Câu 10: C

Câu 13: A

Câu 17:B
Câu 18: A

C

A

B
A

a: ĐKXĐ: \(x\notin\left\{3;-3;-1\right\}\)

\(A=\dfrac{x\left(x-3\right)-2\left(x+3\right)-x^2+1}{x^2-9}:\dfrac{2x+6-x-5}{x+3}\)

\(=\dfrac{x^2-3x-2x-6-x^2+1}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{x+1}\)

\(=\dfrac{-5\left(x+1\right)}{\left(x-3\right)}\cdot\dfrac{1}{x+1}=-\dfrac{5}{x-3}\)

b: \(x^2-x-2=0\)

=>\(\left(x-2\right)\left(x+1\right)=0\)

=>\(\left[{}\begin{matrix}x=2\left(nhận\right)\\x=-1\left(loại\right)\end{matrix}\right.\)

Khi x=2 thì \(A=\dfrac{-5}{2-3}=\dfrac{-5}{-1}=5\)

c: A=1/2

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

=>x-3=-10

=>x=-7(nhận)

Câu 2: 

#include <bits/stdc++.h>

using namespace std;

long long a[100],i,n,dem;

int main()

{

cin>>n;

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

dem=0;

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

if (a[i]<0) dem++;

cout<<dem;

return 0;

}

đăng văn nhé

to go

\(yBLi\)

\(=\dfrac{4x+5-3x}{x\left(x+5\right)}=\dfrac{x+5}{x\left(x+5\right)}=\dfrac{1}{x}\)