1 , ĐKXĐ : \(x\ge0,x\ne1\)

Với điều kiện xác định trên phương trình đã cho thánh :

\(\dfrac{1}{\sqrt{x}+1}-\dfrac{2}{\sqrt{x}-1}+\dfrac{x+3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{\sqrt{x}-1-2\left(\sqrt{x}+1\right)+x+3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{x+\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

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

a) 1/(b+2)*(sqr(a)+c)=5;

b) sqr(k)+sqr(k+1)<>sqr(k+2);

c) (8*x-7)>1;

d) sqr(b)-4*a*c>=0;

đ) (1/n)*(1/(n+1))*(1/(n+2))<0.01;

e) (a-3)*(a+5)=0

a: ĐKXĐ: a<>0; a<>1; a<>-1

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

\(=\dfrac{a+1}{a}\cdot\dfrac{\left(a-1\right)\left(a+1\right)}{a+1}=\dfrac{a^2-1}{a}\)

b: Khi a=1/2 thì K=(1/4-1):1/2=-3/4*2=-3/2

k=6
a,b,c=2

\(\dfrac{1}{1+a}+\dfrac{1}{1+b}+\dfrac{1}{1+c}\ge3\sqrt[3]{\dfrac{1}{\left(1+a\right)\left(1+b\right)\left(1+c\right)}}\)

\(\dfrac{a}{1+a}+\dfrac{b}{1+b}+\dfrac{c}{1+c}\ge3\sqrt[3]{\dfrac{abc}{\left(1+a\right)\left(1+b\right)\left(1+c\right)}}\)

Cộng vế và rút gọn:

\(\Rightarrow1\ge\dfrac{1+\sqrt[3]{abc}}{\sqrt[3]{\left(1+a\right)\left(1+b\right)\left(1+c\right)}}\)

\(\Rightarrow\left(1+a\right)\left(1+b\right)\left(1+c\right)\ge\left(1+\sqrt[3]{abc}\right)^3\)

\(\Rightarrow\dfrac{\left(1+a\right)\left(1+b\right)\left(1+c\right)}{abc}\ge\dfrac{\left(1+\sqrt[3]{abc}\right)^3}{abc}=\left(\dfrac{1}{\sqrt[3]{abc}}+1\right)^3\ge\left(\dfrac{3}{a+b+c}+1\right)^3\ge\left(\dfrac{3}{k}+1\right)^3\)

a) BĐT cần cm tương đương ;

\(a-\dfrac{ab^2}{1+b^2}+b-\dfrac{bc^2}{1+c^2}+a-\dfrac{a^2c}{1+a^2}\ge\dfrac{3}{2}\)

\(\Leftrightarrow3-\left(\dfrac{ab^2}{1+b^2}+\dfrac{bc^2}{1+c^2}+\dfrac{ac^2}{1+c^2}\right)\ge\dfrac{3}{2}\)

\(\Leftrightarrow\left(\dfrac{ab^2}{1+b^2}+\dfrac{bc^2}{1+c^2}+\dfrac{ac^2}{1+c^2}\right)\le\dfrac{3}{2}\)

Áp dụng BĐT Cauchy

\(\Rightarrow\dfrac{ab^2}{1+b^2}\le\dfrac{ab^2}{2b}=\dfrac{ab}{2}\)

tương tự rồi cộng vế theo vế các BĐT lại

\(\Leftrightarrow\dfrac{ab^2}{1+b^2}+\dfrac{bc^2}{1+c^2}+\dfrac{ac^2}{1+c^2}\le\dfrac{ab+bc+ac}{2}\)

mặt khác \(ab+bc+ac\le\dfrac{\left(a+b+c\right)^2}{3}=3\)

\(\Rightarrow\dfrac{ab^2}{1+b^2}+\dfrac{bc^2}{1+c^2}+\dfrac{ac^2}{1+c^2}\le\dfrac{3}{2}\)

ĐPCM

b) tương tự câu a thôi

a) đk: \(a>0;a\ne1\)

b) Xét K = \(\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\dfrac{1}{\sqrt{a}+1}+\dfrac{2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\)

= \(\dfrac{a-1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{\sqrt{a}-1+2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\)

= \(\dfrac{\sqrt{a}+1}{\sqrt{a}}:\dfrac{\sqrt{a}+1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\)

= \(\dfrac{\sqrt{a}+1}{\sqrt{a}}.\left(\sqrt{a}-1\right)\)

= \(\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\sqrt{a}}\)

Xét \(a=3+2\sqrt{2}=\left(1+\sqrt{2}\right)^2\)

<=> \(\sqrt{a}=1+\sqrt{2}\)

<=> K = \(\dfrac{\left(\sqrt{2}+2\right)\sqrt{2}}{\sqrt{2}+1}=2\)

c) Đẻ K < 0

<=> \(\dfrac{a-1}{\sqrt{a}}< 0\)

Mà \(\sqrt{a}>0\)

<=> a < 1

<=> 0 < a < 1

Với cả 3 phần thì dấu "=" xảy ra tại a=b=c=1.

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

(Cosi) \(\ge a-\dfrac{ab^2}{2b}=a-\dfrac{ab}{2}\)

Tương tự : \(\dfrac{b}{1+c^2}\ge b-\dfrac{bc}{2};\dfrac{c}{1+a^2}\ge c-\dfrac{ca}{2}\)

\(\Rightarrow P\ge\left(a+b+c\right)-\dfrac{ab+bc+ca}{2}\ge\left(CS\right)\left(a+b+c\right)-\dfrac{\left(a+b+c\right)^2}{6}=3-\dfrac{3^2}{6}=\dfrac{3}{2}\)

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

Tương tự : \(\dfrac{1}{b^2+1}\ge1-\dfrac{b}{2};\dfrac{1}{c^2+1}\ge1-\dfrac{c}{2}\)

\(\Rightarrow P\ge3-\dfrac{a+b+c}{2}=3-\dfrac{3}{2}=\dfrac{3}{2}\)

c)\(P=\dfrac{a+1}{b^2+1}+\dfrac{b+1}{c^2+1}+\dfrac{c+1}{a^2+1}=\left(\dfrac{a}{b^2+1}+\dfrac{b}{c^2+1}+\dfrac{c}{a^2+1}\right)+\left(\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}+\dfrac{1}{c^2+1}\right)\ge\dfrac{3}{2}+\dfrac{3}{2}=3\)

Ta có \(a+b+c=abc\Leftrightarrow\dfrac{a+b+c}{abc}=1\) \(\Leftrightarrow\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}=1\)

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

\(\Leftrightarrow2^2=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2\)

\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=2\) (đpcm)