#include <bits/stdc++.h>

using namespace std;

double a,b,c,p,s;

int main()

{

cin>>a>>b>>c;

p=(a+b+c)/2;

s=sqrt(p*(p-a)*(p-b)*(p-c));

cout<<fixed<<setprecision(2)<<p;

return 0;

}

1: 

uses crt;

var a,b,c,max,min:longint;

begin

clrscr;

readln(a,b,c);

max=a;

if max<b then max:=b;

if max<c then max:=c;

min:=a;

if min>c then min:=c;

if min>b then min:=b;

writeln(max,' ',min);

readln;

end.

a)\(\dfrac{24}{36}\)=\(\dfrac{8}{12}\)

b)\(\dfrac{14}{56}\)=\(\dfrac{1}{4}\)

c)\(\dfrac{9}{24}\)=\(\dfrac{21}{56}\)

Chúc bạn học tốt!

a: x/12=24/36

nên x/12=2/3

=>x=2/3x12=8

b: 1/x=14/56

nên 1/x=1/14

hay x=14

c: x/56=9/24

=>x/56=3/8

hay x=21

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xét tứ giác AFCD có EA=EC;ED=EF nên tứ giác AFCD là hình bình hành

Đề bài yêu cầu gì?

Bài 3: 

a: \(\sqrt{3x-5}=7\)

\(\Leftrightarrow3x-5=49\)

\(\Leftrightarrow3x=54\)

hay x=18

Bài 3: 

b. \(\sqrt{100-200x}-\dfrac{3}{4}\sqrt{16-32x}+2\sqrt{1-2x}=18\)

<=> \(\sqrt{100\left(1-2x\right)}-\dfrac{3}{4}\sqrt{16\left(1-2x\right)}+2\sqrt{1-2x}=18\)

<=> \(10\sqrt{1-2x}-3\sqrt{1-2x}+2\sqrt{1-2x}=18\)

<=> \(\left(10-3+2\right)\sqrt{1-2x}=18\)

<=> \(11\sqrt{1-2x}=18\)

<=> \(\sqrt{1-2x}=\dfrac{18}{11}\)

<=> 1 - 2x = \(\left(\dfrac{18}{11}\right)^2\)

<=> 1 - 2x = \(\dfrac{324}{121}\)

<=> \(1-\dfrac{324}{121}=2x\)

<=> \(2x=\dfrac{-203}{121}\)

<=> \(x=\dfrac{-203}{242}\)

Bài 4:

Áp dụng hệ thức lượng trong tam giác vuông:

$AH^2=BH.CH$

$3^2=y.5\Rightarrow y=\frac{9}{5}=1,8$ 

Áp dụng định lý Pitago:

$x=\sqrt{3^2+y^2}=\sqrt{9+1,8^2}=\frac{3\sqrt{34}}{5}$ 

$\tan \alpha = \frac{BH}{AH}=\frac{y}{3}=\frac{1,8}{3}=\frac{3}{5}$

$\Rightarrow \alpha \approx 31^0$

b.

$AH\perp BC, CD\perp BC$ nên $AH\parallel CD$

Áp dụng định lý Talet:

$\frac{AH}{CD}=\frac{BH}{BC}$

$\Rightarrow DC=\frac{AH.BC}{BH}=\frac{3(y+5)}{y}=\frac{3(1,8+5)}{1,8}=\frac{34}{3}$ 

$\tan \widehat{HDC}=\frac{HC}{CD}=\frac{5}{\frac{34}{3}}=\frac{15}{34}$

$\Rightarrow \widehat{HDC}=23,8^0$

$\widehat{AHD}=\widehat{HDC}=23,8^0$ (hai góc so le trong)

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