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}`
