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a) \({2^x} > 16 \Leftrightarrow {2^x} > {2^4} \Leftrightarrow x > 4\) (do \(2 > 1\)) .
b) \(0,{1^x} \le 0,001 \Leftrightarrow 0,{1^x} \le 0,{1^3} \Leftrightarrow x \ge 3\) (do \(0 < 0,1 < 1\)).
c) \({\left( {\frac{1}{5}} \right)^{x - 2}} \ge {\left( {\frac{1}{{25}}} \right)^x} \Leftrightarrow {\left( {\frac{1}{5}} \right)^{x - 2}} \ge {\left( {{{\left( {\frac{1}{5}} \right)}^2}} \right)^x} \Leftrightarrow {\left( {\frac{1}{5}} \right)^{x - 2}} \ge {\left( {\frac{1}{5}} \right)^{2x}} \Leftrightarrow x - 2 \le 2{\rm{x}}\) (do \(0 < \frac{1}{5} < 1\))
\( \Leftrightarrow x \ge - 2\).
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
c.
\(\Leftrightarrow cos\left(x+12^0\right)+cos\left(90^0-78^0+x\right)=1\)
\(\Leftrightarrow2cos\left(x+12^0\right)=1\)
\(\Leftrightarrow cos\left(x+12^0\right)=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+12^0=60^0+k360^0\\x+12^0=-60^0+k360^0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=48^0+k360^0\\x=-72^0+k360^0\end{matrix}\right.\)
2.
Do \(-1\le sin\left(3x-27^0\right)\le1\) nên pt có nghiệm khi:
\(\left\{{}\begin{matrix}2m^2+m\ge-1\\2m^2+m\le1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2m^2+m+1\ge0\left(luôn-đúng\right)\\2m^2+m-1\le0\end{matrix}\right.\)
\(\Rightarrow-1\le m\le\dfrac{1}{2}\)
a.
\(\Rightarrow\left[{}\begin{matrix}x+15^0=arccos\left(\dfrac{2}{5}\right)+k360^0\\x+15^0=-arccos\left(\dfrac{2}{5}\right)+k360^0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-15^0+arccos\left(\dfrac{2}{5}\right)+k360^0\\x=-15^0-arccos\left(\dfrac{2}{5}\right)+k360^0\end{matrix}\right.\)
b.
\(2x-10^0=arccot\left(4\right)+k180^0\)
\(\Rightarrow x=5^0+\dfrac{1}{2}arccot\left(4\right)+k90^0\)
a)
ĐK: \(\left\{{}\begin{matrix}2x-4>0\\x-1>0\end{matrix}\right.\Leftrightarrow x>1\)
\(\log_5\left(2x-4\right)+\log_{\dfrac{1}{5}}\left(x-1\right)=0\\ \Leftrightarrow\log_5\left(2x-4\right)-\log_5\left(x-1\right)=0\\ \Leftrightarrow\log_5\left(\dfrac{2x-4}{x-1}\right)=\log_51\\ \Leftrightarrow\dfrac{2x-4}{x-1}=1\\ \Leftrightarrow2x-4=x-1\\ \Leftrightarrow x=3\left(tm\right)\)
Vậy x = 3.
b) ĐK: x > 0
\(\log_2x+\log_4x=3\\ \Leftrightarrow\log_2x+\dfrac{1}{2}\log_2x=3\\ \Leftrightarrow\left(1+\dfrac{1}{2}\right)\log_2x=3\\ \Leftrightarrow\dfrac{3}{2}\log_2x=3\\ \Leftrightarrow\log_2x=2\\ \Leftrightarrow x=4\left(tm\right)\)
Vậy x= 4
\(a,\left(\dfrac{1}{3}\right)^{2x+1}\le9\\ \Leftrightarrow2x+1\ge-2\\ \Leftrightarrow2x\ge-3\\ \Leftrightarrow x\ge-\dfrac{3}{2}\)
\(b,4^x>2^{x-2}\\ \Leftrightarrow2^{2x}>2^{x-2}\\ \Leftrightarrow2x>x-2\\ \Leftrightarrow x>-2\)
a) \({3^{x + 2}} = \sqrt[3]{9} \Leftrightarrow {3^{x + 2}} = {9^{\frac{1}{3}}} \Leftrightarrow {3^{x + 2}} = {\left( {{3^2}} \right)^{\frac{1}{3}}} \Leftrightarrow {3^{x + 2}} = {3^{\frac{2}{3}}} \Leftrightarrow x + 2 = \frac{2}{3} \Leftrightarrow x = - \frac{4}{3}\)
b) \({2.10^{2{\rm{x}}}} = 30 \Leftrightarrow {10^{2{\rm{x}}}} = 15 \Leftrightarrow 2{\rm{x}} = \log 15 \Leftrightarrow x = \frac{1}{2}\log 15\)
c) \({4^{2{\rm{x}}}} = {8^{2{\rm{x}} - 1}} \Leftrightarrow {\left( {{2^2}} \right)^{2{\rm{x}}}} = {\left( {{2^3}} \right)^{2{\rm{x}} - 1}} \Leftrightarrow {2^{4{\rm{x}}}} = {2^{6{\rm{x}} - 3}} \Leftrightarrow 4{\rm{x}} = 6{\rm{x}} - 3 \Leftrightarrow - 2{\rm{x}} = - 3 \Leftrightarrow x = \frac{3}{2}\).
