已知0＜x＜π/4,则sinx>0,cosx>0
sin(π/4-x)=sinπ/4·cosx-cosπ/4·sinx=√2/2(cosx-sinx)=5/13
∴cosx-sinx=5√2/13,
又∵sin²x+cos²x=1
2cosx·sinx=-(cosx-sinx)²+sin²x+cos²x=1-50/169=119/169
(cosx+sinx)²=(cosx-sinx)²+4cosx·sinx=50/169+238/169=288/169
∴cosx+sinx=12√2/13
cos2x/cos(π/4+x)=(cos²x-sin²x)/(cosπ/4·cosx-sinπ/4·sinx)=(cosx+sinx)(cosx-sinx)/[√2/2(cosx-sinx)]=√2(cosx+sinx)=√2×12√2/13=24/13