取倒数(x+2)/(x+3)=√3+√2+1(x+2)/(x+3)-1=√3+√2-1/(x+3)=√3+√2原式=[(x-3)/(2x-4)]÷[5-(x+2)(x-2)]/(x-2)]=[(x-3)/2(x-2)]×[-(x-2)/(x²-9)]=[(x-3)/2(x-2)]×[-(x-2)/(x+3)(x-3)]=-1/(x+3)=√3+√2...