1) X=log1-log2+log2-log3+...+log99-log100

=log1-log100

=0-2

=-2

Đáp án C

2)X=-log₃100=-log₃10²=-2log₃(2.5)=-2log₃2-2log₃5=-2a-2b

Đáp án A

a) \(log_29\cdot log_34=4\)

b) \(log_{25}\cdot\dfrac{1}{\sqrt{5}}=-\dfrac{1}{4}\)

c) \(log_23\cdot log_9\sqrt{5}\cdot log_54=\dfrac{1}{2}\)

a, ĐKXĐ: x≠±3

A=\(\left(\dfrac{3-x}{x+3}.\dfrac{x^2+6x+9}{x^2-9}+\dfrac{x}{x+3}\right):\dfrac{3x^2}{x+3}\)

A=\(\left(\dfrac{3-x}{x+3}.\dfrac{\left(x+3\right)^2}{\left(x+3\right)\left(x-3\right)}+\dfrac{x}{x+3}\right):\dfrac{3x^2}{x+3}\)

A=\(\left(\dfrac{3-x}{x-3}+\dfrac{x}{x+3}\right):\dfrac{3x^2}{x+3}\)

A=\(\left(\dfrac{9-x^2}{x^2-9}+\dfrac{x^2-3x}{x^2-9}\right):\dfrac{3x^2}{x+3}\)

A=\(\left(\dfrac{-3}{x+3}\right):\dfrac{3x^2}{x+3}\)

A=\(\dfrac{-1}{x^2}\)

b, Thay x=\(-\dfrac{1}{2}\) (TMĐKXĐ) vào A ta có:

\(\dfrac{-1}{\left(-\dfrac{1}{2}\right)^2}\)=-4

c, A<0 ⇔ \(\dfrac{-1}{x^2}< 0\) ⇔ x²>0 (Đúng với mọi x)

Vậy để A<0 thì x đúng với mọi giá trị (trừ ±3)

a)     \({3^{{x^2} - 4x + 5}} = 9 \Leftrightarrow {x^2} - 4x + 5 = 2 \Leftrightarrow {x^2} - 4x + 3 = 0 \Leftrightarrow \left( {x - 3} \right)\left( {x - 1} \right) = 0\)

\( \Leftrightarrow \left[ \begin{array}{l}x = 3\\x = 1\end{array} \right.\)

Vậy phương trình có nghiệm là \(x \in \left\{ {1;3} \right\}\)

b)    \(0,{5^{2x - 4}} = 4 \Leftrightarrow 2x - 4 = {\log _{0,5}}4 \Leftrightarrow 2x = 2 \Leftrightarrow x = 1\)

Vậy phương trình có nghiệm là x = 1

c)     \({\log _3}(2x - 1) = 3\)    ĐK: \(2x - 1 > 0 \Leftrightarrow x > \frac{1}{2}\)

\( \Leftrightarrow 2x - 1 = 27 \Leftrightarrow x = 14\) (TMĐK)

Vậy phương trình có nghiệm là x = 14

d)    \(\log x + \log (x - 3) = 1\)  ĐK: \(x - 3 > 0 \Leftrightarrow x > 3\)

\(\begin{array}{l} \Leftrightarrow \log \left( {x.\left( {x - 3} \right)} \right) = 1\\ \Leftrightarrow {x^2} - 3x = 10\\ \Leftrightarrow {x^2} - 3x - 10 = 0\\ \Leftrightarrow \left( {x + 2} \right)\left( {x - 5} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}x =  - 2 (loại) \,\,\,\\x = 5 (TMĐK) \,\,\,\,\,\,\,\end{array} \right.\end{array}\)

Vậy phương trình có nghiệm x = 5

a) \(log_69+log_64=log_636=2\)

b) \(log_52-log_550=log_5\left(2:50\right)=-2\)

c) \(log_3\sqrt{5}-\dfrac{1}{2}log_550=-1,0479\)

C = 59/10 : 3/2 - ( 5/3 x 9/2 x 14/3) : 7/4

= 59/15 - 35 : 7/4

= 59/15 - 20

= -241/15

\(a,A=log_23\cdot log_34\cdot log_45\cdot log_56\cdot log_67\cdot log_78\\ =log_28\\ =log_22^3\\ =3\\ b,B=log_22\cdot log_24...log_22^n\\ =log_22\cdot log_22^2...log_22^n\\ =1\cdot2\cdot...\cdot n\\ =n!\)

\(log_3\sqrt{3}=log_33^{\dfrac{1}{2}}=\dfrac{1}{2}\)

\(lne^3=log_ee^3=3\)

\(log_{27}3=log_{3^3}3=\dfrac{1}{3}\)

\(\log_{\sqrt{3}}3=log_{3^{\dfrac{1}{2}}}3=1:\dfrac{1}{2}=2\)

\(\log_{0,125}2=log_{2^{-3}}2=\dfrac{1}{-3}\)

\(\log_{\sqrt[3]{49}}7=\log_{7^{\dfrac{2}{3}}}7=1:\dfrac{2}{3}=\dfrac{3}{2}\)

\(\log_{\dfrac{1}{125}}5=\log_{5^{-3}}5=-\dfrac{1}{3}\)

\(\log_84=log_{2^3}2^2=\dfrac{1}{3}\cdot2=\dfrac{2}{3}\)

\(\log_{25}\left(\dfrac{1}{5}\right)=\log_{5^2}5^{-1}=\dfrac{1}{2}\cdot\left(-1\right)=-\dfrac{1}{2}\)

\(\log_{\dfrac{1}{5}}\sqrt{5}=\log_{5^{-1}}5^{\dfrac{1}{2}}=\dfrac{1}{-1}\cdot\dfrac{1}{2}=-\dfrac{1}{2}\)

\(log_{\dfrac{1}{7}}\sqrt[5]{49}=\log_{7^{-1}}7^{\dfrac{2}{5}}=\dfrac{1}{-1}\cdot\dfrac{2}{5}=-\dfrac{2}{5}\)

\(\log_4\left(\dfrac{1}{\sqrt{2}}\right)=\log_{2^2}\left(\sqrt{2}\right)^{-1}\)

\(=\log_{2^{-2}}\left(\sqrt{2}\right)^{-\dfrac{1}{2}}=\dfrac{1}{-2}\cdot\dfrac{-1}{2}=\dfrac{1}{4}\)

\(\log_{27}3\sqrt{3}=\log_{3^3}3^{\dfrac{3}{2}}=\dfrac{1}{3}\cdot\dfrac{3}{2}=\dfrac{1}{2}\)

a) \(\frac{1}{5}-\left(-\frac{1}{3}\right)=\frac{1}{5}+\frac{1}{3}=\frac{3}{15}+\frac{5}{15}=\frac{8}{15}\)

b) \(-\frac{1}{2}\cdot\frac{4}{9}+\frac{3}{7}:\frac{5}{14}\)

\(=-\frac{2}{9}+\frac{6}{5}\)

\(=\frac{44}{45}\)