lỗi à b

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 13

a) A = {1; 2; 3; 4; 5}

b) B = {5; 6; 7; 8; 9}

Câu 14:

a) 15 - 4x = 3

4x = 15 - 3

4x = 12

x = 12 : 4

x = 3

b) 5.3ˣ⁻¹ - 110 = 5⁵ : 5³

5.3ˣ⁻¹ - 110 = 5²

5.3ˣ⁻¹ - 110 = 25

5.3ˣ⁻¹ = 25 + 110

5.3ˣ⁻¹ = 135

3ˣ⁻¹ = 135 : 5

3ˣ⁻¹ = 27

3ˣ⁻¹ = 3³

x - 1 = 3

x = 3 + 1

x = 4

Bài 14:

a: ĐKXĐ: x>=-9

\(\sqrt{x+9}=3\)

=>\(x+9=3^2=9\)

=>x=0(nhận)

b: ĐKXĐ: x∈R

\(\sqrt{x^2-2x+4}=x-1\)

=>\(\begin{cases}x-1\ge0\\ x^2-2x+4=\left(x-1\right)^2=x^2-2x+1\end{cases}\)

=>\(\begin{cases}x\ge1\\ 4=1\left(vôlý\right)\end{cases}\)

=>x∈∅

c: ĐKXĐ: x∈R

\(\sqrt{x^2-6x+9}=4-x\)

=>\(\sqrt{\left(x-3\right)^2}=4-x\)

=>|x-3|=4-x

=>\(\begin{cases}4-x\ge0\\ \left(4-x\right)^2=\left(x-3\right)^2\end{cases}\Rightarrow\begin{cases}x\le4\\ x^2-8x+16=x^2-6x+9\end{cases}\)

=>\(\begin{cases}x\le4\\ -2x=-7\end{cases}\Rightarrow x=\frac72\)

d: \(\sqrt{x^2-2x+1}+\sqrt{x^2-4x+4}=3\)

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

=>|x-1|+|x-2|=3(1)

TH1: x<1

(1) sẽ trở thành:

1-x+2-x=3

=>3-2x=3

=>2x=0

=>x=0(nhận)

TH2: 1<=x<2

=>x-1>=0; x-2<0

(1) sẽ trở thành: x-1+2-x=3

=>1=3(vô lý)

TH3: x>=2

=>x-1>0; x-2>=0

(1) sẽ trở thành: x-1+x-2=3

=>2x=6

=>x=3(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

\(yBLi\)

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