Ta có: a + b +  c = 0 => a + b = -c; b + c = -a; a + c = -b

a + b + c = 0 <=> a + b = -c

<=> (a + b)³ = (-c)³ 

<=> a³ + 3a²b + 3ab² + b³ = -c³

<=> a³ + b³ + c³ = -3ab(a + b)

<=> a³ + b³ + c³ = 3abc (vì a + b = -c)

Khi đó: Q = \(\left(\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}\right)\left(\frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a}\right)\)

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

Q = \(3+\left(\frac{a\left(a-b\right)}{c\left(b-c\right)}+\frac{a\left(c-a\right)}{b\left(b-c\right)}\right)+\left(\frac{b\left(a-b\right)}{c\left(c-a\right)}+\frac{b\left(b-c\right)}{a\left(c-a\right)}\right)+\left(\frac{c\left(b-c\right)}{a\left(a-b\right)}+\frac{c\left(c-a\right)}{b\left(a-b\right)}\right)\)

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

Q = \(3+\frac{a\left(ab-b^2+c^2-ac\right)}{bc\left(b-c\right)}+\frac{b\left(a^2-ab+bc-c^2\right)}{ac\left(c-a\right)}+\frac{c\left(b^2-bc+ac-a^2\right)}{ab\left(a-b\right)}\)

Q = \(3+\frac{a\left[a\left(b-c\right)-\left(b-c\right)\left(b+c\right)\right]}{bc\left(b-c\right)}+\frac{b\left[b\left(c-a\right)-\left(c-a\right)\left(c+a\right)\right]}{ac\left(c-a\right)}+\frac{c\left[c\left(a-b\right)-\left(a-b\right)\left(a+b\right)\right]}{ab\left(a-b\right)}\)

Q = \(3+\frac{a\left[a-\left(b+c\right)\right]}{bc}+\frac{b\left(b-\left(c+a\right)\right)}{ac}+\frac{c\left[c-\left(a+b\right)\right]}{ab}\)

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

Q = \(3+\frac{2a^2}{bc}+\frac{2b^2}{ac}+\frac{2c^2}{ab}\)

Q = \(3+\frac{2a^3+2b^3+2c^3}{abc}\)

Q = \(3+\frac{2\left(a^3+b^3+c^3\right)}{abc}\)

Q = \(3+\frac{2.3abc}{abc}=3+6=9\)

Bài làm:

Đặt \(M=\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}\)

\(\Leftrightarrow abc.M=ab\left(a-b\right)+bc\left(b-c\right)+ca\left(c-a\right)\)

\(\Leftrightarrow abc.M=ab\left(a-b\right)+b^2c-bc^2+c^2a-ca^2\)

\(\Leftrightarrow abc.M=ab\left(a-b\right)+c^2\left(a-b\right)-c\left(a-b\right)\left(a+b\right)\)

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

\(\Leftrightarrow abc.M=\left(a-b\right)\left(a-c\right)\left(b-c\right)\)

\(\Rightarrow M=\frac{\left(a-b\right)\left(a-c\right)\left(b-c\right)}{abc}\)

Đặt \(N=\frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a}\)

\(\Rightarrow\left(a-b\right)\left(b-c\right)\left(c-a\right).N=c\left(b-c\right)\left(c-a\right)+a\left(a-b\right)\left(c-a\right)+b\left(a-b\right)\left(b-c\right)\)

Mà \(a+b+c=0\Rightarrow\hept{\begin{cases}a=-b-c\\b=-c-a\\c=-a-b\end{cases}}\)

Thay vào ta được:

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

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

\(N=\frac{c\left(c-a\right)\left(2b-c-a\right)+b\left(a-b\right)\left(a+b-2c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)

\(N=\frac{c\left(c-a\right)\left(2b+b\right)+b\left(a-b\right)\left(-c-2c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)

\(N=\frac{3bc\left(c-a\right)-3bc\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)

\(N=\frac{3bc\left(b+c-2a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)

\(N=\frac{-9abc}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=\frac{9abc}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)

Mà \(Q=M.N=\frac{\left(a-b\right)\left(a-c\right)\left(b-c\right)}{abc}.\frac{9abc}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}=9\)

Vậy Q = 9 

\(A=x^2+y^2-xy^2-x^2y+2xy-5\)

\(=\left(x+y\right)^2-xy\left(y+x\right)-5\)

\(=2^2-2xy-5=-\left(2xy+1\right)\)

Trả lời:

\(A=x^2+y^2-x^2y-xy^2+2xy-5\)

