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Bài 1:
\(A=\log_380=\log_3(2^4.5)=\log_3(2^4)+\log_3(5)\)
\(=4\log_32+\log_35=4a+b\)
\(B=\log_3(37,5)=\log_3(2^{-1}.75)=\log_3(2^{-1}.3.5^2)\)
\(=\log_3(2^{-1})+\log_33+\log_3(5^2)=-\log_32+1+2\log_35\)
\(=-a+1+2b\)
Bài 2:
\(\log_{30}8=\frac{\log 8}{\log 30}=\frac{\log (2^3)}{\log (10.3)}=\frac{3\log2}{\log 10+\log 3}\)
\(=\frac{3\log (\frac{10}{5})}{1+\log 3}=\frac{3(\log 10-\log 5)}{1+\log 3}=\frac{3(1-b)}{1+a}\)
ĐKXĐ: \(x>0\)
\(\Leftrightarrow log_2^22x+log_2\left(\frac{2x}{8}\right)-9< 0\)
\(\Leftrightarrow log^2_22x+log_22x-12< 0\)
\(\Leftrightarrow-4< log_22x< 3\)
\(\Leftrightarrow\frac{1}{32}< x< 4\)
ĐKXĐ: \(6-5^x>0\Rightarrow5^x< 6\)
\(log_5\left(6-5^x\right)=1-x\Leftrightarrow6-5^x=5^{1-x}\)
\(\Leftrightarrow5^x-6+\frac{5}{5^x}=0\Leftrightarrow\left(5^x\right)^2-6.5^x+5=0\)
\(\Rightarrow\left[{}\begin{matrix}5^x=1\\5^x=5\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\) \(\Rightarrow\sum x=0+1=1\)
ĐKXĐ: \(x>0\)
\(\Leftrightarrow log_2^2\left(2x\right)+log_2\left(2x\right)-log_28-9< 0\)
\(\Leftrightarrow log_2^2\left(2x\right)+log_2\left(2x\right)-12< 0\)
\(\Leftrightarrow\left(log_2\left(2x\right)+4\right)\left(log_2\left(2x\right)-3\right)< 0\)
\(\Leftrightarrow-4< log_2\left(2x\right)< 3\)
\(\Leftrightarrow\frac{1}{16}< 2x< 8\Leftrightarrow\frac{1}{32}< x< 4\)
M∈ (S) : (x0 - 2)2 + (y0-1)2 +(z0-1)2 =9.
A=x0+2y0+2z0=(x0-2)+2(y0-1)+2(z0-1)+6
Dùng BĐT bunhiacopski
[(x0-2)+2(y0-1)+2(z0-1)]2 ≤ (1+4+4).[(x0 - 2)2 + (y0-1)2 +(z0-1)2 ]
≤ 81
-9 ≤ (x0-2)+2(y0-1)+2(z0-1) ≤ 9.
-3 ≤ A ≤ 12. vậy GTNN của A = -3.
Dấu bằng xảy ra khi :
x0+2y0+2z0 = -3
và \(\dfrac{x0-2}{1}=\dfrac{y0-1}{1}=\dfrac{z0-1}{1}\)
Giải hệ được x0=1, y0=z0=-1. Suy ra: x0+y0+z0 = -1
Điều kiện xác định : 3\(^x\)>2
Ta có: \(\log_2\left(4.3^x-6\right)=\log_2\left(2\sqrt{2}\right).\log_{2\sqrt{2}}\left(4.3^x-6\right)\)
\(\log_2\left(4.3^x-6\right)-\dfrac{3}{2}\log_{2\sqrt{2}}\left(9^x-6\right)=1\left(1\right)\)\(\Leftrightarrow\log_2\left(2\sqrt{2}\right)\log_{2\sqrt{2}}\left(4.3^x-6\right)-\dfrac{3}{2}\log_{2\sqrt{2}}\left(9^x-6\right)=1\)
\(\Rightarrow\dfrac{3}{2}\log_{2\sqrt{2}}\left(4.3^x-6\right)-\dfrac{3}{2}\log_{2\sqrt{2}}\left(9^x-6\right)=1\)\(\Leftrightarrow\dfrac{3}{2}[\log_{2\sqrt{2}}\left(4.3^x-6\right)-\log_{2\sqrt{2}}\left(9^X-6\right)]=1\)
\(\Leftrightarrow\log_{2\sqrt{2}}\left(\dfrac{4.3^X-6}{9^X-6}\right)=\dfrac{2}{3}\)\(\Leftrightarrow\log_{2\sqrt{2}}\left(\dfrac{4.3^X-6}{9^X-6}\right)=\log_{2\sqrt{2}}\left(2\right)\)
\(\Leftrightarrow\dfrac{4.3^X-6}{9^X-6}=2\Leftrightarrow4.3^X-6=2.9^X-12\)\(\Leftrightarrow2.(3^X)^2-4.3^X-6=0\Rightarrow\left[{}\begin{matrix}3^X=3\left(TM\right)\\3^X=-1\left(loai\right)\end{matrix}\right.\)
\(\Rightarrow x=1.\)Vậy x=1 là nghiệm của phương trình (1)
