GPT: \(5^{2x-1}=125\)

\(\Leftrightarrow2x-1=log_5125\)

\(\Leftrightarrow2x-1=3\)

\(\Leftrightarrow2x=4\)

\(\Leftrightarrow x=2\) 

GPT: \(2^{2x-1}=32\)

\(\Leftrightarrow2x-1=log_232\)

\(\Leftrightarrow2x-1=5\)

\(\Leftrightarrow2x=6\)

\(\Leftrightarrow x=3\)

\(A=log_65\cdot log_76\cdot log_87\)

\(=\dfrac{1}{log_56\cdot log_67\cdot log_78}=\dfrac{1}{log_58}\)

Câu 2:

a: ta có: BD\(\perp\)AC(ABCD là hình vuông)

BD\(\perp\)SA(SA\(\perp\)(ABCD))

SA,AC cùng thuộc mp(SAC)

Do đó: BD\(\perp\)(SAC)

b: \(\widehat{SC;\left(ABCD\right)}=\widehat{CS;CA}=\widehat{SCA}\)

Vì ABCD là hình vuông nên \(AC=AD\cdot\sqrt{2}=a\sqrt{2}\)

Xét ΔSAC vuông tại A có \(tanSCA=\dfrac{SA}{AC}=1\)

nên \(\widehat{SCA}=45^0\)

=>\(\widehat{SC;\left(ABCD\right)}=45^0\)

Câu 3:

a: BC\(\perp\)AB(ΔABC vuông tại B)

BC\(\perp\)SA(SA\(\perp\)(ABC))

SA,AB cùng thuộc mp(SAB)

Do đó:BC\(\perp\)(SAB)

Bài 2:

a: \(3\cdot2^{x+1}-2\cdot2^{x-1}-3>0\)

=>\(6\cdot2^x-2^x-3>0\)

=>\(2^x>\dfrac{3}{5}\)

=>\(x>log_2\left(\dfrac{3}{5}\right)\)

b: \(\left(2+\sqrt{3}\right)^{2x-3}>=\left(2-\sqrt{3}\right)^{x+1}\)

=>2x-3>=x+1

=>x>=4

c: \(\left(\dfrac{3}{4}\right)^{x^2-2}< =\left(\dfrac{3}{4}\right)^x\)

=>\(x^2-2>=x\)

=>\(x^2-x-2>=0\)

=>(x-2)(x+1)>=0

=>\(\left[{}\begin{matrix}x>=2\\x< =-1\end{matrix}\right.\)

Bài 1:

a: \(3^{x+1}>2\)

=>\(x+1>log_32\)

=>\(x>log_32-1\)

b: \(2\cdot5^x< 3\)

=>\(5^x< \dfrac{3}{2}\)

=>\(x< log_5\left(\dfrac{3}{2}\right)\)

c: \(2-3\cdot2^x>=0\)

=>\(3\cdot2^x< =2\)

=>\(2^x< =\dfrac{2}{3}\)

=>\(x< =log_2\left(\dfrac{2}{3}\right)\)

d: \(\left(\dfrac{3}{4}\right)^{x^2-4}>=1\)

=>\(x^2-4< =0\)

=>(x-2)(x+2)<=0

=>-2<=x<=2

e: \(2+3\left(\dfrac{2}{3}\right)^x< =0\)

=>\(3\cdot\left(\dfrac{2}{3}\right)^x< =-2\)

=>\(\left(\dfrac{2}{3}\right)^x< =-\dfrac{2}{3}\)

=>\(x\in\varnothing\)

Bài 2:

a: \(log^2_{\dfrac{1}{3}}x-5\cdot log_3x+4=0\)

=>\(log_{\dfrac{1}{3}}^2x+5\cdot log_{\dfrac{1}{3}}x+4=0\)

=>\(\left(log_{\dfrac{1}{3}}x+1\right)\left(log_{\dfrac{1}{3}}x+4\right)=0\)

=>\(\left[{}\begin{matrix}log_{\dfrac{1}{3}}x=-1\\log_{\dfrac{1}{3}}x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=81\end{matrix}\right.\)

b: \(log_2^24x-log_{\sqrt{2}}2x=5\)

=>\(log_2^24x-2\cdot log_22x=5\)

=>\(\left(log_24x\right)^2-2\cdot log_22x=5\)

=>\(\left(1+log_22x\right)^2-2\cdot log_22x=5\)

=>\(\left(log_22x\right)^2+1=5\)

=>\(\left(log_22x\right)^2=4\)

=>\(\left[{}\begin{matrix}log_22x=2\\log_22x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=4\\2x=\dfrac{1}{4}\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=2\\x=\dfrac{1}{8}\end{matrix}\right.\)

Bài 1:

a:

ĐKXĐ: x>1

 \(log_3\left(2x+1\right)-log_3\left(x-1\right)=1\)

=>\(log_3\left(\dfrac{2x+1}{x-1}\right)=1\)

=>\(\dfrac{2x+1}{x-1}=3\)

=>3(x-1)=2x+1

=>3x-3=2x+1

=>x=4(nhận)

b:

ĐKXĐ: x>2

 \(log_2\left(x-1\right)+log_2\left(x-2\right)=log_5\left(125\right)\)

=>\(log_2\left[\left(x-1\right)\left(x-2\right)\right]=3\)

=>\(\left(x-1\right)\left(x-2\right)=2^3=8\)

=>\(x^2-3x-6=0\)

=>\(\left[{}\begin{matrix}x=\dfrac{3+\sqrt{33}}{2}\left(nhận\right)\\x=\dfrac{3-\sqrt{33}}{2}\left(loại\right)\end{matrix}\right.\)

c: \(log_2\left(sinx\right)+log_2\left(cosx\right)=-2\)

=>\(log_2\left(sinx\cdot cosx\right)=-2\)

=>\(log_2\left(\dfrac{1}{2}\cdot sin2x\right)=-2\)

=>\(\dfrac{1}{2}\cdot sin2x=\dfrac{1}{4}\)

=>\(sin2x=\dfrac{1}{2}\)

=>\(\left[{}\begin{matrix}2x=\dfrac{\Omega}{6}+k2\Omega\\2x=\dfrac{5}{6}\Omega+k2\Omega\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\Omega}{12}+k\Omega\\x=\dfrac{5}{12}\Omega+k\Omega\end{matrix}\right.\)

\(x\in\left(0;2\Omega\right)\)

=>\(\left[{}\begin{matrix}\dfrac{\Omega}{12}+k\Omega\in\left(0;2\Omega\right)\\\dfrac{5}{12}\Omega+k\Omega\in\left(0;2\Omega\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}k+\dfrac{1}{12}\in\left(0;2\right)\\k+\dfrac{5}{12}\in\left(0;2\right)\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}k\in\left(-\dfrac{1}{12};\dfrac{23}{12}\right)\\k\in\left(-\dfrac{5}{12};\dfrac{19}{12}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}k\in\left(0;1\right)\\k\in\left(0;1\right)\end{matrix}\right.\)

=>\(x\in\left\{\dfrac{\Omega}{12};\dfrac{13}{12}\Omega;\dfrac{5}{12}\Omega;\dfrac{17}{12}\Omega\right\}\)

Chọn A