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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...
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.
\(2x-10^0=arccot\left(4\right)+k180^0\)
\(\Rightarrow x=5^0+\dfrac{1}{2}arccot\left(4\right)+k90^0\)
Nhân 2 vế với \(sin4x\) sau đó tách:
\(\frac{sin4x}{cosx}+\frac{sin4x}{sin2x}=\frac{2sin2x.cos2x}{cosx}+\frac{2sin2x.cos2x}{sin2x}=\frac{4sinx.cosx.cos2x}{cosx}+\frac{2sin2x.cos2x}{sin2x}\)
Rồi rút gọn
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) cosx - √3sinx = √2 ⇔ cosx - tan
sinx = √2
⇔ coscosx - sinsinx = √2cos ⇔ cos(x + ) = 
⇔ 
b) 3sin3x - 4cos3x = 5 ⇔ 
sin3x - 
cos3x = 1.
Đặt α = arccos thì phương trình trở thành
cosαsin3x - sinαcos3x = 1 ⇔ sin(3x - α) = 1 ⇔ 3x - α = 
+ k2π
⇔ x = 
, k ∈ Z (trong đó α = arccos).
a: \(6^x=5\)
=>\(x=log_65\)
b: \(7^{3-x}=5\)
=>\(3-x=log_75\)
=>\(x=3-log_75\)
c: \(\left(\dfrac{3}{5}\right)^{x-2}=\dfrac{27}{125}\)
=>\(\left(\dfrac{3}{5}\right)^{x-2}=\left(\dfrac{3}{5}\right)^3\)
=>x-2=3
=>x=5
d: \(\left(\dfrac{4}{5}\right)^x=\dfrac{5}{4}\)
=>\(\left(\dfrac{4}{5}\right)^x=\left(\dfrac{4}{5}\right)^{-1}\)
=>x=-1
a.
\(6^x=5\Rightarrow x=log_65\)
b.
\(7^{3-x}=5\Rightarrow3-x=log_75\)
\(\Rightarrow x=3-log_75\)
c.
\(\left(\dfrac{3}{5}\right)^{x-2}=\dfrac{27}{125}\Rightarrow x-2=log_{\dfrac{3}{5}}\left(\dfrac{27}{125}\right)\)
\(\Rightarrow x-2=3\Rightarrow x=5\)
d.
\(\left(\dfrac{4}{5}\right)^x=\dfrac{5}{4}\Rightarrow\left(\dfrac{4}{5}\right)^x=\left(\dfrac{4}{5}\right)^{-1}\)
\(\Rightarrow x=-1\)
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)
\(16^{x-2}=0,25.2^{-x+4}\\ \Leftrightarrow2^{4\left(x-2\right)}=0,25.2^{-x+4}\\ \Leftrightarrow2^{4x-8+x-4}=0,25\\ \Leftrightarrow2^{5x-12}=0,25\Leftrightarrow5x-12=\log_20,25\\ \Leftrightarrow5x-12=-2\\ \Leftrightarrow x=2\)