trong bộ luật có á

Trong lý thuyết chứ?

\(n_{Fe_3O_4}=\dfrac{8}{232}=\dfrac{1}{29}\left(mol\right)\)

\(Fe_3O_4+8HCl\rightarrow FeCl_2+2FeCl_3+4H_2O\)

\(\dfrac{1}{29}.......\dfrac{8}{29}................\dfrac{2}{29}\)

\(m_{HCl}=\dfrac{8}{29}\cdot36.5=10.06\left(g\right)\)

\(m_{FeCl_3}=\dfrac{2}{29}\cdot162.5=11.21\left(g\right)\)

3. are there any eggs?-No, there aren't some

4.is there any salt?- yes, there is any

5.are there any carrots?-yes, there are some

6.are there any apples?-no there aren't some

7.is there any suger?-yes, there is any

8.are there any cakes?-no there aren't some

9.is there any butter?-no. there isn't any

10. is there any mineral water?- yes. there is any

Câu 5:

1: cos3x-sin 3x=-1

=>\(\sin3x-cos3x=1\)

=>\(\sqrt2\cdot\sin\left(3x-\frac{\pi}{4}\right)=1\)

=>\(\sin\left(3x-\frac{\pi}{4}\right)=\frac{1}{\sqrt2}\)

=>\(\left[\begin{array}{l}3x-\frac{\pi}{4}=\frac{\pi}{4}+k2\pi\\ 3x-\frac{\pi}{4}=\pi-\frac{\pi}{4}+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}3x=\frac{\pi}{2}+k2\pi\\ 3x=\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{\pi}{6}+\frac{k2\pi}{3}\\ x=\frac{\pi}{3}+\frac{k2\pi}{3}\end{array}\right.\)

2: \(\sqrt3\cdot\sin\left(\frac{x}{2}\right)-cos\left(\frac{x}{2}\right)-\sqrt2=0\)

=>\(\sqrt3\cdot\sin\left(\frac{x}{2}\right)-cos\left(\frac{x}{2}\right)=\sqrt2\)

=>\(\frac{\sqrt3}{2}\cdot\sin\left(\frac{x}{2}\right)-\frac12\cdot cos\left(\frac{x}{2}\right)=\frac{\sqrt2}{2}\)

=>\(\sin\left(\frac{x}{2}-\frac{\pi}{6}\right)=\sin\left(\frac{\pi}{4}\right)\)

=>\(\left[\begin{array}{l}\frac{x}{2}-\frac{\pi}{6}=\frac{\pi}{4}+k2\pi\\ \frac{x}{2}-\frac{\pi}{6}=\pi-\frac{\pi}{4}+k2\pi=\frac34\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}\frac{x}{2}=\frac{\pi}{6}+\frac{\pi}{4}+k2\pi=\frac{5}{12}\pi+k2\pi\\ \frac{x}{2}=\frac34\pi+\frac{\pi}{6}+k2\pi=\frac{11}{12}\pi+k2\pi\end{array}\right.\)

=>\(\left[\begin{array}{l}x=\frac56\pi+k4\pi\\ x=\frac{11}{6}\pi+k4\pi\end{array}\right.\)

3: 3*sin 4x+4* cos4x=5

=>\(\frac35\cdot\sin4x+\frac45\cdot cos4x=1\)

=>\(\sin\left(4x+\alpha\right)=1\)

=>\(4x+\alpha=\frac{\pi}{2}+k2\pi\)

=>\(4x=\frac{\pi}{2}-\alpha+k2\pi\)

=>\(x=\frac{\pi}{8}-\frac{\alpha}{4}+\frac{k\pi}{2}\)

Bài 4:

1: \(3\cdot\sin^23x-4\cdot\sin3x+1=0\)

=>\(3\cdot\sin^23x-3\cdot\sin3x-\sin3x+1=0\)

=>(sin 3x-1)(3sin 3x-1)=0

TH1: sin 3x-1=0

=>sin 3x=1

=>\(3x=\frac{\pi}{2}+k2\pi\)

=>\(x=\frac{\pi}{6}+\frac{k2\pi}{3}\)

TH2: 3 sin 3x-1=0

=>3sin 3x=1

=>sin 3x=1/3

=>\(\left[\begin{array}{l}3x=\arcsin\left(\frac13\right)+k2\pi\\ 3x=\pi-\arcsin\left(\frac13\right)+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac13\cdot\arcsin\left(\frac13\right)+\frac{k2\pi}{3}\\ x=\frac{\pi}{3}-\frac13\cdot\arcsin\left(\frac13\right)+\frac{k2\pi}{3}\end{array}\right.\)

