Câu 3

Var st:string;

i,d:integer;

Begin

Write('Nhap xau ky tu ');readln(st);

For i:=1 to length(st) do

If st[i] in ['0'..'9'] then d:=d+1;

Write('Co ',d,' ky tu so');

Readln

End.

Câu 4

Var a:array[1..1000] of integer;

i,n:integer;

Begin

Write('n = ');readln(n);

For i:=1 to n do

Begin

Write('Nhap so thu ',i);readln(a[i]);

End;

Write('Cac so chia het cho 3 hoac 5 la: ');

for i:=1 to n do

If (a[i] mod 3 = 0) or (a[i] mod 5 = 0) then write(a[i]:8);

Readln

End.

Câu 2: 

Ta có: \(x^2-2\left(m+1\right)x+m^2+4=0\)

\(\text{Δ}=\left(2m+2\right)^2-4\left(m^2+4\right)\)

\(=4m^2+8m+4-4m^2-16\)

\(=8m-12\)

Để phương trình có hai nghiệm phân biệt thì Δ>0

\(\Leftrightarrow8m>12\)

hay \(m>\dfrac{3}{2}\)

Áp dụng hệ thức Vi-et, ta được:

\(\left\{{}\begin{matrix}x_1+x_2=2\left(m+1\right)=2m+2\\x_1x_2=m^2+4\end{matrix}\right.\)

Vì x₁ là nghiệm của phương trình nên ta có: \(x_1^2-2\left(m+1\right)x_1+m^2+4=0\)

\(\Leftrightarrow x_1^2=2\left(m+1\right)x_1-m^2-4\)

Ta có: \(x_1^2+2x_2\left(m+1\right)=2m^2+20\)

\(\Leftrightarrow2\left(m+1\right)\cdot x_1-m^2-4+2x_2\left(m+1\right)-2m^2-20=0\)

\(\Leftrightarrow2\left(m+1\right)\left(x_1+x_2\right)-3m^2-24=0\)

\(\Leftrightarrow4\left(m+1\right)\cdot\left(m+1\right)-3m^2-24=0\)

\(\Leftrightarrow4m^2+8m+4-3m^2-24=0\)

\(\Leftrightarrow m^2+8m-20=0\)

\(\Leftrightarrow\left(m+10\right)\left(m-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}m=-10\left(loại\right)\\m=2\left(nhận\right)\end{matrix}\right.\)

P6:

1. big

2. been

3. bought

4. people

P7:

1. T

2. F

3. F

4. F

Câu 16

B

Câu17

C

b=>go

b=>and

1. went => to go 

2. but => and

1 B -> go

2 B => and

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9 THis is a blue car

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13 How wide if the Great wall