f(x)=x的平方+2x+3/x
設(shè)2≤x1分母呢？你的輸入，沒看懂，若有分式得話應(yīng)該用括號(hào)f(x)=x/(x²+2x+3) f(x1)-f(x2)=x1/(x²1+2x1+3)-x2(x²2+x2+3)=[x1(x²2+x2+3)-x2(x²1+2x1+3)]/[(x²1+2x1+3)(x²2+x2+3)]=[x1(x²2+x2+3)-x2(x²1+2x1+3)]/[(x²1+2x1+3)(x²2+x2+3)]=[(x1x²2-x²1x2)+3(x1-x2)]/[(x²1+2x1+3)(x²2+x2+3)] =[x1x2(x1-x2)-3(x1-x2)]/[(x²1+2x1+3)(x²2+x2+3)]=[(x1-x2)(x1x2-3)]/[(x²1+2x1+3)(x²2+x2+3)] ∵ 2≤x10∵(x²1+2x1+3)(x²2+x2+3)>0∴f(x1)-f(x2)<0∴f（x）是增函數(shù)，當(dāng)x=2時(shí)f(x)取最小值2/11哦，謝謝?？梢詥枂杫=3x+5/√x-3的值域嗎？f(x1)-f(x2)=x1/(x²1+2x1+3)-x2(x²2+x2+3)=[x1(x²2+x2+3)-x2(x²1+2x1+3)]/[(x²1+2x1+3)(x²2+x2+3)]=[x1(x²2+x2+3)-x2(x²1+2x1+3)]/[(x²1+2x1+3)(x²2+x2+3)]=[(x1x²2-x²1x2)+3(x1-x2)]/[(x²1+2x1+3)(x²2+x2+3)] =[x1x2(x2-x1)-3(x1-x2)]/[(x²1+2x1+3)(x²2+x2+3)]=[(x2-x1)(x1x2+3)]/[(x²1+2x1+3)(x²2+x2+3)]>0f(x)是減函數(shù)y=（3x+5）/√(x-3) 就怕不加括號(hào)設(shè)√（x-3)=t>0∴x-3=t²x=t²+3∴y=[3(t²+3)+5]/t =(3t²+14)/t =3t+14/t 用均值定理： 3t+14/t≥2√42 ∴y≥2√42值域[2√42,+∞)若解析式理解不對(duì)，再追問確實(shí)沒括號(hào)~~y=3x+5/√x-3x>0可以用三個(gè)數(shù)的均值定理，也可以用導(dǎo)數(shù)3x+5/√x=3x+5/(2√x)+5/(2√x)≥3³√[3x*5/(2√x)*5/(2√x)]=3³√(75/4) y= 3x+5/√x-3≥3/4*³√150-3值域[3/4*³√150-3,+∞) 數(shù)很怪y=3x+5/√(x-3)√（x-3)=t>0∴x-3=t²,x=t²+3t=3t²+5/t+9還可以用均值