$Log_{2x-1}(2x^2+19x-10)+log_{x+10}(2x-1)^2=4$

Đáp án:

Giải thích các bước giải:

`\Log_{2x-1}(2x^2+19x-10)+\log_{x+10}(2x-1)^2=4`

ĐK: \(\begin{cases} 0 < 2x-1 \ne 1\\2x^2+19x-10 > 0\\0 < x+10 \ne 1\\(2x-1)^2 > 0\end{cases}\)

`⇒ 1/2 < x \ne 1`

`⇔ \Log_{2x-1} (2x-1)(x+10)+\log_{x+10}(2x-1)^2=4`

`⇔ 1+\Log_{2x-1} (x+10)+2\log_{x+10} (2x-1)=4`

Đặt `t=\Log_{2x-1} (x+10)`

`⇒ \log_{x+10} (2x-1)=1/(\Log_{2x-1} (x+10))=1/t`

Ta có:

`1+t+2/t=4`

`⇔ t+2/t-3=0`

`⇔ t^2-3t+2=0`

`⇔ (t-1)(t-2)=0`

`⇔` \(\left[ \begin{array}{l}t=1\\t=2\end{array} \right.\) 

+) `t=1`

`⇔ \Log_{2x-1} (x+10)=1`

`⇔ x+10=2x-1`

`⇔ x=11\ (TM)`

+) `t=2`

`⇔ \Log_{2x-1} (x+10)=2`

`⇔ (2x-1)^2=x+10`

`⇔` \(\left[ \begin{array}{l}x=-1\ (L)\\x=\dfrac{9}{4}\ (TM)\end{array} \right.\) 

Vậy `S={11;9/4}`