\(\left(1+\dfrac{1}{2x}\right).lg3+lg2=lg\left(27-3^{\dfrac{1}{x}}\right)\)
\(\Leftrightarrow lg3^{1+\dfrac{1}{2x}}+lg2=lg\left(27-3^{\dfrac{1}{x}}\right)\)
\(\Leftrightarrow lg\left(2.3^{1+\dfrac{1}{2x}}\right)=lg\left(27-3^{\dfrac{1}{x}}\right)\)
\(\Leftrightarrow2.3^{1+\dfrac{1}{2x}}=27-3^{\dfrac{1}{x}}\)
\(\Leftrightarrow2.3.\left(3^{\dfrac{1}{x}}\right)^2=27-3^{\dfrac{1}{x}}\)
Đặt \(3^{\dfrac{1}{x}}=t\left(t>0\right)\) phương trình trở thành:
\(2.3t^2=27-t\)
\(\Leftrightarrow\left[{}\begin{matrix}t_1=\dfrac{-1-\sqrt{649}}{12}\left(l\right)\\t_2=\dfrac{1+\sqrt{649}}{12}\left(tm\right)\end{matrix}\right.\)
Với \(t=\dfrac{-1-\sqrt{649}}{12}\Leftrightarrow3^{\dfrac{1}{x}}=\dfrac{-1-\sqrt{649}}{12}\)
\(\Leftrightarrow\dfrac{1}{x}=log^{\dfrac{-1-\sqrt{649}}{12}}_3\)
\(\Leftrightarrow x=log^3_{\dfrac{-1-\sqrt{649}}{12}}\).

a) Ta có \(\log_32<\log_33=1=\log_22<\log_23\)

b) \(\log_23<\log_24=2=\log_39<\log_311\)

c) Đưa về cùng 1 lôgarit cơ số 10, ta có 

\(\frac{1}{2}+lg3=\frac{1}{2}lg10+lg3=lg3\sqrt{10}\)

\(lg19-lg2=lg\frac{19}{2}\)

So sánh 2 số \(3\sqrt{10}\) và \(\frac{19}{2}\) ta có :

\(\left(3\sqrt{10}\right)^2=9.10=90=\frac{360}{4}<\frac{361}{4}=\left(\frac{19}{2}\right)^2\)

Vì vậy : \(3\sqrt{10}<\frac{19}{2}\)

Từ đó suy ra \(\frac{1}{2}+lg3\)<\(lg19-lg2\)

d) Ta có : \(\frac{lg5+lg\sqrt{7}}{2}=lg\left(5\sqrt{7}\right)^{\frac{1}{2}}=lg\sqrt{5\sqrt{7}}\)

Ta so sánh 2 số : \(\sqrt{5\sqrt{7}}\) và \(\frac{5+\sqrt{7}}{2}\) 

Ta có :

\(\sqrt{5\sqrt{7}}^2=5\sqrt{7}\)

\(\left(\frac{5+\sqrt{7}}{2}\right)^2=\frac{32+10\sqrt{7}}{4}=8+\frac{5}{2}\sqrt{7}\)

\(8+\frac{5}{2}\sqrt{7}-5\sqrt{7}=8-\frac{5}{2}\sqrt{7}=\frac{16-5\sqrt{7}}{2}=\frac{\sqrt{256}-\sqrt{175}}{2}>0\)

Suy ra : \(8+\frac{5}{2}\sqrt{7}>5\sqrt{7}\)

Do đó : \(\frac{5+\sqrt{7}}{2}>\sqrt{5\sqrt{7}}\)

và \(lg\frac{5+\sqrt{7}}{2}>\frac{lg5+lg\sqrt{7}}{2}\)

Chọn C.

Đáp án : A.

Đáp án A

Ta có : \(b=lg2=lg\left(\frac{10}{5}\right)=1-lg5\Rightarrow lg5=1-b\)

                                                       \(\Rightarrow G=\log_{125b}30=\frac{lg30}{lg125}=\frac{lg\left(3.10\right)}{lg\left(5^3\right)}=\frac{1+lg3}{3lg5}=\frac{1+a}{3\left(1-b\right)}\)

Chọn A.

a/\(x:27=3,6\)
\(\Rightarrow x=97,2\)
b/\(\dfrac{2x+1}{-27}=\dfrac{-3}{2x+1}\)
\(\Rightarrow\left(2x+1\right)^2=81\)
\(\Rightarrow\left[{}\begin{matrix}2x+1=9\\2x+1=-9\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=8\\2x=-10\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=4\\x=-5\end{matrix}\right.\)
Vậy \(x\in\left\{4;-5\right\}\)

a) x=3,6\(\times\) 27=97,2

a,x:27=3,6

x=3,6:27

x\(\approx\)0,13

\(\Rightarrow\)x=0,1(3)