bai 11
(2√x/(√x+3)+√x/(√x-3)-(3x+3)/(x-9)]:(2√x-2)/(√x-3)-1]
x≥0;x≠9
a)
P=(2x-6√x+x+3√x-3x-3)/(x-9).(√x-3)/(√x+1)
p=[-3(√x+1).(√x-3)]/[(x-9)(√x+1)]
=-3/(√x+3)
b)
tim x P<1/2
-3/(√x+3)-1/2<0
(-6-√x-3)/(√x-3)<0
(√x+9)/(√x-3)>0
√x <-9; √x>3
x>9
c)
Pmin => √x+3 min=> x=0;
Pmin=-1

bai 16:
[(√a+1)/(√ab+1)+(√ab+√a)/(√ab-1)-1]:[(√a+1)/(√ab+1)-(√ab+√a)/(√ab-1)+1]
<=>
[(√a+1)/(√ab+1)+(√a+1)/(√ab-1)]:[(√a+1)/(√ab+1)-(√a+1)/(√ab-1)
}]
=
[1/(√ab+1)+1/(√ab-1)]:[1/(√ab+1)-1/(√ab-1)
}]
[(√ab-1)+(√ab+1)/(ab-1)]:[(√ab-1)-(√ab+1)/(ab-1)
}]

=2√ab-2

bai 15
P=(a^2+√a)/(a-√a+1)-(2a+√a)/√a+1
a)
P=[√a(√a+1)(a-√a+1)]/(a-√a+1)-[√a(2√a+1)]/√a+1
P=a+√a-(2√a+1)+1
P=a-√a
P=√a(√a-1)
b)a>1=>p>0=>P=|P|

bai 15
P=(a^2+√a)/(a-√a+1)-(2a+√a)/√a+1
a)
P=[√a(√a+1)(a-√a+1)]/(a-√a+1)-[√a(2√a+1)]/√a+1
P=a+√a-(2√a+1)+1
P=a-√a
P=√a(√a-1)
b)a>1=>p>0=>P=|P|
c)
P>2<=>a-√a-2>0
(√a+1)(√a-2)>0
√a>2; a>4
d)
gtnn P
P=(√a-1/2)^2-1/4
P≥-1/4
khi a=1/4

bai 14

P=(2√x/(√x+m)+√x/(√x-m)-m^2/(4x-4m^2)
...x≥0; x≠m
a)
P=[(4.2√x.(√x-m)+4√x(√x+m)-m^2]/(4x-4m^2)
P=(8x-8m√x+4x+4m√x-m^2)/(4(x-m^2)
=(12x-4m√x-m^2)/(4(x-m^2)
=(2√x-m)(6√x+m)/[4(x-m^2)
b)
p=0
6√x+m>0(m>0)

√x=m/2; x=m^2/4

Bạn Nguyễn Mai chép sai đề bài 2 rồi, biểu thức to thế làm sao làm nổi câu b

Câu 4b

4c