已知tanα=3，求值

1）（2sinα-5cosα）/（sinα+cosα）
=(2tana-5)/(tana+1)
=(2*3-5)/(3+1)
=1/4

2）2sin^2α-sinαcosα+cos^2α
=(2tan^2a-tana+1)*cos^2a
=(2*3^2-3+1)*1/(1+tan^2a)
=16/(1+3^2)
=8/5

tanα=3

tan^2 α + 1 = 1/ cos^2 α

cos^2 α = 1/10

sin^2 α = 9/10

(2sinα-5cosα)/(sinα+cosα)

= (2(3根号10 /10) - 5(根号10 /10)) / ((3根号10 /10)+(根号10 /10))

= (6 - 5) / (3+1)

= 1/4

2sin^2α-sinαcosα+cos^2α

= 2*9/10 - (3根号10 /10)(根号10 /10) + 1/10

= 18/10 - 3/10 + 1/10

= 16/10

= 8/5

(1)设tan@=X=3
原式等于(2x-5)/(x+1)=(2*3-5)/(3+1)=1/4
(2)