Lời giải:

Lần sau bạn nhớ ghi đầy đủ đề. $ABC$ là tam giác vuông tại $A$.

$\frac{AB}{AC}=\frac{3}{4}$

$\Rightarrow AC=\frac{4AB}{3}=\frac{4.15}{3}=20$ (cm)

Áp dụng định lý Pitago:

$y=BC=\sqrt{AB^2+AC^2}=\sqrt{15^2+20^2}=25$ (cm) 

$S_{ABC}=AB.AC:2=AH.BC:2$

$\Rightarrow AB.AC=AH.BC$

$\Rightarrow x=AH=\frac{AB.AC}{BC}=\frac{15.20}{25}=12$ (cm)

\(A=16x^2-y^2-16x^2+8x=8x-y^2\\ A=8\cdot3-\left(-1\right)^2=24-1=23\\ B=64x^3-80x-64x^3-1=-80x-1\\ B=-80\cdot\dfrac{1}{5}-1=-16-1=-17\)

C1:

khai báo:

      program  hieu_a_va_b   ("tự đặt tên cái này ko cần thiết)

      uses crt;

      var a, b : integer

phần thân:

       begin

            writeln ("nhap so a");  read(a);

            writeln ("nhap so b");  read(b);

             writeln ("hieu cua hai so la: a - b=" , a - b);

        end.

C2:

khai báo:

      program    tinh_ba_so

      uses crt;

      var a, b, c : integer

phần thân:

       begin

            writeln ("nhap so a");  read(a);

            writeln ("nhap so b");  read(b);

            writeln ("nhap so c");  read(c);

             writeln ("ket qua cua phep tinh a+b-c la" , a+b-c);

        end.

C3:

khai báo:

      program    tinh_ba_so

      uses crt;

      var x, y, z : integer

phần thân:

       begin

            writeln ("nhap so x");  read(x);

            writeln ("nhap so y");  read(y);

            writeln ("nhap so z");  read(z);

             writeln ("ket qua cua phep tinh (x+y)*z la") , (x+y)*z);

        end.

nhớ chạy thử chương trình nha. xem có lỗi ko.

(nhớ tick cho mk nha)

Câu 2: 

#include <bits/stdc++.h>

using namespace std;

long long a,b;

int main()

{

cin>>a>>b;

cout<<a-b;

return 0;

}

a) \(\dfrac{13}{20}+\dfrac{3}{5}+x=\dfrac{5}{6}\)

\(\Rightarrow\dfrac{5}{4}+x=\dfrac{5}{6}\)

\(\Rightarrow x=\dfrac{5}{6}-\dfrac{5}{4}\)

\(\Rightarrow x=\dfrac{-5}{12}\)

b) \(x+\dfrac{1}{3}=\dfrac{2}{5}-\dfrac{-1}{3}\)

\(\Rightarrow x+\dfrac{1}{3}=\dfrac{11}{15}\)

\(\Rightarrow x=\dfrac{11}{15}-\dfrac{1}{3}\)

\(\Rightarrow x=\dfrac{2}{5}\)

c)\(\dfrac{-5}{8}-x=\dfrac{-3}{20}-\dfrac{-1}{6}\)

\(\dfrac{-5}{8}-x=\dfrac{1}{60}\)

\(\Rightarrow x=\dfrac{-5}{8}-\dfrac{1}{60}\)

\(\Rightarrow x=\dfrac{-77}{120}\)

d) \(\dfrac{3}{5}-x=\dfrac{1}{4}+\dfrac{7}{10}\)

\(\Rightarrow\dfrac{3}{5}-x=\dfrac{19}{20}\)

\(\Rightarrow x=\dfrac{3}{5}-\dfrac{19}{20}\)

\(\Rightarrow x=\dfrac{-7}{20}\)

e) \(\dfrac{-3}{7}-x=\dfrac{4}{5}+\dfrac{-2}{3}\)

\(\Rightarrow\dfrac{-3}{7}-x=\dfrac{2}{15}\)

\(\Rightarrow x=\dfrac{-3}{7}-\dfrac{2}{15}\)

\(\Rightarrow x=\dfrac{-59}{105}\)

g) \(\dfrac{-5}{6}-x=\dfrac{7}{12}+\dfrac{-1}{3}\)

\(\Rightarrow\dfrac{-5}{6}-x=\dfrac{1}{4}\)

\(\Rightarrow x=\dfrac{-5}{6}-\dfrac{1}{4}\)

\(\Rightarrow x=\dfrac{-13}{12}\)