\(A=\left(x^2+2xy+y^2\right)-xy.\left(x+y\right)-5\)

\(A=\left(x+y\right)^2-xy.\left(x+y\right)-5\)

\(A=2^2-xy.2-5\)

\(A=4-2xy-5\)

\(A=-1-2xy\)

\(A=-\left(1+2xy\right)\)

Học tốt 

                                    Bài làm :

a) Tổng vận tốc của 2 xe là :

52 + 28 = 80 (km/h)

Vậy 2 xe đi ngược chiều sẽ gặp nhau sau :

60 : 80 = 0,75 (giờ) = 45 phút .

b)Hiệu vận tốc của 2 xe là :

52 - 28 = 24 (km/h)

Vậy 2 xe đi cùng chiều sẽ gặp nhau sau :

60 : 24 = 2,5 (giờ) = 2 giờ 30 phút .

Chúc bạn học tốt !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

bài làm

a) Tổng vận tốc của 2 xe là :

          52+28=80(km/giờ)

    Số giờ 2 xe gặp nhau là:

         60:80=0,75 giờ =45 phút

b) Hiệu vận tốc 2 xe là:

         52-28=24(km/giờ)

   Số giờ 2 xe gặp nhau là:

         60:24=2,5 giờ =2 giờ 30 phút 

                 Đáp số : a) 45 phút

                               b) 2 giờ 30 phút

nhớ k cho mình nha !

\(\frac{3x+2}{x-1}+\frac{2x-4}{x+2}=5\)

\(\Rightarrow\frac{3x-3+5}{x-1}+\frac{2x+4-8}{x+2}=5\)

\(\Rightarrow3+\frac{5}{x-1}+2-\frac{8}{x+2}=5\)

\(\Rightarrow\frac{5}{x-1}-\frac{8}{x+2}=0\)

\(\Leftrightarrow\frac{5}{x-1}=\frac{8}{x+2}\)

\(\Leftrightarrow5\left(x+2\right)=8\left(x-1\right)\)

\(\Leftrightarrow5x+10=8x-8\)

\(\Leftrightarrow8+10=8x-5x\)

\(\Leftrightarrow18=3x\)

\(\Leftrightarrow x=6\)

\(\frac{3x+2}{x-1}+\frac{2x-4}{x+2}=5\)

\(\Rightarrow\frac{3x-3+5}{x-1}+\frac{2x+4-8}{x+2}=5\)

=> \(\frac{3\left(x-1\right)+5}{x-1}+\frac{2\left(x+2\right)-8}{x+2}=5\)

=> \(3+2+\frac{5}{x-1}-\frac{8}{x+2}=5\)

=> \(\frac{5}{x-1}=\frac{8}{x+2}\)

=> 5(x + 2) = 8(x - 1)

=> 5x + 10 = 8x - 8

=> 8x - 5x = 10 + 8

=> 3x = 18

=> x = 6

Vậy x = 6

Dùng BĐT phụ:

\(\left(x+y\right)^2\ge4xy\)

Ta có:\(\left(a+b\right)^2\ge4ab\)

          \(\left(b+c\right)^2\ge4bc\)

           \(\left(c+a\right)^2\ge4ca\)

\(\Rightarrow\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2\ge64\left(abc\right)^2\)

\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge8abc\)

 Dấu “=” xảy ra khi a = b = c
 

Áp dụng BĐT Cauchy - Schwarz :

\(a+b\ge2\sqrt{ab}\)

\(b+c\ge2\sqrt{bc}\)

\(c+a\ge2\sqrt{ca}\)

\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge2\sqrt{ab}.2\sqrt{bc}.2\sqrt{ca}\)

\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge8abc\left(dpcm\right)\)

khongxe

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

Thay vào ta có: \(\sqrt{\frac{bc}{a+bc}}=\sqrt{\frac{bc}{a^2+ab+ca+bc}}=\sqrt{\frac{bc}{\left(a+b\right)\left(a+c\right)}}\)

Áp dụng Cauchy ngược: \(\sqrt{\frac{bc}{a+bc}}=\sqrt{\frac{bc}{a^2+ab+ca+bc}}\le\frac{\frac{b}{a+b}+\frac{c}{a+c}}{2}\)

Tương tự ta CM được: \(\sqrt{\frac{ab}{c+ab}}\le\frac{\frac{a}{c+a}+\frac{b}{c+b}}{2}\)

                                     \(\sqrt{\frac{ca}{b+ca}}\le\frac{\frac{c}{b+c}+\frac{a}{b+a}}{2}\)

Cộng vế 3 BĐT trên ta được:

