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AH
Akai Haruma
Giáo viên
31 tháng 7 2019

Lời giải:
a)

\(ab(a-b)+bc(b-c)+ca(c-a)=ab(a-b)-bc(c-b)+ca(c-a)\)

\(=ab(a-b)-bc[(a-b)+(c-a)]+ca(c-a)\)

\(=ab(a-b)-bc(a-b)-bc(c-a)+ca(c-a)\)

\(=(a-b)(ab-bc)+(c-a)(ca-bc)\)

\(=(a-b)b(a-c)-(a-c).c(a-b)\)

\(=(a-b)(a-c)(b-c)\)

b)

\(a^2(b-c)+b^2(c-a)+c^2(a-b)\)

\(=a^2(b-c)-b^2[(b-c)+(a-b)]+c^2(a-b)\)

\(=a^2(b-c)-b^2(b-c)-b^2(a-b)+c^2(a-b)\)

\(=(b-c)(a^2-b^2)-(b^2-c^2)(a-b)\)

\(=(b-c)(a-b)(a+b)-(b-c)(b+c)(a-b)\)

\(=(b-c)(a-b)(a+b-b-c)=(b-c)(a-b)(a-c)\)

c)

\(a^2(a+1)-b^2(b-1)+ab-3ab(a-b+1)\)

\(=a^3+a^2-b^3+b^2+ab-3ab(a-b)-3ab\)

\(=(a^3-3a^2b+3ab^2-b^3)+(a^2+b^2+ab-3ab)\)

\(=(a-b)^3+(a-b)^2=(a-b)^2(a-b+1)\)

31 tháng 7 2019

#)Giải :

a)\(ab\left(b-a\right)+bc\left(b-c\right)+ca\left(c-a\right)\)

\(=a\left(a-b\right)+b^2c-bc^2+ac^2-a^2c\)

\(=ab\left(a-b\right)-\left(a-b\right)\left(a+b\right)c+c^2\left(a-b\right)\)

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

\(=\left(a-b\right)\left(b-c\right)\left(a-c\right)\)

b) \(a^2\left(b-c\right)-b^2\left(c-a\right)+c^2\left(a-b\right)\)

\(=a^2\left(b-c\right)-b^2\left[\left(b-c\right)+\left(a-b\right)\right]+c^2\left(a-b\right)\)

\(=a^2\left(b-c\right)-b^2\left(b-c\right)-b^2\left(a-b\right)+c^2\left(a-b\right)\)

\(=\left(a^2-b^2\right)\left(b-c\right)-\left(b^2-c^2\right)\left(a-b\right)\)

\(=\left(a-b\right)\left(a+b\right)\left(b-c\right)-\left(b-c\right)\left(b+c\right)\left(a-b\right)\)

\(=\left(a-b\right)\left(b-c\right)\left(a-c\right)\)

từ giả thiết, ta có \(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}=1\) đặt \(\left(\dfrac{1}{xy};\dfrac{1}{yz};\dfrac{1}{zx}\right)=\left(a;b;c\right)\Rightarrow a+b+c=1\) =>\(\left(\dfrac{ac}{b};\dfrac{ab}{c};\dfrac{bc}{a}\right)=\left(\dfrac{1}{x^2};\dfrac{1}{y^2};\dfrac{1}{z^2}\right)\) ta có...
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từ giả thiết, ta có \(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}=1\)

đặt \(\left(\dfrac{1}{xy};\dfrac{1}{yz};\dfrac{1}{zx}\right)=\left(a;b;c\right)\Rightarrow a+b+c=1\) =>\(\left(\dfrac{ac}{b};\dfrac{ab}{c};\dfrac{bc}{a}\right)=\left(\dfrac{1}{x^2};\dfrac{1}{y^2};\dfrac{1}{z^2}\right)\)

ta có VT=\(\dfrac{1}{\sqrt{1+\dfrac{1}{x^2}}}+\dfrac{1}{\sqrt{1+\dfrac{1}{y^2}}}+\dfrac{1}{\sqrt{1+\dfrac{1}{z^1}}}=\sqrt{\dfrac{1}{1+\dfrac{ac}{b}}}+\sqrt{\dfrac{1}{1+\dfrac{ab}{c}}}+\sqrt{\dfrac{1}{1+\dfrac{bc}{a}}}\)

