化简 3x+1/2 (x-1)^2+5ln|x-1|+1/(x-1)-2ln|x|-3

解答

化简

3 x + \frac{1}{2} (x - 1)^{2} + 5 ln ∣ x - 1 ∣ + \frac{1}{x - 1} - 2 ln ∣ x ∣ - 3

\frac{x ^{3} + 3 x ^{2} + 2 x ln ( \frac{∣ x - 1 ∣ ^{5}}{x ^{2}} ) - 3 x - 2 ln ( \frac{∣ x - 1 ∣ ^{5}}{x ^{2}} ) + 1}{2 ( x - 1 )} - 3

求解步骤

3 x + \frac{1}{2} (x - 1)^{2} + 5 ln ∣ x - 1 ∣ + \frac{1}{x - 1} - 2 ln ∣ x ∣ - 3

使用对数计算法则:

a lo g_{c} (b) = lo g_{c} (b^{a})

5 ln ∣ x - 1 ∣ = ln (∣ x - 1 ∣^{5})

= 3 x + \frac{1}{2} (x - 1)^{2} + ln (∣ x - 1 ∣^{5}) + \frac{1}{x - 1} - 2 ln ∣ x ∣ - 3

使用对数计算法则:

a lo g_{c} (b) = lo g_{c} (b^{a})

2 ln ∣ x ∣ = ln (∣ x ∣^{2})

= 3 x + \frac{1}{2} (x - 1)^{2} + ln (∣ x - 1 ∣^{5}) + \frac{1}{x - 1} - ln (∣ x ∣^{2}) - 3

使用对数计算法则:

lo g_{c} (a) - lo g_{c} (b) = lo g_{c} (\frac{a}{b})

ln (∣ x - 1 ∣^{5}) - ln (∣ x ∣^{2}) = ln (\frac{∣ x - 1 ∣ ^{5}}{∣ x ∣ ^{2}})

= 3 x + \frac{1}{2} (x - 1)^{2} + ln (\frac{∣ x - 1 ∣ ^{5}}{∣ x ∣ ^{2}}) + \frac{1}{x - 1} - 3

∣ x ∣^{2} = x^{2}

= 3 x + \frac{1}{2} (x - 1)^{2} + ln (\frac{∣ x - 1 ∣ ^{5}}{x ^{2}}) + \frac{1}{x - 1} - 3

\frac{1}{2} (x - 1)^{2} = \frac{( x - 1 ) ^{2}}{2}

= 3 x + \frac{( x - 1 ) ^{2}}{2} + ln (\frac{∣ x - 1 ∣ ^{5}}{x ^{2}}) + \frac{1}{x - 1} - 3

合并分式

\frac{( x - 1 ) ^{2}}{2} + \frac{1}{x - 1} + 3 x + ln (\frac{∣ x - 1 ∣ ^{5}}{x ^{2}}) : \frac{x ^{3} + 3 x ^{2} + 2 x ln ( \frac{∣ x - 1 ∣ ^{5}}{x ^{2}} ) - 3 x - 2 ln ( \frac{∣ x - 1 ∣ ^{5}}{x ^{2}} ) + 1}{2 ( x - 1 )}

= \frac{x ^{3} + 3 x ^{2} + 2 x ln ( \frac{∣ x - 1 ∣ ^{5}}{x ^{2}} ) - 3 x - 2 ln ( \frac{∣ x - 1 ∣ ^{5}}{x ^{2}} ) + 1}{2 ( x - 1 )} - 3

s im pl i f y 2 lo g_{10} (x) - 6 lo g_{10} (x - 9)

s im pl i f y ln (x) - ln (6)

s im pl i f y \frac{95}{100} \cdot ln (5 \cdot 1 0^{6}) + \frac{5}{100} \cdot ln (1000)

s im pl i f y ln (\frac{x}{y}) - ln (\frac{a}{b})

s im pl i f y ln (\frac{( x + 3 ) ^{2}}{e ^{x} x})