Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1:
a: ĐKXĐ: \(x< >\dfrac{\Omega}{2}+k\Omega\)
=>TXĐ: \(D=R\backslash\left\{\dfrac{\Omega}{2}+k\Omega\right\}\)
b: ĐKXĐ: \(x< >k\Omega\)
=>TXĐ: \(D=R\backslash\left\{k\Omega\right\}\)
c: ĐKXĐ: \(2x< >\dfrac{\Omega}{2}+k\Omega\)
=>\(x< >\dfrac{\Omega}{4}+\dfrac{k\Omega}{2}\)
TXĐ: \(D=R\backslash\left\{\dfrac{\Omega}{4}+\dfrac{k\Omega}{2}\right\}\)
d: ĐKXĐ: \(3x< >\Omega\cdot k\)
=>\(x< >\dfrac{k\Omega}{3}\)
TXĐ: \(D=R\backslash\left\{\dfrac{k\Omega}{3}\right\}\)
e: ĐKXĐ: \(x+\dfrac{\Omega}{3}< >\dfrac{\Omega}{2}+k\Omega\)
=>\(x< >\dfrac{\Omega}{6}+k\Omega\)
TXĐ: \(D=R\backslash\left\{\dfrac{\Omega}{6}+k\Omega\right\}\)
f: ĐKXĐ: \(x-\dfrac{\Omega}{6}< >\Omega\cdot k\)
=>\(x< >k\Omega+\dfrac{\Omega}{6}\)
TXĐ: \(D=R\backslash\left\{k\Omega+\dfrac{\Omega}{6}\right\}\)
d.
\(-1\le sin2x\le1\Rightarrow2\le y\le1+\sqrt{3}\)
\(y_{min}=2\) khi \(sin2x=-1\)
\(y_{max}=1+\sqrt{3}\) khi \(sin2x=1\)
e.
\(0\le sin^2x\le1\Rightarrow\frac{4}{3}\le y\le2\)
\(y_{min}=\frac{4}{3}\) khi \(sin^2x=1\)
\(y_{max}=2\) khi \(sinx=0\)
a.
\(0\le cos^2x\le1\Rightarrow2\le y\le1+\sqrt{3}\)
\(y_{min}=2\) khi \(cosx=0\)
\(y_{max}=1+\sqrt{3}\) khi \(cos^2x=1\)
b.
\(-1\le sin\left(2x-\frac{\pi}{4}\right)\le1\Rightarrow-2\le y\le4\)
\(y_{min}=-2\) khi \(sin\left(2x-\frac{\pi}{4}\right)=-1\)
\(y_{max}=4\) khi \(sin\left(2x-\frac{\pi}{4}\right)=1\)
c.
\(0\le cos^23x\le1\Rightarrow1\le y\le3\)
\(y_{min}=1\) khi \(cos^23x=1\)
\(y_{max}=3\) khi \(cos3x=0\)
e/
Đề câu này chắc chắn đúng chứ bạn?
f/
\(sin^4x+cos^4x=\frac{3}{4}\)
\(\Leftrightarrow\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x=\frac{3}{4}\)
\(\Leftrightarrow1-\frac{1}{2}\left(2sinx.cosx\right)^2=\frac{3}{4}\)
\(\Leftrightarrow\frac{1}{4}-\frac{1}{2}sin^22x=0\)
\(\Leftrightarrow1-2sin^22x=0\)
\(\Leftrightarrow cos4x=0\)
\(\Leftrightarrow x=\frac{\pi}{8}+\frac{k\pi}{4}\)
c/
\(y=sin\left(4x-\frac{\pi}{3}\right)+sin\left(\frac{\pi}{3}\right)+5\)
\(=sin\left(4x-\frac{\pi}{3}\right)+\frac{\sqrt{3}}{2}+5\)
Do \(-1\le sin\left(4x-\frac{\pi}{3}\right)\le1\)
\(\Rightarrow4+\frac{\sqrt{3}}{2}\le y\le6+\frac{\sqrt{3}}{2}\)
d/
\(y=\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)+3sin2x+5\)
\(y=6-3sin^2x.cos^2x+3sin2x\)
\(y=-\frac{3}{4}sin^22x+3sin2x+6\)
\(y=\frac{3}{4}\left(sin2x+1\right)\left(5-sin2x\right)+\frac{9}{4}\ge\frac{9}{4}\)
\(y_{min}=\frac{9}{4}\) khi \(sin2x=-1\)
\(y=\frac{3}{4}\left(sin2x-1\right)\left(3-sin2x\right)+\frac{33}{4}\le\frac{33}{4}\)
\(y_{max}=\frac{33}{4}\) khi \(sin2x=1\)
a.
\(-1\le sinx\le1\Rightarrow-7\le y\le-3\)
\(y_{min}=-7\) khi \(sinx=-1\)
\(y_{max}=-3\) khi \(sinx=1\)
b.
