Chương 2: HÀM SỐ LŨY THỪA. HÀM SỐ MŨ VÀ HÀM SỐ LÔGARIT

\(y=ln\left(a^{1000}\right)+ln\left(b^{1000}\right)=ln\left(ab\right)^{100}\)

Mà \(a^2-ab+b^2-ab=\left(a-b\right)^2\ge0;\forall a;b\)

\(\Rightarrow a^2-ab+b^2\ge ab\Rightarrow ln\left(a^2-ab+b^2\right)^{1000}\ge ln\left(ab\right)^{1000}\)

\(\Rightarrow x\ge y\)

Đặt \(log_4x=log_6y=log_9\left(x+y\right)=k\)

\(\Rightarrow\left\{{}\begin{matrix}x=4^k\\y=6^k\\x+y=9^k\end{matrix}\right.\)

\(\Rightarrow4^k+6^k=9^k\)

\(\Leftrightarrow\left(\dfrac{4}{9}\right)^k+\left(\dfrac{6}{9}\right)^k=1\)

\(\Leftrightarrow\left(\dfrac{2}{3}\right)^{2k}+\left(\dfrac{2}{3}\right)^k-1=0\)

Đặt \(\left(\dfrac{2}{3}\right)^k=t>0\)

\(\Rightarrow t^2+t-1=0\Rightarrow t=\dfrac{-1+\sqrt{5}}{2}\)

\(\dfrac{x}{y}=\dfrac{4^k}{6^k}=\left(\dfrac{2}{3}\right)^k=t=\dfrac{-1+\sqrt{5}}{2}\)

\(f\left(x\right)=-x^2-mx+4m-5>0\); \(\forall x\in(-1;2]\)

\(\Leftrightarrow\min\limits_{(-1;2]}f\left(x\right)>0\)

TH1: \(-\dfrac{m}{2}\le-1\Rightarrow m\ge2\)

Khi đó hàm nghịch biến trên khoảng xét nên \(f\left(x\right)_{min}=f\left(2\right)=2m-9>0\Rightarrow m>\dfrac{9}{2}\)

TH2: \(-\dfrac{m}{2}\ge2\Rightarrow m\le-4\)

Khi đó hàm đồng biến trên khoảng xét nên \(f\left(x\right)_{min}=f\left(-1\right)=5m-6>0\Rightarrow m>\dfrac{6}{5}\left(ktm\right)\)

TH3: \(-1< -\dfrac{m}{2}< 2\Rightarrow-4< m< 2\)

Khi đó \(f\left(x\right)_{min}=min\left(f\left(-1\right);f\left(2\right)\right)=min\left(2m-9;5m-6\right)\)

- TH3.1: \(2m-9\ge5m-6>0\) (ko tồn tại m thỏa mãn)

- TH3.2: \(5m-6\ge2m-9>0\Rightarrow m>\dfrac{9}{2}\) (ktm)

Vậy \(m>\dfrac{9}{2}\)

x;y;z lập thành cấp số nhân nên \(y^2=xz\)

3 số kia lập thành cấp số cộng nên \(2log_{\sqrt{a}}y=log_ax+log_{\sqrt[3]{a}}z\)

\(\Leftrightarrow4log_ay=log_ax+3log_az\)

\(\Leftrightarrow log_ay^4=log_a\left(xz^3\right)\)

\(\Rightarrow y^4=xz^3\)

\(\Rightarrow\left(xz\right)^2=xz^3\)

\(\Rightarrow x=z\)

\(\Rightarrow y^2=x^2\Rightarrow y=x\)

\(\Rightarrow P=1959+2019+60\)

ĐKXĐ: \(x\ge-1\)

\(log_2\left(\sqrt{x^2-x+1}+\sqrt{x+1}\right)=log_2\left(x+2\right)\)

\(\Rightarrow\sqrt{x^2-x+1}+\sqrt{x+1}=x+2\)

\(\Leftrightarrow\sqrt{x^2-x+1}=x+2-\sqrt{x+1}\)

\(\Leftrightarrow x^2-x+1=x^2+4x+4+x+1-2\left(x+2\right)\sqrt{x+1}\)

\(\Leftrightarrow\left(x+2\right)\sqrt{x+1}=3x+2\) \(\left(x\ge-\dfrac{2}{3}\right)\)

\(\Leftrightarrow\left(x+2\right)^2\left(x+1\right)=\left(3x+2\right)^2\)

\(\Leftrightarrow x\left(x^2-4x-4\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=2+2\sqrt{2}\\x=2-2\sqrt{2}\left(loại\right)\end{matrix}\right.\)

Bài này dùng BBT được:

\(x^2-4mx+m-1>0\) ; \(\forall x\in\left[-3;2\right]\)

\(\Leftrightarrow\min\limits_{\left[-3;2\right]}f\left(x\right)>0\)

Xét \(f\left(x\right)=x^2-4mx+m-1\Rightarrow f'\left(x\right)=2x-4m=0\Rightarrow x=2m\)

BBT:

TH1: \(2m\le-3\Rightarrow m\le-\dfrac{3}{2}\)

Khi đó  \(\min\limits_{\left[-3;2\right]}f\left(x\right)=f\left(-3\right)=13m+8>0\Rightarrow m>-\dfrac{8}{13}\) (ktm)

TH2: \(2m\ge2\Rightarrow m\ge1\)

Khi đó \(\min\limits_{\left[-3;2\right]}f\left(x\right)=f\left(2\right)=3-7m>0\Rightarrow m< \dfrac{3}{7}\) (ktm)

TH3: \(-3< 2m< 2\Rightarrow-\dfrac{3}{2}< m< 1\)

Khi đó \(\min\limits_{\left[-3;2\right]}f\left(x\right)=f\left(2m\right)=-4m^2+m-1>0\Rightarrow\) ko tồn tại m thỏa mãn

Vậy ko tồn tại m thỏa mãn yêu cầu

\(\left\{{}\begin{matrix}mx+3>0\\3m-5-x>0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}mx>-3\\x< 3m-5\end{matrix}\right.\)

- Với \(m=0\Rightarrow x< -5\) ko thỏa mãn

- Với \(m>0\Rightarrow\left\{{}\begin{matrix}x>-\dfrac{3}{m}\\x< 3m-5\end{matrix}\right.\)

\(\Rightarrow-\dfrac{3}{m}< x< 3m-5\)

\(\Rightarrow\left\{{}\begin{matrix}-1>-\dfrac{3}{m}\\3m-5\ge2\\\end{matrix}\right.\) \(\Rightarrow\dfrac{7}{3}< m< 3\)

- Với \(m< 0\Rightarrow\left\{{}\begin{matrix}x< -\dfrac{3}{m}\\x< 3m-5\end{matrix}\right.\) \(\Rightarrow x< 3m-5\)

\(\Rightarrow3m-5\ge2\Rightarrow m\ge\dfrac{7}{3}\) (ktm)

Đặt \(log_4\left(2a+3b\right)=log_{10}a=log_{25}b=k\)

\(\Rightarrow\left\{{}\begin{matrix}a=10^k\\b=25^k\\2a+3b=4^k\end{matrix}\right.\) \(\Rightarrow2.10^k+3.25^k=4^k\)

\(\Rightarrow2.\left(\dfrac{5}{2}\right)^k+3.\left(\dfrac{25}{4}\right)^k=1\)

Đặt \(\left(\dfrac{5}{2}\right)^k=x>0\)

\(\Rightarrow3x^2+2x-1=0\Rightarrow\left[{}\begin{matrix}x=-1\left(loại\right)\\x=\dfrac{1}{3}\end{matrix}\right.\)

\(\Rightarrow\left(\dfrac{5}{2}\right)^k=\dfrac{1}{3}\) \(\Rightarrow\dfrac{b}{a}=\left(\dfrac{25}{10}\right)^k=\left(\dfrac{5}{2}\right)^k=\dfrac{1}{3}\)

\(P=\dfrac{1-\left(\dfrac{b}{a}\right)^2+\left(\dfrac{b}{a}\right)^3}{1+\left(\dfrac{b}{a}\right)^2-3\left(\dfrac{b}{a}\right)^3}=...\)

ĐKXĐ: \(mx+3m-5>0\)

\(\Leftrightarrow mx>-3m+5\)

- Với \(m=0\) ko thỏa mãn

- Với \(m< 0\Rightarrow x< \dfrac{-3m+5}{m}\)

\(\Rightarrow\left\{{}\begin{matrix}m< 0\\\dfrac{-3m+5}{m}\ge3\end{matrix}\right.\) \(\Rightarrow\) ko tồn tại m thỏa mãn

- Với \(m>0\Rightarrow x>\dfrac{-3m+5}{m}\)

\(\Rightarrow\left\{{}\begin{matrix}m>0\\\dfrac{-3m+5}{m}< -1\end{matrix}\right.\) \(\Rightarrow m>\dfrac{5}{2}\)

Cách 2:

\(mx+3m-5>0\Leftrightarrow m\left(x+3\right)>5\)

\(\Rightarrow m>\dfrac{5}{x+3}\) (do \(x+3>0\) trên khoảng xét)

\(\Rightarrow m>\max\limits_{[-1;3)}\dfrac{5}{x+3}=\dfrac{5}{2}\)

a/

Đặt $\log_2b=m\Rightarrow b=2^m$

Đặt $\log_ba=n\Rightarrow a=b^n$

$\Rightarrow \log_2b.\log_ba=mn(1)$

$\log_2a=\log_2(b^n)=n\log_2b=nm(2)$

Từ $(1); (2)$ ta có đpcm

b/

$\log_a(\sqrt{a})=\log_a(a^{\frac{1}{2}})=\frac{1}{2}$

$\log_{b^2}(b)=\log_{b^2}[(b^2)^{\frac{1}{2}}]=\frac{1}{2}$

$\Rightarrow \log_a(\sqrt{a})=\log_{b^2}(b)$