Ta có : 

\(a\left(b+1\right)+b\left(a+1\right)\)

\(=ab+a+ba+b\)

\(=2ab+a+b\)

\(=2.1+a+b\)

\(=2+a+b\left(1\right)\)

Lại có : 

\(\left(a+1\right)\left(b+1\right)\)

\(=ab+b+a+1\)

\(=1+b+a+1\)

\(=2+a+b\left(2\right)\)

Từ \(\left(1\right);\left(2\right)\)

\(\Rightarrow a\left(b+1\right)+b\left(a+1\right)=\left(a+1\right)\left(b+1\right)\left(đpcm\right)\)

a(b+1) + b(a+1)

= ab+a+b+ab

= ab+a+b+1

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

<=> \(a^2+b^2+4ab-a^2b^2-1=0\)

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

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

+)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 !

1.

a.\(A=1+2^1+2^2+2^3+...+2^{2007}\)

\(2A=2+2^2+2^3+....+2^{2008}\)

b. \(A=\left(2+2^2+2^3+...+2^{2008}\right)-\left(1+2^1+2^2+..+2^{2007}\right)\)

\(=2^{2008}-1\) (bạn xem lại đề)

2.

\(A=1+3+3^1+3^2+...+3^7\)

a. \(2A=2+2.3+2.3^2+...+2.3^7\)

b.\(3A=3+3^2+3^3+...+3^8\)

\(2A=3^8-1\)

\(=>A=\dfrac{2^8-1}{2}\)

3

.\(B=1+3+3^2+..+3^{2006}\)

a. \(3B=3+3^2+3^3+...+3^{2007}\)

b. \(3B-B=2^{2007}-1\)

\(B=\dfrac{2^{2007}-1}{2}\)

4.

Sửa: \(C=1+4+4^2+4^3+4^4+4^5+4^6\)

a.\(4C=4+4^2+4^3+4^4+4^5+4^6+4^7\)

b.\(4C-C=4^7-1\)

\(C=\dfrac{4^7-1}{3}\)

5.

\(S=1+2+2^2+2^3+...+2^{2017}\)

\(2S=2+2^2+2^3+2^4+...+2^{2018}\)

\(S=2^{2018}-1\)

4:

a:Sửa đề: C=1+4+4^2+4^3+4^4+4^5+4^6

=>4*C=4+4^2+...+4^7

b: 4*C=4+4^2+...+4^7

C=1+4+...+4^6

=>3C=4^7-1

=>\(C=\dfrac{4^7-1}{3}\)

5:

2S=2+2^2+2^3+...+2^2018

=>2S-S=2^2018-1

=>S=2^2018-1

Áp dụng BĐT Bunhiacopxki:

\(\left(1+ab\right)\left(1+\dfrac{a}{b}\right)\ge\left(1+a\right)^2\)

\(\Rightarrow\dfrac{1}{\left(1+a\right)^2}\ge\dfrac{1}{\left(1+ab\right)\left(1+\dfrac{a}{b}\right)}=\dfrac{b}{\left(a+b\right)\left(1+ab\right)}\)

Tương tự:

\(\dfrac{1}{\left(1+b\right)^2}\ge\dfrac{a}{\left(a+b\right)\left(1+ab\right)}\)

Cộng vế:

\(\dfrac{1}{\left(1+a\right)^2}+\dfrac{1}{\left(1+b\right)^2}\ge\dfrac{a+b}{\left(a+b\right)\left(1+ab\right)}=\dfrac{1}{1+ab}\)

Dấu "=" xảy ra khi \(a=b=1\)

xí câu 1:))

Áp dụng bất đẳng thức Cauchy-Schwarz dạng Engel ta có :

\(\frac{x^2}{y-1}+\frac{y^2}{x-1}\ge\frac{\left(x+y\right)^2}{x+y-2}\)(1)

Đặt a = x + y - 2 => a > 0 ( vì x,y > 1 )

Khi đó \(\left(1\right)=\frac{\left(a+2\right)^2}{a}=\frac{a^2+4a+4}{a}=\left(a+\frac{4}{a}\right)+4\ge2\sqrt{a\cdot\frac{4}{a}}+4=8\)( AM-GM )

Vậy ta có đpcm

Đẳng thức xảy ra <=> a=2 => x=y=2