(x+y)^2  =a^2

x^2 +2xy +y^2 =a^2

x^2+y^2 =a^2-2xy =a^2 -2b

x^3 +y^3 = (x+y)(x^2 -xy +y^2)

             =a(a^2-2b-b)

            =a(a^2-3b)

            =a^3- 3ab

(x^2 +y^2)^2=(a^2-2b)^2  ( cái này tính cho x^4 + y^4)

tương tự như câu đầu tiên 

x^5+ y^5 (cái đó mình không biết)

sai con khi

\(1.\)

\(a)\)

\(x^2+y^2\)

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

\(=a^2-2b\)

\(b)\)

\(x^3+y^3\)

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

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

\(=a\left(a^2-3b\right)\)

\(=a^3-3ab\)

\(c)\)

\(x^4+y^4\)

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

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

\(=a^4-4a^2b+2b^2\)

\(d)\)

\(x^5+y^5\)

\(=\left(x^2+y^2\right)\left(x^3+y^3\right)-x^2y^2\left(x+y\right)\)

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

\(=\left(a^2-2b\right)\left(a^3-3ab\right)-ab^2\)

\(=a^5-3a^3b-2a^3b+6ab^2-ab^2\)

\(=a^5-5a^3b+5ab^2\)

Bài 1:

a: =>x^3-x-6x-6=0

=>x(x-1)(x+1)-6(x+1)=0

=>(x+1)(x-3)(x+2)=0

hay \(x\in\left\{-1;3;-2\right\}\)

b: \(\Leftrightarrow x^2-6x+9+y^2+6y+9=0\)

=>(x-3)^2+(y+3)^2=0

=>x=3 và y=-3

a) 

A=\(x^2+y^2=\left(x^2+2xy+y^2\right)-2xy=\left(x+y\right)^2-2xy=a^2-2b\)

\(B=x^3+y^3=\left(x^3+3x^2y+3xy^2+y^3\right)-3x^2y-3xy^2=\left(x+y\right)^3-3xy\left(x+y\right)=a^3-3ab\)

\(C=x^5+y^5=\left(x^5+y^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4\right)-5x^4y-10x^3y^2-10x^2y^3-5xy^4\)

\(=\left(x+y\right)^5-5xy\left(x^3+2xy^2+2x^2y+y^3\right)=\left(x+y\right)^5-5xy\left(x^3+3xy^2+3x^2y+y^3-xy^2-x^2y\right)\)

\(=\left(x+y\right)^5-5xy\left(\left(x+y\right)^3-xy\left(x+y\right)\right)=a^5-5b\left(a^3-ab\right)\)

giup minh cau b o tren nha

a) Ta có : \(\left(x+y\right)^3=1^3=1\)

\(\Leftrightarrow x^3+y^3+3xy\left(x+y\right)=1\)

\(\Leftrightarrow x^3+y^3+3xy=1\) ( do x + y = 1 )

Áp dụng BĐT Cauchy-Schwarz ta có:

\(\left(\dfrac{x^3}{y^2}+\dfrac{y^3}{z^2}+\dfrac{z^3}{x^2}\right)\left(x+y+z\right)\ge\left(\dfrac{x^2}{y}+\dfrac{y^2}{z}+\dfrac{z^2}{x}\right)^2\)

Cần chứng minh \(\dfrac{x^2}{y}+\dfrac{y^2}{z}+\dfrac{z^2}{x}\ge x+y+z\)

Dễ thấy;\(VT=\dfrac{x^2}{y}+\dfrac{y^2}{z}+\dfrac{z^2}{x}\ge\dfrac{\left(x+y+z\right)^2}{x+y+z}=x+y+z\)

BĐT được chứng minh

\("="\Leftrightarrow x=y=z\)

Bài 2:

\(M=x^2-2xy+y^2=\left(x-y\right)^2=\left(-3\right)^2=9\)

\(N=x^2+y^2=\left(x-y\right)^2+2xy=9+2.10=29\)

\(P=x^3-3x^2y+3xy^2-y^3=\left(x-y\right)^3=\left(-3\right)^3=-27\)

\(Q=x^3-y^3=\left(x-y\right)^3+3xy\left(x-y\right)=\left(-3\right)^3+3.10.\left(-3\right)=-117\)

Bài 1:

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

b)  \(B=x^2+y^2=\left(x+y\right)^2-2xy=\left(-1\right)^2-2.\left(-12\right)=25\)

c)  \(C=x^3+3x^2y+3xy^2+y^3=\left(x+y\right)^3=\left(-1\right)^3=-1\)

d)  \(D=x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=\left(-1\right)^3-3.\left(-12\right).\left(-1\right)=-37\)

Có (a+b+c)² = 3(ab+bc+ac)

\(\Rightarrow a^2+b^2+c^2+2ab+2bc+2ac=3ab+3bc+3ac\)

\(\Rightarrow a^2+b^2+c^2+2ab+2bc+2ac-3ab-3bc-3ac\)\(=0\)

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

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

\(\Rightarrow2a^2+2b^2+2c^2=2ab+2bc+2ac\)

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

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

\(\Rightarrow a=b=c\)

1) Áp dụng bđt Cauchy-Schwarz:

\(A\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{x^3+x^2y+xy^2+y^3+y^2z+yz^2+z^3+z^2x+x^2z}\)

\(=\dfrac{\left(x^2+y^2+z^2\right)^2}{x\left(x^2+y^2+z^2\right)+y\left(x^2+y^2+z^2\right)+z\left(x^2+y^2+z^2\right)}=\dfrac{\left(x^2+y^2+z^2\right)^2}{\left(x+y+z\right)\left(x^2+y^2+z^2\right)}=\dfrac{x^2+y^2+z^2}{x+y+z}\ge\dfrac{\dfrac{\left(x+y+z\right)^2}{3}}{x+y+z}=\dfrac{x+y+z}{3}=\dfrac{2012}{3}\)

\("="\Leftrightarrow x=y=z=\dfrac{2012}{3}\)

2)

Áp dụng bđt AM-GM:

\(\dfrac{x^3}{x^2+y^2}=x-\dfrac{xy^2}{x^2+y^2}\ge x-\dfrac{xy^2}{2xy}=x-\dfrac{y}{2}\)

Chứng minh tương tự và cộng theo vế:

\(S\ge x-\dfrac{y}{2}+y-\dfrac{z}{2}+z-\dfrac{x}{2}=\dfrac{2015}{2}\)

\("="\Leftrightarrow x=y=z=\dfrac{2015}{3}\)

Mk vừa nghĩ ra 1 cách xem thử nhé :v

AM-GM:

\(\left\{{}\begin{matrix}xy\le\dfrac{x^2+y^2}{2}\\yz\le\dfrac{y^2+z^2}{2}\\xz\le\dfrac{x^2+z^2}{2}\end{matrix}\right.\Leftrightarrow A\ge\dfrac{x^3}{x^2+\dfrac{x^2+y^2}{2}+y^2}+\dfrac{y^3}{y^2+\dfrac{y^2+z^2}{2}+z^2}+\dfrac{z^3}{z^2+\dfrac{x^2+z^2}{2}+x^2}\)

\(=\dfrac{x^3}{\dfrac{3}{2}\left(x^2+y^2\right)}+\dfrac{y^3}{\dfrac{3}{2}\left(y^2+z^2\right)}+\dfrac{z^3}{\dfrac{3}{2}\left(x^2+z^2\right)}\)

Rút mẫu ra rồi làm như bài 2 thôi :>