\(P\le\frac{\frac{a}{c+a}+\frac{b}{c+b}+\frac{b}{a+b}+\frac{c}{a+c}+\frac{c}{b+c}+\frac{a}{b+a}}{2}\)

\(=\frac{\left(\frac{a}{c+a}+\frac{c}{a+c}\right)+\left(\frac{b}{c+b}+\frac{c}{b+c}\right)+\left(\frac{a}{b+a}+\frac{b}{a+b}\right)}{2}\)

\(=\frac{1+1+1}{2}=\frac{3}{2}\)

Dấu "=" xảy ra khi: \(a=b=c=\frac{1}{3}\)

Vậy \(Max_P=\frac{3}{2}\Leftrightarrow a=b=c=\frac{1}{3}\)

Ta có :

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

Tương tự :  \(a+bc=\left(a+b\right)\left(a+c\right);c+ab=\left(c+b\right)\left(c+a\right)\)

 \(\Rightarrow P=\sqrt{\frac{ab}{\left(a+c\right)\left(b+c\right)}}+\sqrt{\frac{bc}{\left(a+b\right)\left(a+c\right)}}+\sqrt{\frac{ca}{\left(c+a\right)\left(c+b\right)}}\)

Áp dụng BĐT cauchy :

\(\sqrt{\frac{ab}{\left(a+c\right)\left(b+c\right)}}\le\frac{1}{2}\left(\frac{a}{a+c}+\frac{b}{b+c}\right)\)

\(\sqrt{\frac{bc}{\left(a+b\right)\left(a+c\right)}}\le\frac{1}{2}\left(\frac{b}{a+b}+\frac{c}{a+c}\right)\)

\(\sqrt{\frac{ca}{\left(c+b\right)\left(c+a\right)}}\le\frac{1}{2}\left(\frac{c}{c+b}+\frac{a}{c+a}\right)\)

Cộng vế với vế :

\(\Rightarrow P\le\frac{1}{2}\left(\frac{a}{a+c}+\frac{b}{b+c}+\frac{b}{a+b}+\frac{c}{a+c}+\frac{c}{c+b}+\frac{a}{c+a}\right)\)

\(\Leftrightarrow P\le\frac{1}{2}\left(\frac{a+c}{a+b}+\frac{b+c}{b+c}+\frac{a+b}{a+b}\right)=\frac{1}{2}.3=\frac{3}{2}\)

Dấu "=" xảy ra \(\Leftrightarrow a=b=c=\frac{1}{3}\)

Đặt \(A=x^2+y^2+z^2+xy+yz+zx\)

Áp dụng BĐT Bunyakovsky dạng phân thức, ta được: \(2A=x^2+y^2+z^2+\left(x+y+z\right)^2\ge\frac{\left(x+y+z\right)^2}{3}+\left(x+y+z\right)^2\)

\(=\frac{4\left(x+y+z\right)^2}{3}=12\Rightarrow A\ge6\)

Đẳng thức xảy ra khi x = y = z = 1

Bài làm:

Ta có: \(x^2+y^2\ge2xy\Leftrightarrow xy\le\frac{x^2+y^2}{2}=\frac{1}{2}\)

Thay vào A ta được: 

\(A=\frac{-2xy}{1+xy}\ge\frac{-2.\frac{1}{2}}{1+\frac{1}{2}}=-\frac{2}{3}\)

Dấu "=" xảy ra khi: \(x=y=\frac{1}{\sqrt{2}}\)

E = \(\frac{x^4+1}{\left(x^2+1\right)^2}\)

để E lớn nhất 

thì \(\left(x^2+1\right)^2\) phải nhỏ nhất

mà \(\left(x^2+1\right)^2\)> 0 và khác 0 ( vì là mẫu số )

=> \(\left(x^2+1\right)^2=1\)

=> \(x^2+1=1\)

=> \(x^2=0\)

=> x = 0

để E đạt giá trị lớn nhất thì x = 0

\(E=\frac{x^4+1}{\left(x^2+1\right)^2}=\frac{x^4+1}{x^4+2x^2+1}\le\frac{x^4+1}{x^4+1}=1\\ \Rightarrow maxE=1\Leftrightarrow x=0\)

\(E=\frac{x^4+1}{\left(x^2+1\right)^2}=\frac{x^4+1}{x^4+2x^2+1}=1-\frac{2x^2}{x^4+2x^2+1}\\ \ge1-\frac{2x^2}{2x^2+2x^2}=\frac{1}{2}\\ \Rightarrow minE=\frac{1}{2}\Leftrightarrow x=1\)