Ôn tập cuối năm môn Đại số 11

\(\left[log_24x\right]^2-log_{\sqrt{2}}2x=5\)

=>\(\left[log_2\left(2\cdot2x\right)\right]^2-log_{2^{\dfrac{1}{2}}}2x=5\)

=>\(\left[1+log_22x\right]^2-1:\dfrac{1}{2}\cdot log_22x=5\)

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

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

=>\(\left[{}\begin{matrix}log_22x=2\\log_22x=-2\left(loại\right)\end{matrix}\right.\Leftrightarrow log_22x=2\)

=>\(2x=2^2=4\)

=>x=2

ĐKXĐ: x>0

Đặt \(a=log_x2\left(a>=0\right)\)

BPT sẽ trở thành \(a+\dfrac{1}{a}>=\dfrac{5}{2}\)

=>\(\dfrac{a^2+1}{a}>=\dfrac{5}{2}\)

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

=>\(2a^2-5a+2>=0\)

=>\(\left(2a-1\right)\left(a-2\right)>=0\)

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

=>\(\left[{}\begin{matrix}a>=2\\0< a< =\dfrac{1}{2}\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}log_x2>=2\\0< log_x2< =\dfrac{1}{2}\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x>=\sqrt{2}\\0< x< =4\end{matrix}\right.\)

\(25^x-20\cdot5^{x-1}+3=0\)

=>\(\left(5^x\right)^2-20\cdot\dfrac{5^x}{5}+3=0\)

=>\(\left(5^x\right)^2-4\cdot5^x+3=0\)

=>\(\left(5^x-3\right)\left(5^x-1\right)=0\)

=>\(\left[{}\begin{matrix}5^x-3=0\\5^x-1=0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}5^x=3\\5^x=1\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=log_53\\x=0\end{matrix}\right.\)

10: \(\left(\sqrt{2}-1\right)^x+\left(\sqrt{2}+1\right)^x-2\sqrt{2}=0\)

=>\(\dfrac{1}{\left(\sqrt{2}+1\right)^x}+\left(\sqrt{2}+1\right)^x-2\sqrt{2}=0\)(1)

Đặt \(a=\left(\sqrt{2}+1\right)^x\)(a>0)

Phương trình (1) sẽ trở thành \(\dfrac{1}{a}+a-2\sqrt{2}=0\)

=>\(\dfrac{a^2+1-2\sqrt{2}\cdot a}{a}=0\)

=>\(a^2-2\sqrt{2}\cdot a+1=0\)

=>\(a^2-2\sqrt{2}a+2-1=0\)

=>\(\left(a-\sqrt{2}\right)^2=1\)

=>\(\left[{}\begin{matrix}a=\sqrt{2}+1\left(nhận\right)\\a=\sqrt{2}-1\left(nhận\right)\end{matrix}\right.\)

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

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

\(log_3\left(2^x-1\right)=4\)

=>\(2^x-1=3^4=81\)

=>\(2^x=82\)

=>\(x=log_282\)

\(4^x+2^{x+1}-3=0\)

=>\(\left(2^x\right)^2+2\cdot2^x-3=0\)

=>\(\left(2^x+3\right)\left(2^x-1\right)=0\)

mà \(2^x+3>0\forall x\)

nên \(2^x-1=0\)

=>\(2^x=1\)

=>x=0

\(3^x+9\cdot3^{-x}=10\)

=>\(3^x+\dfrac{9}{3^x}=10\)

=>\(\dfrac{\left(3^x\right)^2+9}{3^x}=10\)

=>\(\left(3^x\right)^2+9=10\cdot3^x\)

=>\(\left(3^x\right)^2-10\cdot3^x+9=0\)

=>\(\left(3^x-1\right)\left(3^x-9\right)=0\)

=>\(\left[{}\begin{matrix}3^x-1=0\\3^x-9=0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}3^x=1\\3^x=9\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

a: \(\left(\sqrt{3}\right)^x=243\)

=>\(3^{\dfrac{1}{2}\cdot x}=3^5\)

=>\(\dfrac{1}{2}\cdot x=5\)

=>x=10

b: \(0,1^x=1000\)

=>\(\left(\dfrac{1}{10}\right)^x=1000\)

=>\(10^{-x}=10^3\)

=>-x=3

=>x=-3

c: \(\left(0,2\right)^{x+3}< \dfrac{1}{5}\)

=>\(\left(0,2\right)^{x+3}< 0,2\)

=>x+3>1

=>x>-2

d: \(\left(\dfrac{3}{5}\right)^{2x+1}>\left(\dfrac{5}{3}\right)^2\)

=>\(\left(\dfrac{3}{5}\right)^{2x+1}>\left(\dfrac{3}{5}\right)^{-2}\)

=>2x+1<-2

=>2x<-3

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

e: \(5^{x-1}+5^{x+2}=3\)

=>\(5^x\cdot\dfrac{1}{5}+5^x\cdot25=3\)

=>\(5^x=\dfrac{3}{25,2}=\dfrac{1}{8,4}=\dfrac{10}{84}=\dfrac{5}{42}\)

=>\(x=log_5\left(\dfrac{5}{42}\right)=1-log_542\)

\(A=\left(lna+log_{\alpha}e\right)^2+ln^2a-\log_a^2e\)

\(=ln^2a+\log_{\alpha}^2e+2\cdot lna\cdot\log_{\alpha}e+ln^2a-\log_{\alpha}^2e\)

\(=2\cdot\log_e^2\alpha+2\cdot\log_e\alpha\cdot\log_{\alpha}e\)

\(=2\cdot ln^2\alpha+2\)