Giúp mình giải bài này với ạ

Đáp án:

Giải thích các bước giải:

 1) `\log_{5}\ (x-1)-\log_{1/5}\ (x+2)=0`

ĐK: `x > 1`

`⇔ \log_{5}\ (x-1)-\log_{5^{-1}}\ (x+2)=0`

`⇔ \log_{5}\ (x-1)+\log_{5}\ (x+2)=0`

`⇔ \log_{5}\ [(x-1)(x+2)]=0`

`⇔ (x-1)(x+2)=1`

`⇔ x^2+x-2=1`

`⇔ x^2+x-3=0`

`⇔` \(\left[ \begin{array}{l}x=\dfrac{-1+\sqrt{13}}{2}\ (TM)\\x=\dfrac{-1-\sqrt{13}}{2}\ (L)\end{array} \right.\) 

Vậy `S={\frac{-1+\sqrt{13}}{2}}`

2) `\log_{9}\ (x+8)-\log_{3}\ (x+26)+2=0`

ĐK: `x > -8`

`⇔ \log_{3^2}\ (x+8)-\log_{3}\ (x+26)+2=0`

`⇔ 1/2 \log_{3}\ (x+8)-\log_{3}\ (x+26)+2=0`

`⇔ \log_{3}\ (x+8)^{1/2}-\log_{3}\ (x+26)+2=0`

`⇔ \log_{3}\ [\frac{(x+8)^{1/2}}{x+26}]=-2`

`⇔ \frac{(x+8)^{1/2}}{x+26}=3^{-2}`

`⇔ \frac{(x+8)^{1/2}}{x+26}=1/9`

`⇔ 9\sqrt{x+8}=x+26`

`⇔ 81x+648=x^2+52x+676`

`⇔ x^2-29x+28=0`

`⇔ (x-28)(x+1)=0`

`⇔` \(\left[ \begin{array}{l}x=28\ (TM)\\x=1\ (TM)\end{array} \right.\) 

Vậy `S={1;28}`

3) `\log_{2} x=1/2`

ĐK: `x > 0`

`⇔ x=2^{1/2}`

`⇔ x=\sqrt{2}\ (TM)`

Vậy `S={\sqrt{2}}`

4) `\log\ x+\log\ (x-9)=1`

ĐK: `x > 9`

`⇔ \log\ [x(x-9)]=1`

`⇔ x(x-9)=10`

`⇔ x^2-9x-10=0`

`⇔` \(\left[ \begin{array}{l}x=10\ (TM)\\x=-1\ (L)\end{array} \right.\) 

Vậy `S={10}`

5) `\log_{3}\ (x^2-x-5)=\log_{3}\ (2x+5)`

TXĐ: `D=(-5/2;\frac{1-\sqrt{21}}{2})∪(\frac{1+\sqrt{21}}{2};+∞)`

`⇔ x^2-x-5=2x+5`

`⇔ x^2-3x-10=0`

`⇔` \(\left[ \begin{array}{l}x=5\\x=-2\end{array} \right.\) (TM)

Vậy `S={-2;5}`