a)

b)

*Ta thấy x = 4 thì ta có (4 – 4).f(4) = (4– 5).f(4 + 2) suy ra f(6) = 0 hay x = 6 là nghiệm của f(x)

* Với x = 5 thì ta có (5 – 4).f(5) = (5– 5).f(5 + 2)suy ra f(5) = 0 hay x = 5 là nghiệm của f(x)

Vậy f(x) có ít nhất hai nghiệm.

Câu 1: B

Câu 2: C

Câu 3: B

b, Ta có A+B=a+b-5-b-c+1

=a+(b-b)-5+1-c

=a-c-4(1)

Lại có C-D=b-c-4-(b-a)

=b-c-4-b+a

=(b-b)+a-c-4

=a-c-4(2)

Từ (1) và (2) ta có A+B=C-D

Bài 1:

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

\(=\dfrac{-x-1+2x-2-x+5}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{\left(x-1\right)\left(x+1\right)}{1-2x}\)

\(=\dfrac{2}{1-2x}\)

b: Để A>0 thì 1-2x>0

=>2x<1

=>x<1/2

uses crt;

var a:array[1..100]of integer;

n,i,s,max,k:integer;

begin

clrscr;

write('Nhap n='); readln(n);

for i:=1 to n do 

begin

write('A[',i,']='); readln(a[i]);

end;

max:=0;

for i:=1 to n do 

  for j:=1 to n do 

if i<=j then 

begin

s:=0;

for k:=i to j do 

  s:=s+a[k];

if max<s then max:=s;

end;

writeln(max);

readln;

end.

uses crt;
var n,i:integer;
a,b,c:array[1..100]of integer;
begin
clrscr;
write('n='); readln(n);
for i:=1 to n do
begin
write('a[',i,']='); readln(a[i]);
end;
for i:=1 to n do
begin
write('b[',i,']='); readln(b[i]);
end;
for i:=1 to n do
c[i]:=a[i]+b[i];
for i:=1 to n do
write(c[i]:4);
readln;
end.

uses crt;
var n,i:integer;
a,b,c: array [1..100] of integer;
begin
clrscr;
write('So phan tu cua day: ');readln(n);
for i:= 1 to n do
begin
write('Phan tu ',i,' cua day a: ');
readln(a[i]);
end;
writeln;
for i:= 1 to n do
begin
write('Phan tu ',i,' cua day b: ');
readln(b[i]);
end;
for i:= 1 to n do
c[i]:= a[i] + b[i];
writeln('Mang C:');
for i:= 1 to n do
write(c[i],' ');
readln
end.

a) \(P=\left[\frac{2}{3a}-\frac{2}{a+1}\cdot\left(\frac{a+1}{3a}-a-1\right)\right]:\frac{a-1}{a}\)

\(P=\left[\frac{2}{3a}-\frac{2}{a+1}\cdot\left(\frac{a+1-3a^2-3a}{3a}\right)\right]\cdot\frac{a}{a-1}\)

\(P=\left[\frac{2}{3a}-\frac{2}{a+1}\cdot\frac{-3a^2-2a+1}{3a}\right]\cdot\frac{a}{a-1}\)

\(P=\left[\frac{2}{3a}-\frac{2}{a+1}\cdot\frac{\left(a+1\right)\left(-3a+1\right)}{3a}\right]\cdot\frac{a}{a-1}\)

\(P=\left[\frac{2-2\left(-3a+1\right)}{3a}\right]\cdot\frac{a}{a-1}\)

\(P=\frac{6a}{3a}\cdot\frac{a}{a-1}=\frac{2a}{a-1}\)

Vậy...

b) \(2a⋮\left(a-1\right)\)

\(\Leftrightarrow2\left(a-1\right)+2⋮\left(a-1\right)\)

Do đó \(2⋮\left(a-1\right)\)

\(\Rightarrow\left(a-1\right)\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

\(\Leftrightarrow a\in\left\{2;0;3;-1\right\}\)

Mà \(a\ne0;1\) \(\Rightarrow a\in\left\{-1;2;3\right\}\)

Vậy...

c) \(\frac{2a}{a-1}\le1\)

\(\Leftrightarrow\frac{2a}{a-1}-1\le0\)

\(\Leftrightarrow\frac{2a-a+1}{a-1}\le0\)

\(\Leftrightarrow\frac{a+1}{a-1}\le0\)

\(\Leftrightarrow\left(a+1\right)\left(a-1\right)\le0\)

\(\Leftrightarrow-1\le a\le1\)

Vậy....