%7C%20%20%5C%5C" title="\int \frac{dx}{cosx};\\ tg\frac{x}{2}=t => cosx = \frac{1-t^2}{1+t^2} => dx = \frac{2}{1+t^2}dt\\ \int \frac{2(1+t^2)}{(1+t^2)(1-t^2)} dt = \int \frac{2}{(1-t)(1+t)}dt = \int ( \frac{1}{1-t} + \frac{1}{1+t})dt = ln(1-t)+ln( 1+t) = ln|1-t^2| = ln|1-tg^2\frac{x}{2}| \\">
Возвращаемся к исходному:
