22x+1+ 22x+2=3.211
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NQ
1
26 tháng 7 2023
Đặt \(A=2^{2x}+2^{2x+1}+...+2^{2x+1918}\)
=>\(2\cdot A=2^{2x+1}+2^{2x+2}+...+2^{2x+1919}\)
=>\(A=2^{2x+1919}-2^{2x}\)
Theo đề, ta có; \(2^{2x+1919}-2^{2x}=2^{1923}-2^4\)
=>\(2^{2x}\cdot\left(2^{2019}-1\right)=2^4\left(2^{2019}-1\right)\)
=>2x=4
=>x=2
PT
1
NA
2
PN
1
15 tháng 7 2022
Câu 1:
\(\Leftrightarrow2\cdot\left(1-\sin^22x\right)+5\sin2x-4=0\)
\(\Leftrightarrow-2\sin^22x+5\sin2x-2=0\)
\(\Leftrightarrow\left(\sin2x-2\right)\left(2\sin2x-1\right)=0\)
\(\Leftrightarrow\sin2x=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{\Pi}{6}+k2\Pi\\2x=\dfrac{5}{6}\Pi+k2\Pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\Pi}{12}+k\Pi\\x=\dfrac{5}{12}\Pi+k\Pi\end{matrix}\right.\)
VM
0
HP
3 tháng 9 2021
1.
\(3sin^22x-2sin2x.cos2x-4cos^22x=2\)
\(\Leftrightarrow-\dfrac{3}{2}\left(1-2sin^22x\right)-2sin2x.cos2x-2\left(2cos^22x-1\right)=\dfrac{5}{2}\)
\(\Leftrightarrow sin4x+\dfrac{7}{2}cos4x=-\dfrac{5}{2}\)
\(\Leftrightarrow\dfrac{\sqrt{53}}{2}\left(\dfrac{2}{\sqrt{53}}sin4x+\dfrac{7}{\sqrt{53}}cos4x\right)=-\dfrac{5}{2}\)
\(\Leftrightarrow sin\left(4x+arccos\dfrac{2}{\sqrt{53}}\right)=-\dfrac{5}{\sqrt{53}}\)
\(\Leftrightarrow\left[{}\begin{matrix}4x+arccos\dfrac{2}{\sqrt{53}}=arcsin\left(-\dfrac{5}{\sqrt{53}}\right)+k2\pi\\4x+arccos\dfrac{2}{\sqrt{53}}=\pi-arcsin\left(-\dfrac{5}{\sqrt{53}}\right)+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{4}arccos\dfrac{2}{\sqrt{53}}+\dfrac{1}{4}arcsin\left(-\dfrac{5}{\sqrt{53}}\right)+\dfrac{k\pi}{2}\\x=\dfrac{\pi}{4}-\dfrac{1}{4}arccos\dfrac{2}{\sqrt{53}}-\dfrac{1}{4}arcsin\left(-\dfrac{5}{\sqrt{53}}\right)+\dfrac{k\pi}{2}\end{matrix}\right.\)
HP
3 tháng 9 2021
2.
\(2\sqrt{3}cos^2x+6sinx.cosx=3+\sqrt{3}\)
\(\Leftrightarrow\sqrt{3}\left(2cos^2x-1\right)+6sinx.cosx=3\)
\(\Leftrightarrow\sqrt{3}cos2x+3sin2x=3\)
\(\Leftrightarrow2\sqrt{3}\left(\dfrac{1}{2}cos2x+\dfrac{\sqrt{3}}{2}sin2x\right)=3\)
\(\Leftrightarrow cos\left(2x-\dfrac{\pi}{3}\right)=\dfrac{\sqrt{3}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{\pi}{3}=\dfrac{\pi}{6}+k2\pi\\2x-\dfrac{\pi}{3}=-\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\x=\dfrac{\pi}{12}+k\pi\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
17 tháng 9 2020
ĐKXĐ: \(x\ne\frac{k\pi}{2}\)
\(\frac{4sin^2x.cos^2x-4sin^2x}{4sin^2x.cos^2x+4sin^2x}+1=2tan^2x\)
\(\Leftrightarrow\frac{4sin^2x\left(cos^2x-1\right)}{4sin^2x\left(cos^2x+1\right)}+1=\frac{2sin^2x}{cos^2x}\)
\(\Leftrightarrow\frac{cos^2x}{cos^2x+1}=\frac{1-cos^2x}{cos^2x}\)
Đặt \(cos^2x=t\Rightarrow0< t< 1\)
\(\Rightarrow\frac{t}{t+1}=\frac{1-t}{t}\Leftrightarrow t^2=1-t^2\Leftrightarrow t^2=\frac{1}{2}\)
\(\Leftrightarrow t=\frac{\sqrt{2}}{2}\Leftrightarrow cos^2x=\frac{\sqrt{2}}{2}\)
SB
0
x = 6
22x+1 + 22x+2 = 3.211
22x+1 (1 +2) = 3.211
22x+1..3 = 3,211
22x+1 = 211
2x + `1 = 11
2x = 10
x = 10: 2
x = 5