\(a,3^{x-1}=27\\ \Leftrightarrow3^{x-1}=3^3\\ \Leftrightarrow x-1=3\\ \Leftrightarrow x=4\\ b,100^{2x^2-3}=0,1^{2x^2-18}\\ \Leftrightarrow10^{4x^2-6}=10^{-2x^2+18}\\ \Leftrightarrow4x^2-6=-2x^2+18\\ \Leftrightarrow6x^2=24\\ \Leftrightarrow x^2=4\\ \Leftrightarrow x=\pm2\)
\(c,\sqrt{3}e^{3x}=1\\ \Leftrightarrow e^{3x}=\dfrac{1}{\sqrt{3}}\\ \Leftrightarrow3x=ln\left(\dfrac{1}{\sqrt{3}}\right)\\ \Leftrightarrow x=\dfrac{1}{3}ln\left(\dfrac{1}{\sqrt{3}}\right)\)
\(d,5^x=3^{2x-1}\\ \Leftrightarrow2x-1=log_35^x\\ \Leftrightarrow2x-1-xlog_35=0\\ \Leftrightarrow x\left(2-log_35\right)=1\\ \Leftrightarrow x=\dfrac{1}{2-log_35}\)
\(a,3^{1-2x}=4^x\\ \Leftrightarrow1-2x=log_34^x\\ \Leftrightarrow1-2x=xlog_34\\ \Leftrightarrow2x+xlog_34=1\\ \Leftrightarrow x\left(2+log_34\right)=1\\ \Leftrightarrow x=\dfrac{1}{2+log_34}=\dfrac{1}{log_39+log_34}=\dfrac{1}{log_336}=log_{36}3\)
b, ĐK: \(x>-1\)
\(log_3\left(x+1\right)+log_3\left(x+4\right)=2\\ \Leftrightarrow log_3\left(x^2+5x+4\right)=2\\ \Leftrightarrow x^2+5x+4=9\\ \Leftrightarrow x^2+5x-5=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-5+3\sqrt{5}}{2}\left(tm\right)\\x=\dfrac{-5-3\sqrt{5}}{2}\left(ktm\right)\end{matrix}\right.\)
Đề bài: Giải hệ phương trình:
\(\left\{{}\begin{matrix}y^3-12y-x^3+6x^2-16=0\left(1\right)\\4y^2+2\sqrt{4-y^2}-5\sqrt{4x-x^2}+6=0\left(2\right)\end{matrix}\right.\).
Giải:
ĐKXĐ: \(\left\{{}\begin{matrix}0\le x\le4\\-2\le y\le2\end{matrix}\right.\).
\(\left(1\right)\Leftrightarrow y^3-12y=\left(x-2\right)^3-12\left(x-2\right)\)
\(\Leftrightarrow\left(x-2-y\right)\left[\left(x-2\right)^2+\left(x-2\right)y+y^2-12\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=y+2\\x^2+xy+y^2-4x-2y-8=0\end{matrix}\right.\).
+) TH1: \(x=y+2\): Thay vào (2) ta được:
\(4y^2+2\sqrt{4-y^2}-5\sqrt{4\left(y+2\right)-\left(y+2\right)^2}+6=0\)
\(\Leftrightarrow4y^2+2\sqrt{4-y^2}-5\sqrt{4-y^2}+6=0\)
\(\Leftrightarrow4y^2+6=3\sqrt{4-y^2}\)
\(\Leftrightarrow\left(4y^2+6\right)^2=9\left(4-y^2\right)\)
\(\Leftrightarrow16y^4+57y^2=0\)
\(\Leftrightarrow y=0\Rightarrow x=2\) (TMĐK).
+) TH2: \(x^2+xy+y^2-4x-2y-8=0\):
\(\Leftrightarrow\left(x-2\right)^2+y^2+\left(x-2\right)y=12\).
Do VT \(\le12\) (Đẳng thức xảy ra khi và chỉ khi x = 4; y = 2 hoặc x = 0; y = -2).
Do đó \(\left[{}\begin{matrix}x=4;y=2\\x=0;y=-2\end{matrix}\right.\).
Thử lại không có gt nào thỏa mãn.
Vậy...
a)
\(9^{16-x}=27^{x+4}\\ \Leftrightarrow3^{2.\left(16-x\right)}=3^{3.\left(x+4\right)}\\ \Leftrightarrow2.\left(16-x\right)=3.\left(x+4\right)\\ \Leftrightarrow32-2x-3x-12=0\\ \Leftrightarrow-5x=-20\Leftrightarrow x=4\)
b)
\(16^{x-2}=0,25.2^{-x+4}\\ \Leftrightarrow2^{4\left(x-2\right)}=0,25.2^{-x+4}\\ \Leftrightarrow2^{4x-8+x-4}=0,25\\ \Leftrightarrow2^{5x-12}=0,25\Leftrightarrow5x-12=\log_20,25\\ \Leftrightarrow5x-12=-2\\ \Leftrightarrow x=2\)