\(a^2+\frac{1}{a^2}\ge2\sqrt{a^2+\frac{1}{a^2}}=2\\ \)(do Bđt cosi)=> \(a^2+b^2+c^2+\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge6\\ \)

Dấu "=" xảy ra <=> a=b=c=1

=>B=3

Bất đẳng thức cosi mình chưa học

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)

\(\Leftrightarrow\frac{bc+ca+ab}{abc}=0\)

\(\Leftrightarrow bc+ca+ab=0\)

\(\Leftrightarrow\hept{\begin{cases}bc=-ab-ca\\ca=-ab-bc\\ab=-ca-bc\end{cases}}\)

Ta có : \(A=\frac{a^2}{a^2+2bc}+\frac{b^2}{b^2+2ac}+\frac{c^2}{c^2+2ab}\)

\(\Leftrightarrow A=\frac{a^2}{a^2+bc-ab-ca}+\frac{b^2}{b^2+ac-ab-bc}+\frac{c^2}{c^2+ab-ca-bc}\)

\(\Leftrightarrow A=\frac{a^2}{\left(a-b\right)\left(a-c\right)}+\frac{b^2}{\left(b-a\right)\left(b-c\right)}+\frac{c^2}{\left(c-a\right)\left(c-b\right)}\)

\(\Leftrightarrow A=\frac{a^2}{\left(a-b\right)\left(a-c\right)}-\frac{b^2}{\left(b-c\right)\left(a-b\right)}+\frac{c^2}{\left(a-c\right)\left(b-c\right)}\)

\(\Leftrightarrow A=\frac{a^2\left(b-c\right)-b^2\left(a-c\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)

\(\Leftrightarrow A=\frac{a^2\left(b-c\right)-b^2\left[\left(b-c\right)+\left(a-b\right)\right]+c^2\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)

\(\Leftrightarrow A=\frac{a^2\left(b-c\right)-b^2\left(b-c\right)-b^2\left(a-b\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)

\(\Leftrightarrow A=\frac{\left(a^2-b^2\right)\left(b-c\right)-\left(b^2-c^2\right)\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)

\(\Leftrightarrow A=\frac{\left(a+b\right)\left(a-b\right)\left(b-c\right)-\left(b+c\right)\left(b-c\right)\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)

\(\Leftrightarrow A=\frac{\left(a-b\right)\left(b-c\right)\left[\left(a+b\right)-\left(b+c\right)\right]}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)

\(\Leftrightarrow A=\frac{\left(a-b\right)\left(a-c\right)\left(b-c\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}=1\)

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Leftrightarrow\frac{ab+bc+ca}{abc}=0\Rightarrow ab+bc+ca=0\\ \)

\(\Rightarrow bc=-ab-ac,ca=-ab-bc,ab=-bc-ca\)

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

     Làm tương tự. có: \(\frac{b^2+ca}{b^2+2ca}=\frac{b^2+ca}{b^2+ca-ab-bc}=\frac{b^2+ca}{\left(a-b\right).\left(c-b\right)}\)

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

\(\Rightarrow A=\frac{a^2+bc}{\left(a-b\right).\left(a-c\right)}+\frac{b^2+ca}{\left(a-b\right).\left(c-b\right)}+\frac{c^2+ab}{\left(b-c\right).\left(a-c\right)}\)

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

Sau đó bạn thực hiện tiếp nhé.

Bài 1: Cho \(a,b,c\ge0:a^2+b^2+c^2=3\). CMR: \(a^4b^4+b^4c^4+c^4a^4\le3\)

Bài 2: Cho \(a,b,c\ge0\). CMR: \(a^2+b^2+c^2+2abc+1\ge2\left(ab+bc+ca\right)\)

Bài 3: Cho \(a,b,c\ge0:a^2+b^2+c^2=a+b+c\). CMR: \(a^2b^2+b^2c^2+c^2a^2\le ab+bc+ca\)

Bài 4: Cho \(a,b,c\ge0\). CMR: \(4\left(a+b+c\right)^3\ge27\left(ab^2+bc^2+ca^2+abc\right)\)

Bài 5: Cho \(a,b,c\ge0:a+b+c=3\).CMR: \(\frac{1}{2bc^2+1}+\frac{1}{2ca^2+1}+\frac{1}{2ab^2+1}\ge1\)

dat bt tren la A . ap bdt bunhiacopxki ta co                                                                                                                                    (a+b+c)^2 = ( a/(can1+c^2) . (can1+c^2) + b/(can1+a^2) . (can1+a^2) +c/(can1+b^2) . (can1+b^2) )^2 <= A(1 + c^2 + 1 + a^2 +1 + b^2)      ...  0 <= A(3+a^2+b^2+c^2) ...nen 0<=A  vì a,b,c>0 nen(3+a^2+b^2+c^2)>0                                                                                          vay minA=0 khi a=b=c=0

1) Thay xyz = 1  , ta có : 

 \(\frac{1}{1+x+xy}+\frac{1}{1+y+yz}+\frac{1}{1+z+xz}=\frac{z}{z+xz+xyz}+\frac{xz}{xz+xyz+xyz^2}+\frac{1}{1+z+xz}\)

\(=\frac{z}{z+xz+1}+\frac{xz}{xz+1+z}+\frac{1}{z+xz+1}=\frac{z+xz+1}{z+xz+1}=1\)

2) Phân tích A thành nhân tử được \(A=\left(a-b\right)\left(b-c\right)\left(a-c\right)\left(a+b+c\right)\)

