Lời giải:

$\log(8.5^x+20^x)=x+\log 25$

$\Rightarrow 8.5^x+20^x=10^{x+\log 25}=10^x.25$

$\Rightarrow \frac{8.5^x+20^x}{10^x}=25$

$\Leftrightarrow \frac{8}{2^x}+2^x=25$

Đặt $2^x=t$ thì $\frac{8}{t}+t=25$

$\Leftrightarrow t^2-25t+8=0$

Dễ thấy PT trên luôn có 2 nghiệm dương $t_1,t_2$ nên kéo theo PT ban đầu có 2 nghiệm $x_1,x_2$

Tổng các nghiệm $x_1+x_2=\log_2(t_1)+\log_2(t_2)=\log_2(t_1t_2)=\log_2(8)=3$

Đáp án đúng : A

tham khảo

a)Điều kiện \(3-x>0\) hay \(x< 3\)

\(4-log\left(3-x\right)=3log\left(3-x\right)=1\Leftrightarrow10^1=3-x\)

Vậy nghiệm của phương trình là \(x=2\) thỏa mãn điều kiện

b) Điều kiện \(x+2>0\) và \(x-1>0\) tức là \(x>1\)

\(\left(x+2\right)\left(x-1\right)=2\Rightarrow x^2+x-4=0\)

Vậy pt có nghiệm \(x=\dfrac{-1+\sqrt{17}}{2}\)

a, ĐK: \(x+1>0\Leftrightarrow x>-1\)

\(log\left(x+1\right)=2\\ \Leftrightarrow x+1=10^2\\ \Leftrightarrow x+1=100\\ \Leftrightarrow x=99\left(tm\right)\)

b, ĐK: \(\left\{{}\begin{matrix}x-3>0\\x>0\end{matrix}\right.\Rightarrow x>3\)

\(2log_4x+log_2\left(x-3\right)=2\\ \Leftrightarrow log_2x+log_2\left(x-3\right)=2\\ \Leftrightarrow log_2\left(x^2-3x\right)=2\\ \Leftrightarrow x^2-3x=4\\ \Leftrightarrow x^2-3x-4=0\\ \Leftrightarrow\left(x+1\right)\left(x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\left(ktm\right)\\x=4\left(tm\right)\end{matrix}\right.\)

c, ĐK: \(x>1\)

\(lnx+ln\left(x-1\right)=ln4x\\ \Leftrightarrow ln\left[x\left(x-1\right)\right]-ln4x=0\\ \Leftrightarrow ln\left(\dfrac{x-1}{4}\right)=0\\ \Leftrightarrow\dfrac{x-1}{4}=1\\ \Leftrightarrow x-1=4\\ \Leftrightarrow x=5\left(tm\right)\)

d, ĐK: \(\left\{{}\begin{matrix}x^2-3x+2>0\\2x-4>0\end{matrix}\right.\Rightarrow x>2\)

\(log_3\left(x^2-3x+2\right)=log_3\left(2x-4\right)\\ \Leftrightarrow x^2-3x+2=2x-4\\ \Leftrightarrow x^2-5x+6=0\\ \Leftrightarrow\left(x-2\right)\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\left(ktm\right)\\x=3\left(tm\right)\end{matrix}\right.\)

a) \({\log _{\frac{1}{7}}}\left( {x + 1} \right) > {\log _7}\left( {2 - x} \right)\)               (ĐK: \(x + 1 > 0;2 - x > 0 \Leftrightarrow  - 1 < x < 2\))

\(\begin{array}{l} \Leftrightarrow {\log _{{7^{ - 1}}}}\left( {x + 1} \right) > {\log _7}\left( {2 - x} \right)\\ \Leftrightarrow  - {\log _7}\left( {x + 1} \right) > {\log _7}\left( {2 - x} \right)\\ \Leftrightarrow {\log _7}{\left( {x + 1} \right)^{ - 1}} > {\log _7}\left( {2 - x} \right)\\ \Leftrightarrow {\left( {x + 1} \right)^{ - 1}} > 2 - x\\ \Leftrightarrow \frac{1}{{x + 1}} - 2 + x > 0\\ \Leftrightarrow \frac{{1 + \left( {x - 2} \right)\left( {x + 1} \right)}}{{x + 1}} > 0\\ \Leftrightarrow \frac{{1 + {x^2} - x - 2}}{{x + 1}} > 0 \Leftrightarrow \frac{{{x^2} - x - 1}}{{x + 1}} > 0\end{array}\)

Mà – 1 < x < 2 nên x + 1 > 0

\( \Leftrightarrow {x^2} - x - 1 > 0 \Leftrightarrow \left[ \begin{array}{l}x < \frac{{1 - \sqrt 5 }}{2}\\x > \frac{{1 + \sqrt 5 }}{2}\end{array} \right.\)

KHĐK ta có \(\left[ \begin{array}{l} - 1 < x < \frac{{1 - \sqrt 5 }}{2}\\\frac{{1 + \sqrt 5 }}{2} < x < 2\end{array} \right.\)

b) \(2\log \left( {2x + 1} \right) > 3\)              (ĐK: \(2x + 1 > 0 \Leftrightarrow x > \frac{{ - 1}}{2}\))

\(\begin{array}{l} \Leftrightarrow \log \left( {2x + 1} \right) > \frac{3}{2}\\ \Leftrightarrow 2x + 1 > {10^{\frac{3}{2}}} = 10\sqrt {10} \\ \Leftrightarrow x > \frac{{10\sqrt {10}  - 1}}{2}\end{array}\)

KHĐK ta có \(x > \frac{{10\sqrt {10}  - 1}}{2}\)

\(a,3^{1-2x}=4^x\\ \Leftrightarrow1-2x=log_34^x\\ \Leftrightarrow1-2x=xlog_34\\ \Leftrightarrow2x+xlog_34=1\\ \Leftrightarrow x\left(2+log_34\right)=1\\ \Leftrightarrow x=\dfrac{1}{2+log_34}=\dfrac{1}{log_39+log_34}=\dfrac{1}{log_336}=log_{36}3\)

b, ĐK: \(x>-1\)

\(log_3\left(x+1\right)+log_3\left(x+4\right)=2\\ \Leftrightarrow log_3\left(x^2+5x+4\right)=2\\ \Leftrightarrow x^2+5x+4=9\\ \Leftrightarrow x^2+5x-5=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-5+3\sqrt{5}}{2}\left(tm\right)\\x=\dfrac{-5-3\sqrt{5}}{2}\left(ktm\right)\end{matrix}\right.\)