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:
3x²-12=0
x²-4=0
x²=4
х=√4
x=± 2
.
2x²+6x=0
x²+3x=0
x(x+3)=0
x1=0
x+3=0
x2= -3
.
1.8x²=0
x²=0
x=0
.
x²+25=0
x²= -25 < 0 -- решений нет
.
1/7 x² - 6/7=0
1/7 x²=6/7
x²=6/7 : 1/7
x²=6/7 * 7
x²=6
x=±√6
.
x²=3x
x²-3x=0
x(x-3)=0
x1=0
x-3=0
x2=3
.
x²+2x-3=2x+6
x²+2x-3-2x-6=0
x²-9=0
(х-3)(х+3)
x-3=0
х1=3
х+3=0
х2= -3
x= ±3
.
x²=3,6
x=±√3,6
.
2x²-18=0
x²-9=0
(x-3)(x+3)=0
x-3=0
x1=3
x+3=0
x2= -3
x=±3
.
3x²-12x=0
x²-4x=0
x(x-4)=0
x1=0
x-4=0
x2=4
.
2.7x²=0
x²=0
x=0
.
x²+16=0
x²= -16 < 0 --- нет решений
.
1/6 x² - 5/6=0
1/6 x²=5/6
x²=5/6 : 1/6
x²=5/6 * 6
x²=5
x=±√5
.
x²=7x
x²-7x=0
x(x-7)=0
x₁=0
x-7=0
x₂=7
.
x²-3x-5=11-3x
x²-3x-5-11+3x=0
x²-16=0
x²=16
x=√16
x=±4
.
x²=2,5
x=±√2,5
