1: \(y=cos^2x+2\cdot cos2x\)

\(=cos^2x+2\left(2\cdot cos^2x-1\right)=cos^2x+4\cdot cos^2x-2=5\cdot cos^2x-2\)

Ta có: \(0\le cos^2x\le1\)

=>\(0\le5\cdot cos^2x\le5\)

=>0-2<=\(5\cdot cos^2x-2\le5-2\)

=>-2<=y<=3

y min =-2 khi \(cos^2x=0\)

=>cosx=0

=>\(x=\frac{\pi}{2}+k\pi\)

y max=3 khi \(cos^2x=1\)

=>\(1-cos^2x=1-1\)

=>\(\sin^2x=0\)

=>sin x=0

=>\(x=k\pi\)

2: \(y=2\cdot\sin^2x-cos2x\)

\(=2\cdot\frac{1-cos2x}{2}-cos2x=1-cos2x-cos2x=1-2\cdot cos2x\)

Ta có: \(-1\le cos2x\le1\)

=>\(-2\le2\cdot cos2x\le2\)

=>\(2\ge-2\cdot cos2x\ge-2\)

=>\(2+1\ge-2\cdot cos2x+1\ge-2+1\)

=>3>=y>=-1

y min=-1 khi cos2x=1

=>\(2x=k2\pi\)

=>\(x=k\pi\)

y max=3 khi cos2x=-1

=>\(2x=\pi+k2\pi\)

=>\(x=\frac{\pi}{2}+k\pi\)

4: \(y=3\cdot\sin^2x+5\cdot cos^2x-4\cdot cos2x\)

\(=3\cdot\frac{1-cos2x}{2}+\frac52\cdot\left(1+cos2x\right)-4\cdot cos2x\)

\(=\frac32-\frac32\cdot cos2x+\frac52+\frac52\cdot cos2x-4\cdot cos2x=-3\cdot cos2x+4\)

Ta có: \(-1\le cos2x\le1\)

=>\(-3\le3\cdot cos2x\le3\)

=>\(3\ge-3\cdot cos2x\ge-3\)

=>\(3+4\ge-3\cdot cos2x+4\ge-3+4\)

=>7>=y>=1

y min=1 khi cos2x=1

=>\(2x=k2\pi\)

=>\(x=k\pi\)

y max=7 khi cos2x=-1

=>\(2x=\pi+k2\pi\)

=>\(x=\frac{\pi}{2}+k\pi\)

1: TXĐ là D=R

KHi x∈D thì -x∈D

\(f\left(x\right)=-2\cdot cos^3\left(3x+\frac{\pi}{2}\right)\)

\(=-2\cdot\left\lbrack-\sin3x\right\rbrack^3=2\cdot\sin^33x\)
\(f\left(-x\right)=2\cdot\sin^3\left(-3x\right)=-2\cdot\sin^33x=-f\left(x\right)\)

=>f(x) là hàm số lẻ

2: ĐKXĐ: \(2x-7\pi<>k\pi\)

=>\(2x<>7\pi+k\pi\)

=>\(x<>\frac72\pi+\frac{k\pi}{2}\)

=>\(x<>\frac{k\pi}{2}\)

=>TXĐ là D=R\{\(\frac{k\pi}{2}\) }

Khi x∈D thì -x∈D

\(f\left(x\right)=\sin^3\left(3x+5\pi\right)+\cot\left(2x-7\pi\right)\)

\(=\sin^3\left(3x+\pi\right)+\cot2x=-\sin^33x+\cot2x\)

\(f\left(-x\right)=-\sin^3\left(-3x\right)+\cot\left(-2x\right)=\sin^33x-\cot2x=-f\left(x\right)\)

=>f(x) là hàm số lẻ

3: ĐKXĐ: \(\begin{cases}4x+5\pi<>k\pi\\ 2x-3\pi<>\frac{\pi}{2}+k\pi\end{cases}\Rightarrow\begin{cases}4x<>k\pi-5\pi\\ 2x<>\frac72\pi+k\pi\end{cases}\Rightarrow\begin{cases}x<>\frac{k\pi}{4}-\frac{5\pi}{4}\\ x<>\frac74\pi+\frac{k\pi}{2}\end{cases}\)

