Giải:

a) \(2^5=4^x\) 

\(\Rightarrow2^5=\left(2^2\right)^x\) 

\(\Rightarrow2^5=2^{2x}\) 

\(\Rightarrow2x=5\) 

\(\Rightarrow x=\dfrac{5}{2}\) 

b) \(2.4^2.8^3.16^4=8^x\) 

\(\Rightarrow2.\left(2^2\right)^2.\left(2^3\right)^3.\left(2^4\right)^4=\left(2^3\right)^x\) 

\(\Rightarrow2.2^4.2^9.2^{16}=2^{3x}\) 

\(\Rightarrow2^{30}=2^{3x}\) 

\(\Rightarrow3x=30\) 

\(\Rightarrow x=30:3\) 

\(\Rightarrow x=10\) 

c) \(3^3:3^5=9^x\) 

\(\Rightarrow3^{-2}=\left(3^2\right)^x\) 

\(\Rightarrow3^{-2}=3^{2x}\) 

\(\Rightarrow2x=-2\) 

\(\Rightarrow x=-2:2\)

\(\Rightarrow x=-1\) 

Chúc bạn học tốt!

a) Ta có: \(2^5=4^x\)

nên \(2^{2x}=2^5\)

\(\Leftrightarrow2x=5\)

hay \(x=\dfrac{5}{2}\)

b) Ta có: \(2\cdot4^2\cdot8^3\cdot16^4=8^x\)

\(\Leftrightarrow2^{3x}=2\cdot2^5\cdot2^9\cdot2^{16}=2^{31}\)

\(\Leftrightarrow3x=31\)

hay \(x=\dfrac{31}{3}\)

c) Ta có: \(3^3:3^5=9^x\)

\(\Leftrightarrow3^{-2}=3^{2x}\)

\(\Leftrightarrow2x=-2\)

hay x=-1

const fi='tamgiac.dat'

      fo='tamgiac.out'

var f1,f2:text;

    a,b,c,d,e,f:array[1..100]of integer;

    i,n,dem1,dem2,dem3:integer;

    ab,bc,ac:real;

begin

assign(f1,fi); reset(f1);

assign(f2,fo); rewrite(f2);

readln(f1,n);

for i:=1 to n do

  readln(f1,a[i],b[i],c[i],d[i],e[i],f[i]);

dem1:=0;

dem2:=0;

dem3:=0;

for i:=1 to n do

  begin

     ab:=sqrt(sqr(a[i]-c[i])+sqr(b[i]-d[i]));

     ac:=sqrt(sqr(a[i]-e[i])+sqr(b[i]-f[i]));

     bc:=sqrt(sqr(c[i]-e[i])+sqr(d[i]-f[i]));

     if (ab>0) and (ac>0) and (bc>0) and (ab+ac>bc) and (ab+bc>ac) and

(ac+bc>ab) then

        begin

           if (ab=ac) or (ac=bc) then inc(dem1);

           if ((ab=ac) and (ab<>bc) and (ac<>bc)) then inc(dem2);

           if ((ac=bc) and (bc<>ab) and (ac<>ab)) then inc(dem2);

           if ((ac=bc) and (ac<>ab) and (bc<>ab)) then inc(dem2);

           if sqr(ab)=sqr(ac)+sqr(bc) then inc(dem3);

           if sqr(ac)=sqr(bc)+sqr(ab) then inc(dem3);

           if sqr(bc)=sqr(ab)+sqr(ac) then inc(dem3);

        end;

  end;

writeln(f2,dem1);

writeln(f2,dem2);

writeln(f2,dem3);

close(f1);

close(f2);

end.

Chọn C.

Phương trình mặt phẳng theo đoạn chắn (ABC) là:

giúp mình với mik cần nhanh

a,k= 2/3

b,y=2/3.x

c,y=20/3 nếu x=10

cho mình cách giải chi tiết đc ko 

uses crt;

var xa,ya,xb,yb,xc,yc:integer;

ab,ac,bc,cv,p,s:real;

begin

clrscr;

write('Nhap toa do diem A:'); readln(xa,ya);

write('Nhap toa do diem B:'); readln(xb,yb);

write('Nhap toa do diem C:'); readln(xc,yc);

ab:=sqrt(sqr(xa-xb)+sqr(ya-yb));

ac:=sqrt(sqr(xa-xc)+sqr(ya-yc));

bc:=sqrt(sqr(xb-xc)+sqr(yb-yc));

if (ab+bc>ac) and (ab+ac>bc) and (ac+bc>ab) then

begin

cv:=ab+bc+ac;

p:=cv/2;

s:=sqrt(p*(p-ab)*(p-ac)*(p-bc));

writeln('Chu vi la: ',cv:4:2);

writeln('Dien tich la: ',dt:4:2);

kt:=0;

if (ab=ac) and (ab<>bc) and (ac<>bc) then kt:=1;

if (ac=bc) and (ac<>ab) and (bc<>ab) then kt:=1;

if (ab=bc) and (ab<>ac) and (bc<>ac) then kt:=1;

if (ab=ac) and (ac=bc) then kt:=2;

if sqr(ab)=sqr(ac)+sqr(bc) then kt:=3;

if sqr(ac)=sqr(ab)+sqr(bc) then kt:=3;

if sqr(bc)=sqr(ab)+sqr(ac) then kt:=3;

if kt=0 then writeln('Day la tam giac thuong');

if kt=1 then writeln('Day la tam giac can');

if kt=2 then writeln('Day la tam giac deu');

if kt=3 then writeln('Day la tam giac vuong');

end

else writeln('Day khong phai la toa do 3 dinh cua mot tam giac');

readln;

end.