2: \(4\cdot cos^2\left(\frac{x}{2}\right)-1=0\)

=>\(4\cdot cos^2\left(\frac{x}{2}\right)=1\)

=>\(cos^2\left(\frac{x}{2}\right)=\frac14\)

=>\(\left[\begin{array}{l}cos\left(\frac{x}{2}\right)=\frac12\\ cos\left(\frac{x}{2}\right)=-\frac12\end{array}\right.\Rightarrow\left[\begin{array}{l}\frac{x}{2}=\frac{\pi}{3}+k2\pi\\ \frac{x}{2}=-\frac{\pi}{3}+k2\pi\\ \frac{x}{2}=\frac23\pi+k2\pi\\ \frac{x}{2}=-\frac23\pi+k2\pi\end{array}\right.\)

=>\(\left[\begin{array}{l}x=\frac{2\pi}{3}+k4\pi\\ x=-\frac23\pi+k4\pi\\ x=\frac43\pi+k4\pi\\ x=-\frac43\pi+k4\pi\end{array}\right.\)

3: \(3\cdot\tan^24x-\sqrt3\cdot\tan4x=0\)

=>\(\sqrt3\cdot\tan4x\left(\sqrt3\cdot\tan4x-1\right)=0\)

TH1: tan 4x=0

=>\(4x=k\pi\)

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

TH2: \(\sqrt3\cdot\tan4x-1=0\)

=>\(\tan4x=\frac{1}{\sqrt3}\)

=>\(4x=\frac{\pi}{6}+k\pi\)

=>\(x=\frac{\pi}{24}+\frac{k\pi}{4}\)

5: \(\sin^2x+cosx-1=0\)

=>\(1-cos^2x+cosx-1=0\)

=>\(-cos^2x+cosx=0\)

=>cosx(cosx-1)=0

TH1: cosx=0

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

TH2: cos x-1=0

=>cosx =1

=>\(x=k2\pi\)

6: \(\cot^22x-2\cdot\cot2x-3=0\)

=>(cot 2x-3)(cot 2x+1)=0

TH1: cot 2x-3=0

=>cot 2x=3

=>\(2x=arc\cot\left(3\right)+k\pi\)

=>\(x=\frac12\cdot arc\cot\left(3\right)+\frac{k\pi}{2}\)

TH2: cot 2x+1=0

=>cot 2x=-1

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

=>\(x=-\frac{\pi}{8}+\frac{k\pi}{2}\)

Bài 1:

a: Xét ΔIQB vuông tại Q và ΔIQC vuông tại Q có

IQ chung

QB=QC

Do đó: ΔIQB=ΔIQC

=>IB=IC

Xét ΔIPA vuông tại P và ΔIPD vuông tại P có

IP chung

PA=PD

Do đó: ΔIPA=ΔIPD

=>IA=ID

Xét ΔAIB và ΔDIC có

AI=DI

AB=DC

IB=IC

Do đó: ΔAIB=ΔDIC

b: ΔAIB=ΔDIC

=>\(\hat{IAB}=\hat{IDC}\)

mà \(\hat{IDC}=\hat{IAC}\) (ΔIPA=ΔIPD)

nên \(\hat{IAB}=\hat{IAC}\)

=>AI là phân giác của góc BAC
c: Xét ΔAEI vuông tại E và ΔAPI vuông tại P có

AI chung

\(\hat{EAI}=\hat{PAI}\)

Do đó: ΔAEI=ΔAPI

=>AE=AP

mà \(AP=\frac{AD}{2}\)

nên \(AE=\frac{AD}{2}\)

a: \(A=\sqrt{x^2+2x+5}=\sqrt{x^2+2x+1+4}\)

=>\(A=\sqrt{\left(x+1\right)^2+4}>=\sqrt{4}=2\)

b: \(B=\sqrt{x^2-4x+4+1}=\sqrt{\left(x-2\right)^2+1}>=1\)

a,

S S' I G

b,

S M N I G

program bai_toan;

var

      S, S1: string;

      i, j, n: integer;

begin

      write('Nhap xau S: ');

      readln(S);

      n := length(S);

      for i := n downto 1 do

      begin

            S1 := S1 + S[i];

      end;

      for i := n downto 1 do

      begin

            if S = S1 then break;

            S := S + S1[i];

      end;

      writeln('Xau doi xung ngan nhat la: ', S);

      readln;

end.

Hả... 1+1=2

=2