F(π/4)=π/4*1=π/4
f`(x)=tgx+x/cos²x
f`(π/4)=1+π/4:1/2=1+π/2
8а-(2а-4+3а)=8а-2а+4-3а=3а+4
= ( а + 2b ) * ( a - 3b + 5b ) = ( a + 2b ) * ( a + 2b ) = ( a + 2b ) ^2
1)х²-4х-3√(х²-4х+20)+10=0
х²-4х+20-3√(х²-4х+20)-10=0
√(х²-4х+20)=t
t²-3t-10=0
D=9+40=7²
t1=(3-7):2=-2
t2=(3+7):2=5
√(x²-4x+20)=-2. √(x²-4x+20)=5
x²-4x+20=4. x²-4x-5=0
x²-4x+16=0. D=16+20=6²
D=16-4•16<0. x1=-1
x2=5