khó quá ak

ừ, bạn bik làm thì giúp mình nha ^^

a) vì ab > 0 nên chia cả hai vế Bất đẳng thức cho \(\sqrt{ab}\) ta được

\(\sqrt{\dfrac{c\left(a-c\right)}{ab}}+\sqrt{\dfrac{c\left(b-c\right)}{ab}}\le1\)

Áp dụng Bất đẳng thức Cauchy cho hai số

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

vậy nên ta có đpcm

\(\frac{2005}{\sqrt{2006} }+\frac{2006}{\sqrt{2005} }>\sqrt{2005}+\sqrt{2006} \)

<=>\(2005\sqrt{2005}+2006\sqrt{2006}>2005\sqrt{2006}+2006\sqrt{2005} \)

<=>\(\sqrt{2006}<\sqrt{2005} \)

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

\(\Leftrightarrow\left(a+b+c\right)\left(ab+ac+bc\right)=abc\)

\(\Leftrightarrow a\left(ab+ac+bc\right)+\left(b+c\right)\left(ab+ac+bc\right)-abc=0\)

\(\Leftrightarrow a\left(ab+ac+bc-bc\right)+\left(b+c\right)\left(ab+ac+bc\right)=0\)

\(\Leftrightarrow a^2\left(b+c\right)+\left(b+c\right)\left(ab+ac+bc\right)=0\)

\(\Leftrightarrow\left(a^2+ab+ac+bc\right)\left(b+c\right)=0\)

\(\Leftrightarrow\left(a+b\right)\left(a+c\right)\left(b+c\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}a=-c\\a=-b\\b=-c\end{matrix}\right.\)

- Nếu \(a=-c\Rightarrow a^{2006}=c^{2006}\Rightarrow c^{2006}-a^{2006}=0\Rightarrow P=0\)

- Nếu \(a=-b\Rightarrow a^{2004}=b^{2004}\Rightarrow a^{2004}-b^{2004}=0\Rightarrow P=0\)

- Nếu \(b=-c\Rightarrow b^{2005}=-c^{2005}\Rightarrow b^{2005}+c^{2005}=0\Rightarrow P=0\)

Vậy \(P=0\)

Đúng là câu b sai, nhầm dấu đoạn đầu, phải là \(\frac{2006.2006-\left(2005.2006+2005\right)}{2006.\left(2007-2005\right)}\)

Phá ngoặc thì thành trừ nhưng cô của em bạn lại sót=> sai luôn cả tính chất bài toán.

P/s: Thử lại bằng casio là thấy rõ bạn đúng.

Bạn tham khảo :

Ta có :

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

\(\Rightarrow\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+3=1\)

\(\Rightarrow\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+2=0\)

\(\Rightarrow abc\left(\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+2\right)=abc.0\)

\(\Rightarrow a^2b+b^2c+a^2c+b^2a+c^2a+c^2b+2abc=0\)

\(\Rightarrow\left(a^2b+ab^2\right)+\left(b^2c+abc\right)+\left(a^2c+abc\right)+\left(c^2a+c^2b\right)=0\)

\(\Rightarrow ab\left(a+b\right)+bc\left(a+b\right)+ac\left(a+b\right)+c^2\left(a+b\right)=0\)

\(\Rightarrow\left(ab+bc+ac+c^2\right)\left(a+b\right)=0\)

\(\Rightarrow\left[\left(ab+bc\right)+\left(ac+c^2\right)\right]\left(a+b\right)=0\)

\(\Rightarrow\left[b\left(a+c\right)+c\left(a+c\right)\right]\left(a+b\right)=0\)

\(\Rightarrow\left(a+c\right)\left(b+c\right)\left(a+b\right)=0\)

TH1 : \(a+c=0\)

\(\Rightarrow a=-c\)

\(\Rightarrow c^{2006}=a^{2006}\)

\(\Rightarrow P=\left(a^{2004}-b^{2004}\right)\left(b^{2005}+c^{2005}\right)\left(c^{2006}-a^{2006}\right)\)

\(=\left(a^{2004}-b^{2004}\right)\left(b^{2005}+c^{2005}\right)0\)

\(=0\)

CMTT đều có \(P=0\)

Vậy ...

hay quá cảm ơn nha nhưng có cách nào gọn hơn ko

Đặt \(a=\sqrt{2006}-\sqrt{2005};b=\sqrt{2006}+\sqrt{2005}\)

Ta có 

\(a=\sqrt{2006}-\sqrt{2005}=\dfrac{\left(\sqrt{2006}-\sqrt{2005}\right)\left(\sqrt{2006}+\sqrt{2005}\right)}{\sqrt{2006}+\sqrt{2005}}=\dfrac{1}{b}\)

\(\RightarrowĐfcm\)

\(\sqrt{2006}-\sqrt{2005}=\dfrac{1}{\sqrt{2006}+\sqrt{2005}}\)

Do đó: \(\sqrt{2006}-\sqrt{2005};\sqrt{2006}+\sqrt{2005}\) là hai số nghịch đảo

Từ giả thiết suy ra: \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}\)

\(\Rightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}\right)+\left(\dfrac{1}{c}-\dfrac{1}{a+b+c}\right)=0\)

\(\Rightarrow\dfrac{a+b}{ab}+\dfrac{a+b}{c\left(a+b+c\right)}=0\)

\(\Rightarrow\) (a + b)[c(a + b + c) + ab] = 0

\(\Rightarrow\) (a + b)(ac + ab + bc + c²) = 0

\(\Rightarrow\) (a + b)(b + c)(a + c) = 0

P = (a²⁰⁰⁴ - b²⁰⁰⁴)(b²⁰⁰⁵ + c²⁰⁰⁵)(c²⁰⁰⁶ - a²⁰⁰⁶)

= (a + b)(b + c)(a + c) = 0

TA CÓ  A= \(\left(\frac{2006-2005}{2006+2005}\right)^2\)=\(\frac{1}{4011^2}\)

            B=\(\frac{2006^2-2005^2}{2006^2+2005^2}\) = \(\frac{\left(2006-2005\right)\left(2006+2005\right)}{\left(2006+2005\right)^2-2.2005.2006}\) = \(\frac{4011}{4011^2-2.2006.2005}\)

VÌ 1.(\(4011^2\)-2.200.2005)<\(4011^2\).4011                       (DO \(4011^2\)>\(4011^2\)-2.2006.2005)

\(\Rightarrow\)\(\frac{1}{4011^2}\)< \(\frac{4011}{4011^2-2.2005.2006}\) .HAY A<B

                                                                    VẬY A<B