\(-1\le cos\left(x+\frac{\pi}{3}\right)\le1\Rightarrow1\le y\le5\)
\(y_{min}=1\) khi \(cos\left(x+\frac{\pi}{3}\right)=-1\)
\(y_{max}=5\) khi \(cos\left(x+\frac{\pi}{3}\right)=1\)
c.
\(0\le1-cosx\le2\Rightarrow-5\le y\le3\sqrt{2}-5\)
\(y_{min}=-5\) khi \(cosx=1\)
\(y_{max}=3\sqrt{2}-5\) khi \(cosx=-1\)
d.
ĐKXĐ: \(0\le sinx\Rightarrow0\le sinx\le1\Rightarrow1\le y\le3\)
\(y_{min}=1\) khi \(sinx=0\)
\(y_{max}=3\) khi \(sinx=1\)
\(\text{1) Đ}K:\left\{{}\begin{matrix}sinx\ne0\\1-sinx\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne m\pi\\x\ne\frac{\pi}{2}+n2\pi\end{matrix}\right.\)
\(2\text{) }ĐK:\left\{{}\begin{matrix}cos\left(2x+\frac{\pi}{3}\right)\ne0\\sinx\ne0\end{matrix}\right.\Leftrightarrow\\ \left\{{}\begin{matrix}2x+\frac{\pi}{3}\ne\frac{\pi}{2}+m\pi\\x\ne n\pi\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne\frac{\pi}{12}+\frac{m\pi}{2}\\x\ne n\pi\end{matrix}\right.\)
\(3\text{) }ĐK:\left\{{}\begin{matrix}\frac{5-3cos2x}{1+sin\left(2x-\frac{\pi}{2}\right)}\ge0\\1+sin\left(2x-\frac{\pi}{2}\right)\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5-3cos2x\ge0\\sin\left(2x-\frac{\pi}{2}\right)\ne-1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}cos2x\le\frac{5}{3}\left(T/m\right)\\2x-\frac{\pi}{2}\ne\frac{3\pi}{2}+k2\pi\end{matrix}\right.\Leftrightarrow x\ne\pi+k\pi\)
\(4\text{) }ĐK:\left\{{}\begin{matrix}sin\left(x+\frac{\pi}{3}\right)\ne0\\cos\left(3x-\frac{\pi}{4}\right)\ne0\\tan\left(3x-\frac{\pi}{4}\right)\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+\frac{\pi}{3}\ne a\pi\\3x-\frac{\pi}{4}\ne\frac{\pi}{2}+b\pi\\3x-\frac{\pi}{4}\ne c\pi\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne-\frac{\pi}{3}+a\pi\\x\ne\frac{\pi}{4}+\frac{b\pi}{3}\\x\ne\frac{\pi}{12}+\frac{c\pi}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne-\frac{\pi}{3}+a\pi\\x\ne\frac{\pi}{12}+\frac{k\pi}{6}\end{matrix}\right.\)
a: ĐKXĐ: 2*sin x+1<>0
=>sin x<>-1/2
=>x<>-pi/6+k2pi và x<>7/6pi+k2pi
b: ĐKXĐ: \(\dfrac{1+cosx}{2-cosx}>=0\)
mà 1+cosx>=0
nên 2-cosx>=0
=>cosx<=2(luôn đúng)
c ĐKXĐ: tan x>0
=>kpi<x<pi/2+kpi
d: ĐKXĐ: \(2\cdot cos\left(x-\dfrac{pi}{4}\right)-1< >0\)
=>cos(x-pi/4)<>1/2
=>x-pi/4<>pi/3+k2pi và x-pi/4<>-pi/3+k2pi
=>x<>7/12pi+k2pi và x<>-pi/12+k2pi
e: ĐKXĐ: x-pi/3<>pi/2+kpi và x+pi/4<>kpi
=>x<>5/6pi+kpi và x<>kpi-pi/4
f: ĐKXĐ: cos^2x-sin^2x<>0
=>cos2x<>0
=>2x<>pi/2+kpi
=>x<>pi/4+kpi/2
`TXĐ: R`
Ta có: `-1 <= sin(x+ \pi/3) <= 1`
`<=>0 <= sin^4 (x+\pi/3) <= 1`
`<=>2 <= y <= 3`
`=>y_[mi n]=2<=>sin(x +\pi/3)=0<=>x= -\pi/3+k\pi` `(k in ZZ)`
`y_[max]=3<=>sin(x +\pi/3)=1<=>x=\pi/6 +k2\pi` `(k in ZZ)`
a/ \(-1\le sin2x\le1\Rightarrow-7\le y\le-1\)
b/ \(-1\le cos2x\le1\Rightarrow1\le y\le7\)
c/ \(-1\le sin2x\le1\Rightarrow3\le y\le11\)
d/ \(-1\le cos\left(3x+\frac{\pi}{3}\right)\le1\Rightarrow8\le y\le12\)