Vì a + b + c = 0 nên A = 0

3) Phân tích  A thành  \(\frac{\left(b-a\right)\left(a-c\right)\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=1\)

Đặt  \(u=\frac{x}{a};\)  và  \(v=\frac{y}{b}\)  \(\Rightarrow\)  \(\hept{\begin{cases}u,v\in Z\\u+v=1\\uv=-2\end{cases}}\)

Khi đó, ta có:

\(u+v=1\)

nên  \(\left(u+v\right)^3=1\)  \(\Leftrightarrow\)  \(u^3+v^3+3uv\left(u+v\right)=1\)

Do đó,  \(u^3+v^3=1-3uv\left(u+v\right)=1+6=7\)

Vậy,  \(\frac{x^3}{a^3}+\frac{y^3}{b^3}=7\)

\(ĐK:\)  \(a,b,c\ne0\)

Ta có: 

\(a+b+c=0\)

\(\Leftrightarrow\) \(a+b=-c\)

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

\(\Leftrightarrow\)  \(a^2+b^2+2ab=c^2\)

nên    \(a^2+b^2-c^2=-2ab\)

Tương tự với vòng hoán vị  \(b\rightarrow c\rightarrow a\)  ta cũng suy ra được:

\(\hept{\begin{cases}b^2+c^2-a^2=-2bc\\c^2+a^2-b^2=-2ca\end{cases}}\)

Khi đó, biểu thức  \(P\)  được viết lại dưới dạng:

\(P=-\frac{1}{2bc}-\frac{1}{2ca}-\frac{1}{2ab}=-\frac{1}{2}\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=-\frac{1}{2}\left(\frac{a+b+c}{abc}\right)=0\) (do \(a,b,c\ne0\)  )

1. Ta có: \(\frac{x}{a}+\frac{y}{b}=1\)

\(\Rightarrow\left(\frac{x}{a}+\frac{y}{b}\right)^3=1\)

\(\Rightarrow\left(\frac{x}{a}\right)^3+3.\frac{x}{a}.\frac{y}{b}\left(\frac{x}{a}+\frac{y}{b}\right)+\left(\frac{y}{b}\right)^3=1\)

\(\Rightarrow\left(\frac{x}{a}\right)^3+\left(\frac{y}{b}\right)^3+3.\left(-2\right).1=1\)

\(\Rightarrow\left(\frac{x}{a}\right)^3+\left(\frac{y}{b}\right)^3=1+6=7\)

2.Do \(a+b+c=0\)

Ta có:

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

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

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

CM tương tự: \(a^2+b^2-c^2=-2ab\)

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

Vậy \(P=\frac{1}{-2bc}+\frac{1}{-2ca}+\frac{1}{-2ab}=\frac{a+b+c}{-2abc}=0\)

3.

a)Ta có : \(x^2+y^2=1\Rightarrow x^4+2x^2y^2+y^4=1\Rightarrow x^4+y^4=1-2x^2y^2\)

Ta có :

\(\frac{x^4}{a}+\frac{y^4}{b}=\frac{1}{a+b}\)

\(\Leftrightarrow\frac{bx^4+ay^4}{ab}=\frac{1}{a+b}\)

\(\Leftrightarrow\left(a+b\right)\left(bx^4+ay^4\right)=ab\)

\(\Leftrightarrow\left(a+b\right)\left(bx^4+ay^4\right)-ab=0\)

\(\Leftrightarrow abx^4+a^2y^4+b^2x^4+aby^4-ab=0\)

\(\Leftrightarrow\left(ay^2\right)^2+\left(bx^2\right)^2+ab\left(x^4+y^4\right)-ab=0\)

\(\Leftrightarrow\left(ay^2\right)^2+\left(bx^2\right)^2+ab-2abx^2y^2-ab=0\)(Do \(x^4+y^4=1-2x^2y^2\))

\(\Leftrightarrow\left(ay^2-bx^2\right)^2=0\)

\(\Leftrightarrow ay^2=bx^2\)

b) Ta có : \(x^2+y^2=1\Rightarrow-x^2=y^2-1\)

Xét \(ay^2\left(a+b\right)-ab\)

\(\Leftrightarrow\left(ay\right)^2+aby^2-ab\)

\(\Leftrightarrow\left(ay\right)^2-abx^2\)

\(\Leftrightarrow a\left(ay^2-bx^2\right)=0\)(Do \(ay^2=bx^2\))

\(\Rightarrow ay^2\left(a+b\right)-ab=0\)

\(\Rightarrow ay^2\left(a+b\right)=ab\)

\(\Rightarrow\frac{ay^2}{ab}=\frac{1}{a+b}\)

\(\Rightarrow\frac{\left(ay^2\right)^{1004}}{\left(ab\right)^{1004}}=\frac{1}{\left(a+b\right)^{1004}}\)

\(\Rightarrow\frac{\left(ay^2\right)^{1004}+\left(bx^2\right)^{1004}}{\left(ab\right)^{1004}}=\frac{2}{\left(a+b\right)^{1004}}\)

\(\Rightarrow\frac{x^{2008}}{a^{1004}}+\frac{y^{2008}}{b^{1004}}=\frac{2}{\left(a+b\right)^{1004}}\)

Bạn ơi, mk làm câu a), các câu sau bạn tự làm dc k ???

các câu sau giải dựa vào câu a) nhé ^^

ok Milky Way