5,6,7,8题求解,要具体过程~~

来源:学生作业帮助网 编辑:作业帮 时间:2024/11/17 04:35:54
5,6,7,8题求解,要具体过程~~
xSn@QPQ"'tUՌ6&1 TT  !6H E(nêQhs;4VGُOO>?m|YX'\Mn(ւw:5mZ?k~G(jKxwhO in؄(v)>37tG8risUQU4t=}6@]J/&xI/ޫ^GQ0L^厢^=D S,2V[QTJV0'q~"aV ̊MRj3w+ԲkIڲZI}tuE*ha0Z*KWhQ\"),A2*9}0;XmvnN4Ou *dN^~ C?EjB)GAZɭ 8Sb߭

5,6,7,8题求解,要具体过程~~
5,6,7,8题求解,要具体过程~~

5,6,7,8题求解,要具体过程~~
(5) [√(5x-4) - √x ] / (x-1)
= [√(5x-4) - √x ][√(5x-4) + √x ] / (x-1)[√(5x-4) + √x ]
= (5x-4 - x) / (x-1)[√(5x-4) + √x ]
= 4 / [√(5x-4) + √x ]
∴原式 = 2
(6)和差化积
( sinx - sina ) / (x-a)
= cos[(x+a)/2] sin[(x-a)/2] / [(x-a)/2]
令 t = (x-a)/2
原式 = lim(t→0) cos[(x+a)/2] ( sint / t ) = cosa
(7)√(x²+x) - √(x² -x)
= [ (x²+x) - (x² - x)] / [ √(x²+x) +√(x² -x) ]
= 2x / [ √(x²+x) +√(x² -x) ]
= 2 / [ √(1+1/x) +√(1 - 1/x) ]
1/x →0
所以原式 = 1
(8)√x[√(x+a) - √x ]
= √x [ (x+a) - x ] / [√(x+a) + √x ]
= a √x / [√(x+a) + √x ]
= a / [√(1+a/x) + 1 ]
a/x→0
原式 = a/2