=\(\dfrac{1}{\sqrt{\dfrac{b+ac}{b}}}+\dfrac{1}{\sqrt{\dfrac{a+bc}{a}}}+\dfrac{1}{\sqrt{\dfrac{c+ab}{c}}}=\sqrt{\dfrac{a}{\left(a+b\right)\left(a+c\right)}}+\sqrt{\dfrac{b}{\left(b+c\right)\left(b+a\right)}}+\sqrt{\dfrac{c}{\left(c+a\right)\left(c+b\right)}}\)

\(\le\sqrt{3}\sqrt{\dfrac{ac+ab+bc+ba+ca+cb}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}=\sqrt{3}.\sqrt{\dfrac{2\left(ab+bc+ca\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}\)

ta cần chứng minh \(\sqrt{\dfrac{2\left(ab+bc+ca\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}\le\dfrac{3}{2}\Leftrightarrow\dfrac{2\left(ab+bc+ca\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\le\dfrac{9}{4}\Leftrightarrow8\left(ab+bc+ca\right)\le9\left(a+b\right)\left(b+c\right)\left(c+a\right)\)

<=>\(8\left(a+b+c\right)\left(ab+bc+ca\right)\le9\left(a+b\right)\left(b+c\right)\left(c+a\right)\) (luôn đúng )

^_^

0
31 tháng 1 2023

\(Ta\) \(có:\) \(1+a^2=ab+bc+ca+a^2=b\left(a+c\right)+a\left(a+c\right)=\left(a+b\right)\left(c+a\right)\)

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

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

\(Khi\) \(đó:\) \(A=\dfrac{\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2}{\left(a+b\right)\left(a+c\right)\left(a+b\right)\left(b+c\right)\left(a+c\right)\left(c+b\right)}\)

\(\Rightarrow A=1\)

NV
23 tháng 12 2020

\(P=\dfrac{\left(a^2+abc\right)^2}{a^2b^2+2abc^2}+\dfrac{\left(b^2+abc\right)^2}{b^2c^2+2a^2bc}+\dfrac{\left(c^2+abc\right)}{a^2c^2+2ab^2c}\)

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

\(P\ge\dfrac{\left[a^2+b^2+c^2+3abc\right]^2}{\left(ab+bc+ca\right)^2}\)

Do đó ta chỉ cần chứng minh \(\dfrac{a^2+b^2+c^2+3abc}{ab+bc+ca}\ge2\)

Ta có: \(abc\ge\left(a+b-c\right)\left(b+c-a\right)\left(c+a-b\right)\)

\(\Leftrightarrow abc\ge\left(3-2a\right)\left(3-2b\right)\left(3-2c\right)\)

\(\Leftrightarrow3abc\ge4\left(ab+bc+ca\right)-9\)

\(\Rightarrow\dfrac{a^2+b^2+c^2+3abc}{ab+bc+ca}\ge\dfrac{a^2+b^2+c^2+4\left(ab+bc+ca\right)-9}{ab+bc+ca}\)

\(=\dfrac{\left(a+b+c\right)^2-9+2\left(ab+bc+ca\right)}{ab+bc+ca}=2\) (đpcm)

5 tháng 6 2021

sai cơ bản rồi bạn ơi : a(a+bc)^2 không bằng dc (a^2+abc)^2

6 tháng 10 2019

\(VT=\frac{ab+bc+ca}{ab}+\frac{ab+bc+ca}{bc}+\frac{ab+bc+ca}{ca}\)

\(=3+\frac{c\left(a+b\right)}{ab}+\frac{a\left(b+c\right)}{bc}+\frac{b\left(c+a\right)}{ca}\)(1)

Theo BĐT AM-GM: \(\frac{1}{2}\left[\frac{c\left(a+b\right)}{ab}+\frac{a\left(b+c\right)}{bc}\right]\ge\sqrt{\frac{\left(a+b\right)\left(b+c\right)}{b^2}}\)

Tương tự: \(\frac{1}{2}\left[\frac{a\left(b+c\right)}{bc}+\frac{b\left(c+a\right)}{ca}\right]\ge\sqrt{\frac{\left(a+c\right)\left(b+c\right)}{c^2}}\)

\(\frac{1}{2}\left[\frac{c\left(a+b\right)}{ab}+\frac{b\left(c+a\right)}{ca}\right]\ge\sqrt{\frac{\left(a+c\right)\left(a+b\right)}{a^2}}\)

Cộng theo vế 3 BĐT trên rồi thay vào 1 ta sẽ thu được đpcm.

6 tháng 10 2019

Ý em là thay vào (1) !!