để sai phải không ạ ? tìm Max chứ 

VT = a³ + b³ + c³ - 3abc = (a + b)(a² - ab + b²) + c³ - 3abc

= (a + b)(a² + 2ab + b² - 3ab) + c³ - 3abc

= (a + b)³ - 3ab(a + b) + c³ - 3abc

= (a + b+ c)[(a + b)² - c(a + b) + c²] - 3ab(a + b+  c)

= (a + b + c))(a² + 2ab + b² - ac - bc + c² - 3abc)

= (a + b + c)(a² + b² + c² - ab - ac - bc) = VP

=> ĐPCM

Sửa đề :

VP= (a+b+c)(a²+b²+c²-ab-bc-ca)

     =a³+ab²+ac²-a²b-abc-ca²+ba²+b³+bc²-ab²-b²c-abc+ca²+cb²+c³-abc-bc²-c²a

     =a³+b³+c³-3abc

Cách này đỡ phức tạp hơn cách của edogawa conan

Bài làm:

Ta có: \(x^2+4y^2+z^2-2x-6z+8y+15\)

\(=\left(x^2-2x+1\right)+\left(4y^2+8y+4\right)+\left(z^2-6z+9\right)+1\)

\(=\left(x-1\right)^2+4\left(y+1\right)^2+\left(z-3\right)^2+1\ge1>0\left(\forall x,y,z\right)\)

x² + 4y² + z² - 2x - 6z + 8y + 15 

= ( x² - 2x + 1 ) + ( 4y² + 8y + 4 ) + ( z² - 6z + 9 ) + 1

= ( x - 1 )² + ( 2y + 2 )² + ( z - 3 )² + 1 ≥ 1 > 0 ∀ x,y,z ( đpcm )

Bài làm:

a) \(x^2-6x+4=\left(x^2-6x+9\right)-5=\left(x-3\right)^2-\left(\sqrt{5}\right)^2\)

\(=\left(x-3-\sqrt{5}\right)\left(x-3+\sqrt{5}\right)\)

b) \(x^2-4x+3=x^2-x-3x+3=\left(x-1\right)\left(x-3\right)\)

c) \(6x^2-5x+1=6x^2-3x-2x+1=\left(2x-1\right)\left(3x-1\right)\)

d) \(3x^2+13x-10=3x^2+15x-2x-10=\left(x-5\right)\left(3x-2\right)\)

Ta có: xy = 3 (1)

x + 2y = 7 <=> x = 7 - 2y (2)

Thay (2) vào (1) => (7 - 2y)y = 3

<=> -2y²  + 7y = 3

<=> 2y² - 7y + 3 = 0

<=> 2y² - 6y - y + 3 = 0

<=> (2y - 1)(y - 3) = 0

<=> \(\orbr{\begin{cases}2y-1=0\\y-3=0\end{cases}}\)

<=> \(\orbr{\begin{cases}y=\frac{1}{2}\\y=3\end{cases}}\)

Với y = 1/2 => x = 7 - 2.1/2 = 7 (tm vì x = 7 > 2y = 1/2.2 = 1)

Với y = 3=> x = 7  - 2.3/2 = 4 (ktm: vì 2y = 6; x < 2y)

Khi đó: x⁵ - 32y⁵ = 7⁵ - 32. (1/2)⁵ = 16806

Bài làm:

Ta có: \(a^2+2b^2+3=\left(a^2+b^2\right)+\left(b^2+1\right)+2\ge2ab+2b+2=2\left(ab+b+1\right)\)

\(\Rightarrow\frac{1}{a^2+2b^2+3}\le\frac{1}{2\left(ab+b+1\right)}\)

Tương tự ta CM được:

\(\frac{1}{b^2+2c^2+3}\le\frac{1}{2\left(bc+c+1\right)}\)

\(\frac{1}{c^2+2a^2+3}\le\frac{1}{2\left(ca+a+1\right)}\)

Cộng vế 3 BĐT trên ta được:

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

\(=\frac{1}{2}\left(\frac{1}{ab+b+1}+\frac{ab}{ab^2c+abc+ab}+\frac{b}{abc+ab+b}\right)\)

\(=\frac{1}{2}\left(\frac{1}{ab+b+1}+\frac{ab}{b+1+ab}+\frac{b}{1+ab+b}\right)\)

\(=\frac{1}{2}.\frac{ab+b+1}{ab+b+1}=\frac{1}{2}.1=\frac{1}{2}\)

Dấu "=" xảy ra khi: \(a=b=c=1\)

p/s : đéo biết làm thì câm mẹ mồm lại , loại súc vật như bạn ý thì cút khỏi olm cho sạch ạ !

Theo Cauchy ta dễ có : \(b^2+1\ge2\sqrt{b^2}=2b\)

\(a^2+b^2\ge2\sqrt{a^2b^2}=2ab\)

Khi đó  : \(\frac{1}{a^2+2b^2+3}\le\frac{1}{2+2b+2ab}=\frac{1}{2\left(ab+b+1\right)}\)

Bằng cách chứng minh tương tự rồi cộng theo vế các bđt cùng chiều thì ta được : 

\(VT\le\frac{1}{2}.\frac{1}{ab+b+1}+\frac{1}{2}.\frac{1}{bc+c+1}+\frac{1}{2}.\frac{1}{ca+a+1}=\frac{1}{2}.\left(\frac{1}{ab+b+1}+\frac{1}{bc+c+1}+\frac{1}{ca+a+1}\right)\)

Đặt \(A=\frac{1}{ab+b+1}+\frac{1}{bc+c+1}+\frac{1}{ca+a+1}=\frac{ac}{abc.c+abc+ac}+\frac{a}{abc+ca+1}+\frac{1}{ca+a+1}=1\)

Từ đó ta thu được \(VT\le\frac{1}{2}.1=\frac{1}{2}\)hay \(\frac{1}{a^2+2b^2+3}+\frac{1}{b^2+2c^2+3}+\frac{1}{c^2+2a^2+3}\le1\)

Đẳng thức xảy ra khi và chỉ khi \(a=b=c=1\)

Vậy ta có điều phải chứng minh

P/s: Ko chắc lắm.

\(A=x^3+y^3+6xy-3x-3y+1\)

\(A=\left(x+y\right)\left(x^2-xy+y^2\right)-3\left(x+y\right)+6xy+1\)

\(A=\left(x+y\right)\left(x^2+2xy+y^2-2xy-xy\right)-3\left(x+y\right)+6xy+1\)

\(A=\left(x+y\right)\left[\left(x+y\right)^2-3xy\right]-3\left(x+y\right)+6xy+1\)

\(A=\left(x+y\right)\left[\left(x+y\right)^2-3xy-3\right]+6xy+1\)

Thay x+y=2 vào biểu thức, ta có:

\(A=2\left(2^2-3xy-3\right)+6xy+1\)

\(A=2\left(1-3xy\right)+6xy+1\)

\(A=2-6xy+6xy+1\)

\(A=3\)

\(B=x^2-y^2+4y+1\)

\(B=\left(x-y\right)\left(x+y\right)+4y+1\)

\(B=2\left(x-y\right)+4y+1\)

\(B=2x-2y+4y+1\)

\(B=2x+2y+1\)

\(B=2\left(x+y\right)+1=2.2+1=5\)

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:

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

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

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

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

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

\(A=-1-2xy\)

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

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