B, C và D

B, C và D không phải hằng đẳng thức

a, \(VT=\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}.\left(\sqrt{a}-\sqrt{b}\right)=a-b=VP\) đpcm

b,\(VT=1-\dfrac{a-\sqrt{a}}{\sqrt{a}-1}+\dfrac{a+\sqrt{a}}{\sqrt{a}+1}-\dfrac{a^2-a}{a-1}=1-\sqrt{a}+\sqrt{a}-a=1-a=VP\) đpcm

Bài 1 : Theo BĐT Cô - Si cho các số không âm ta có :

\(\left\{{}\begin{matrix}a+b\ge2\sqrt{ab}\\b+c\ge2\sqrt{bc}\\c+d\ge2\sqrt{cd}\\d+a\ge2\sqrt{da}\end{matrix}\right.\)

Nhân từng vế của BĐT ta được :

\(\left(a+b\right)\left(b+c\right)\left(c+d\right)\left(d+a\right)\ge16\sqrt{a^2b^2c^2d^2}=16abcd\)

Dấu \("="\) xảy ra khi \(a=b=c=d\)

Bài 2 : Theo BĐT Cô Si cho các số không âm ta có :

\(\left\{{}\begin{matrix}x+y\ge2\sqrt{xy}\\\dfrac{1}{x}+\dfrac{1}{y}\ge2\sqrt{\dfrac{1}{xy}}\end{matrix}\right.\)

Nhân vế theo vế ta được :

\(\left(x+y\right)\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\ge4\sqrt{xy.\dfrac{1}{xy}}=4\)

Dấu \("="\) xảy ra khi \(x=y\)

a(b+1) + b(a+1) = ab + a + ab + b = 2ab + a + b = a + b + 2 (1)

(a+ 1)(b+1) = ab + a + b + 1 = 1 + a + b + 1 = a + b + 2 (2)

Từ (1) (2) => a(b+1) + b(a+1) = (a+1)(b+1)

Ta có:

\(VT=\sqrt{\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}}\)

\(=\sqrt{\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)-2\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)}\)

\(=\sqrt{\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2-2\left(\dfrac{c}{abc}+\dfrac{a}{abc}+\dfrac{b}{bca}\right)}\)

\(=\sqrt{\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2-2\left(\dfrac{a+b+c}{abc}\right)}\)

\(=\sqrt{\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2}\)

\(=\left|\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right|\)

\(\Rightarrow VT=VP\)

Vậy \(\sqrt{\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}}=\left|\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right|\) (Đpcm)

\(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ca}\)

\(=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2\left(a+b+c\right)}{abc}=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\)

\(\Rightarrow\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}=\sqrt{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}=\left|\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right|\)

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 !