=>TXĐ là D=R\{\(\frac{k\pi}{4}-\frac{5\pi}{4};\frac74\pi+\frac{k\pi}{2}\) }

Khi x∈D thì -x∈D

\(F\left(x\right)=\cot\left(4x+5\pi\right)\cdot\tan\left(2x-3\pi\right)\)

\(=\cot4x\cdot\tan2x\)

\(F\left(-x\right)=\cot\left(-4x\right)\cdot\tan\left(-2x\right)=\cot4x\cdot\tan2x=f\left(x\right)\)

=>F(x) là hàm số chẵn

4: ĐKXĐ: \(9-x^2\ge0\)

=>\(x^2\le9\)

=>-3<=x<=3

=>TXĐ là D=[-3;3]

Khi x∈D thì -x∈D

\(f\left(-x\right)=\sin\sqrt{9-\left(-x\right)^2}=\sin\sqrt{9-x^2}=f\left(x\right)\)

=>f(x) là hàm số chẵn

5: ĐKXĐ: x∈R

=>TXĐ là D=R

Khi x∈D thì -x∈D

\(f\left(-x\right)=\sin^2\left\lbrack2\cdot\left(-x\right)\right\rbrack+cos\left\lbrack3\cdot\left(-x\right)\right\rbrack=\sin^22x+cos3x=f\left(x\right)\)

=>f(x) là hàm số chẵn

câu a cho nồng độ rồi ấy em

\(b,n_{H_2SO_4}=\dfrac{4,9}{98}=0,05\left(mol\right)\\ C_{MddH_2SO_4}=\dfrac{0,05}{0,2}=0,25\left(M\right)\\ c,n_{HCl}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\\ C_{MddHCl}=\dfrac{0,4}{0,25}=1,6\left(M\right)\\ d,C_{MddHCl}=\dfrac{C\%_{ddHCl}.10.d_{ddHCl}}{M_{HCl}}=\dfrac{7,3.10.1,25}{36,5}=2,5\left(M\right)\)

1: ĐKXĐ: \(\frac{2+\sin x}{cosx+1}\ge0\)

mà \(1\le\sin x+2\le3\)

nên cosx+1>0

=>cosx>-1

=>\(x<>\pi+k2\pi\)

=>TXĐ là D=R\{\(\pi+k2\pi\) }

2: ĐKXĐ: \(2x<>k\pi;1-cos^2x>0\)

=>\(x<>\frac{k\pi}{2};\sin^2x>0\)

=>\(x<>\frac{k\pi}{2};\sin x<>0\)

=>\(x<>\frac{k\pi}{2};x<>k\pi\)

=>\(x<>\frac{k\pi}{2}\)

=>TXĐ là D=R\{\(\frac{k\pi}{2}\) }

3: ĐKXĐ: \(\frac{1-\sin x}{1+cosx}\ge0\) và 1+cosx<>0

mà \(0\le cosx+1\le2\)

nên cosx<>-1 và 1-sin x>=0

=>\(x<>\pi+k2\pi;\sin x\le1\)

=>\(x<>\pi+k2\pi;\sin x<>1\)

=>\(x<>\pi+k2\pi;x<>\frac{\pi}{2}+k2\pi\)

=>TXĐ là D=R\{\(\pi+k2\pi;\frac{\pi}{2}+k2\pi\) }

5: ĐKXĐ: 1-sin x<>0 và \(x<>\frac{\pi}{2}+k\pi\)

=>sin x<>1 và \(x<>\frac{\pi}{2}+k\pi\)

=>\(x<>\frac{\pi}{2}+k2\pi;x<>\frac{\pi}{2}+k\pi\)

=>\(x<>\frac{\pi}{2}+k\pi\)

=>TXĐ là D=R\{\(\frac{\pi}{2}+k\pi\) }

1: ĐKXĐ: \(\begin{cases}\frac{\pi}{4}-x<>\frac{\pi}{2}+k\pi\\ cosx-2<>0\end{cases}\)

