4:

a: =>2/5x+7/20-2/20=1/10

=>2/5x+5/20=1/10

=>2/5x=1/10-1/4=4/40-10/40=-6/40=-3/20

=>x=-3/20:2/5=-3/20*5/2=-15/40=-3/8

b: 3/2-1/2x=-1/3+3=8/3

=>1/2x=3/2-8/3=9/6-16/6=-7/6

=>x=-7/6*2=-7/3

c: 15/8-1/8:(1/4x-0,5)=5/4

=>1/8:(1/4x-1/2)=15/8-5/4=15/8-10/8=5/8

=>1/4x-1/2=1/8:5/8=1/5

=>1/4x=1/5+1/2=7/10

=>x=7/10*4=28/10=2,8

d: \(\Leftrightarrow\left[\left(x+\dfrac{1}{2}\right)^3-\dfrac{5}{4}\right]=\dfrac{11}{4}-\dfrac{5}{8}=\dfrac{22-5}{8}=\dfrac{17}{8}\)

=>\(\left(x+\dfrac{1}{2}\right)^3=\dfrac{17}{8}+\dfrac{5}{4}=\dfrac{27}{8}\)

=>x+1/2=3/2

=>x=1

uses crt;

var st:string;

d,i,t,x,y,a,b:integer;

begin

clrscr;

readln(st);

d:=length(st);

for i:=1 to d do write(st[i]:4);

writeln;

t:=0;

for i:=1 to d do

begin

val(st[i],x,y);

t:=t+x;

end;

writeln(t);

val(st[d],a,b);

if (a mod 2=0) then write(1)

else write(-1);

readln;

end.

#include <bits/stdc++.h>

using namespace std;

long long a[1000],i,n,t,dem,t1;

int main()

{

cin>>n;

for (i=1; i<=n; i++) cin>>a[i];

t=0;

for (i=1; i<=n; i++) if (a[i]%2==0) t+=a[i];

cout<<t<<endl;

t1=0;

dem1=0;

for (i=1; i<=n; i++)

if (a[i]<0)

{

cout<<a[i]<<" ";

t1+=a[i];

dem1++;

}

cout<<endl;

cout<<fixed<<setprecision(1)<<(t1*1.0)/(dem1*1.0);

return 0;

}

#include <bits/stdc++.h>
using namespace std;
long long a,b;
//chuongtrinhcon
long long gcd(long long a,long long b)
{
    if (b==0) return(a);
    return gcd(b,a%b);
}
//chuongtrinhchinh
int main()
{
    cin>>a>>b;

cout<<max(a,b)<<endl;

cout<<gcd(a,b)<<endl;
    if ((a>0 && b>0) or (a<0 && b<0)) cout<<a/gcd(a,b)<<" "<<b/gcd(a,b);
    else cout<<"-"<<-a/gcd(-a,b)<<" "<<b/gcd(-a,b);
    return 0;
}

b: Xét ΔABE vuông tại A có AH là đường cao ứng với cạnh huyền BE

nên \(BH\cdot BE=AB^2\left(1\right)\)

Xét ΔABC vuông tại B có BH là đường cao ứng với cạnh huyền AC

nên \(AH\cdot AC=AB^2\left(2\right)\)

Từ (1) và (2) suy ra \(BH\cdot BE=AH\cdot AC\)

???

Ta có:

\(\dfrac{1}{cos^2x-sin^2x}+\dfrac{2tanx}{1-tan^2x}=\dfrac{1}{cos2x}+tan2x=\dfrac{1}{cos2x}+\dfrac{sin2x}{cos2x}=\dfrac{1+sin2x}{cos2x}=\dfrac{cos2x}{1-sin2x}\)

\(\Rightarrow P=a+b=2+1=3\)