Question

Cho ab+a+b=1.Chứng minh (a^2 + 1)(b^2 + 1)=2(a+b)^2

Answer 1

Có  \(ab+a+b=1\)

=> (1-a)(b-1) + 2ab = 0 

=> 2(1-a)(b-1) + 4ab = 0   (1)

Có ab+a+b=1

=> (a+1)(b+1) = 2             (2) 

Thay (2) vào (1) ta có \(\left(1-a^2\right)\left(b^2-1\right)+4ab=0\)

<=> \(a^2+b^2+4ab-a^2b^2-1=0\)

<=> \(2a^2+2b^2+4ab=a^2b^2+a^2+b^2+1\)

<=> \(2\left(a+b\right)^2=\left(a^2+1\right)\left(b^2+1\right)\)

Answer 2

+)ta có ab+a+b=1

<=>ab=1-a-b

+)(a²+1).(b²+1)=2(a+b)²

<=>a²b²+a²+b²+1-2(a²+2ab+b²)=0

<=>a²b²+a²+b²+1-2a²-4ab-2b²=00

<=>-3ab-a²-b²+1=0

<=>-ab-2ab-a²-b²+1=0

<=>-(a²+2ab+b²)+1-ab=0

<=>1-(a+b)²-ab=0

<=>(1-a-b)(1+a+b)-ab=0

Mà ab+a+b=1=>ab=1-a-b

<=>ab(1+a+b)-ab=0

<=>ab(1+a+b-1)=0

<=>ab(a+b)=0

Mà ab+a+b=1=>ab=1-a-b

=>(1-a-b)(a+b)=0

Tự giải pt sẽ ra !