=>\(\begin{cases}x<>\frac{\pi}{4}-\frac{\pi}{2}-k\pi=-\frac{\pi}{4}-k\pi\\ cosx<>2\end{cases}\Rightarrow x<>-\frac{\pi}{4}-k\pi\)

Vậy: TXĐ là D=R\{\(-\frac{\pi}{4}-k\pi\) }

2: ĐKXĐ: 3-sin 4x>=0 và cosx+1<>0

=>cosx<>-1

=>\(x<>\pi+k2\pi\)

=>TXĐ là D=R\{\(\pi+k2\pi\) }

3: ĐKXĐ: cosx-cos3x<>0

=>cos3x<>cosx

=>\(\begin{cases}3x<>x+k2\pi\\ 3x<>-x+k2\pi\end{cases}\Rightarrow\begin{cases}2x<>k2\pi\\ 4x<>k2\pi\end{cases}\Rightarrow x<>\frac{k\pi}{2}\)

=>TXĐ là D=R\{\(\frac{k\pi}{2}\) }

4: ĐKXĐ: \(2x+\frac{\pi}{3}<>k\pi\) và \(2x<>\frac{\pi}{2}+k\pi\)

=>\(2x<>-\frac{\pi}{3}+k\pi;2x<>\frac{\pi}{2}+k\pi\)

=>\(x<>-\frac{\pi}{6}+\frac{k\pi}{2};x<>\frac{\pi}{4}+\frac{k\pi}{2}\)

=>TXĐ là D=R\{\(-\frac{\pi}{6}+\frac{k\pi}{2};\frac{\pi}{4}+\frac{k\pi}{2}\) }

5: ĐKXĐ: sin x+2>=0 và tan 2x-1<>0 và \(2x<>\frac{\pi}{2}+k\pi\)

=>tan 2x<>1 à \(x<>\frac{\pi}{4}+\frac{k\pi}{2}\)

=>\(2x<>\frac{\pi}{4}+k\pi;x<>\frac{\pi}{4}+\frac{k\pi}{2}\)

=>\(x<>\frac{\pi}{8}+\frac{k\pi}{2};x<>\frac{\pi}{4}+\frac{k\pi}{2}\)

=>TXĐ là D=R\{\(\frac{\pi}{8}+\frac{k\pi}{2};\frac{\pi}{4}+\frac{k\pi}{2}\) }

6: ĐKXĐ: \(\sin^2x-cos^2x<>0\)

=>\(cos^2x-\sin^2x<>0\)

=>cos2x<>0

=>\(2x<>\frac{\pi}{2}+k\pi\)

=>\(x<>\frac{\pi}{4}+\frac{k\pi}{2}\)

=>TXĐ là D=R\{\(\frac{\pi}{4}+\frac{k\pi}{2}\) }

tìm n nguyên để gtri bth nguyên hả bạn ? 

\(B=\dfrac{2n-6}{n-1}=\dfrac{2\left(n-1\right)-4}{n-1}=2-\dfrac{4}{n-1}\Rightarrow n-1\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)

Ta có 2n-6\(\in Z\)

         n-1\(\in\)Z

         n-1\(\ne0\)

\(\Rightarrow\) \(\dfrac{2n-6}{n-1}\)là phân số

  Để B có giá trị nguyên thì 2n-6\(⋮\)n-1

       2n-6\(⋮\)n-1

       n-1\(⋮\)n-1\(\Rightarrow\)2(n-1)\(⋮\)n-1\(\Rightarrow\)2n-2\(⋮\)n-1

\(\Rightarrow\)  (2n-2)\(-\left(2n-6\right)\)\(⋮\)n-1

\(\Rightarrow\)2n-2-2n+6\(⋮\)n-1

\(\Rightarrow\)(2n-2n)+(6-2) \(⋮\)n-1

\(\Rightarrow\)           4         \(⋮\)n-1

\(\Rightarrow n-1\) là ước của 4

\(\Rightarrow\)n-1\(\in\){1;-1;2;-2;4;-4}

\(\Rightarrow\)n\(\in\){2;0;3;-1;5;-3}

B

B

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