<span>(tgx+)(2cosx-1)=0
ОДЗ sinx>0⇒x∈[πn+π+πn]
tgx+√3=0
tgx=-√3
x=-π/6+πn∉ОДЗ
2cosx-1=0
cosx=1/2
x=π/3+2πn
x=-π/3+2πn∉ОДЗ</span>
6_3/5 *10 =33/5 *10 =33* 2 =66
<span>(k-3)(2k+1)=k*2k+1*k-3*2k-3*1=2k</span>²+k-6k-3=2k²-5k-3
( ab / c ) + d = xb - c
( ab + dc ) / c = xb - c
xb = ( ( ab + dc ) / c ) + c
xb = ( ab + dc + c^2 ) / c
x = ( ( ab + dc + c^2 ) / c ) : b
x = ( ab + dc + c^2 ) / bc
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ab + d = b( x - c )
x - c = ( ab + d ) / b
x = ( ( ab + d ) / b ) + c
x = ( ab + d + cb ) / b
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a + d = d( x + c )
x + c = ( a + d ) / d
x = ( ( a + d ) / d ) - c
x = ( a + d - cd ) / d
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ab / c = xd / c
ab = xd
x = ab / d