Ta có:

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=k\)

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

\(\Rightarrow\dfrac{1}{a^2}+\dfrac{2}{ab}+\dfrac{1}{b^2}+\dfrac{2}{bc}+\dfrac{1}{c^2}+\dfrac{2}{ac}=k^2\)

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

\(\Rightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=k^2-k\)

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

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

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

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

Cộng vế:

\(P\ge a+b+c-\dfrac{1}{2}\left(ab+bc+ca\right)\ge a+b+c-\dfrac{1}{6}\left(a+b+c\right)^2=\dfrac{3}{2}\)

\(P_{min}=\dfrac{3}{2}\) khi \(a=b=c=1\)

c) Để P=3 thì \(\dfrac{x+1}{2x}=3\)

\(\Leftrightarrow x+1=6x\)

\(\Leftrightarrow x-6x=-1\)

\(\Leftrightarrow-5x=-1\)

hay \(x=\dfrac{1}{5}\)(thỏa ĐK)

Vậy: Để P=3 thì \(x=\dfrac{1}{5}\)

a) Ta có: \(A=\dfrac{1}{x^2+x}+\dfrac{1}{x+1}\)

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

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

b) Ta có: \(P=\dfrac{A}{B}\)

\(=\dfrac{1}{x}:\dfrac{2}{x+1}\)

\(=\dfrac{1}{x}\cdot\dfrac{x+1}{2}\)

\(=\dfrac{x+1}{2x}\)

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

=
\(\dfrac{a+b+c}{\left(b^2+c^2-a^2\right)\left(a+b+c\right)}+\dfrac{a+b+c}{\left(a^2+c^2-b^2\right)\left(a+b+c\right)}+\dfrac{a+b+c}{\left(a^2+b^2-c^2\right)\left(a+b+c\right)}\)
= 0+0+0 = 0
Vậy P= 0 
Ngu vãi ko bt đúng không nx

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

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

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

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

\(=\dfrac{1}{-2bc}+\dfrac{1}{-2ac}+\dfrac{1}{-2ab}\)

\(=\dfrac{a}{-2bca}+\dfrac{b}{-2acb}+\dfrac{c}{-2abc}\)

\(=\dfrac{a+b+c}{-2abc}=\dfrac{0}{-2abc}=0\)

Đề sai rồi bạn

cái cuối là \(\dfrac{1}{\sqrt{c^2-ca+a^2}}\)  nha

\(a^2+b^2-ab\ge\dfrac{1}{2}\left(a+b\right)^2-\dfrac{1}{4}\left(a+b\right)^2=\dfrac{1}{4}\left(a+b\right)^2\)

\(\Rightarrow\dfrac{1}{\sqrt{a^2-ab+b^2}}\le\dfrac{1}{\sqrt{\dfrac{1}{4}\left(a+b\right)^2}}=\dfrac{2}{a+b}\le\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\)

Tương tự:

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

Cộng vế:

\(P\le\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=3\)

Dấu "=" xảy ra khi \(a=b=c=1\)