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

1) Cho x+y=2 và x^2+y^2=10. Tính x^3+y^3. Giải

(x+y)^2=x^2+y^2+2xy => xy= -3 
x^3+y^3=(x+y)^3-3xy(x+y) = 26

2) Ta có: x^3+y^3 = (x+y)(x^2-xy+y^2) (1)

(x+y)^2=a^2

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

=> b+2xy=a^2

=> xy=\(\frac{a^2-b}{2}\)

Thay (1) vào đó ta có:

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

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

Ta có: \(x^2+y^2=\left(x+y\right)^2-2xy=2^2-2xy=4-2xy=10\Rightarrow2xy=-6\Rightarrow xy=-3\)

Vậy: \(x^3+y^3=2\left(10+3\right)=2.13=26\)

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

a) Ta có: A = (x + y)³ + 2x² + 4xy + 2y²

    A = 7³ + 2(x² + 2xy + y²)

   A = 343 + 2(x + y)²

 A = 343 + 2. 7²

  A = 343 + 98 = 441

b) B = (x - y)³ - x² + 2xy - y²

=> B = (-5)³ - (x² - 2xy + y²)

=> B = -125 - (x - y)²

=> B = -125 - (-5)²

=> B = -125 - 25 = -150

a) Theo đầu bài ta có:
\(x+y=2\Rightarrow x=2-y\)
\(x^2+y^2=10\)
\(\Rightarrow\left(2-y\right)^2+y^2=10\)
\(\Rightarrow4+y^2-4y+y^2=10\)
\(\Rightarrow2y^2-4y=6\)
\(\Rightarrow2\left(y^2-2y\right)=6\)
\(\Rightarrow y\left(y-2\right)=3\)
Mà \(\hept{\begin{cases}y-\left(y-2\right)=2\\y+\left(y-2\right)=k\end{cases}\Rightarrow\hept{\begin{cases}y=\frac{k+2}{2}\\y-2=\frac{k-2}{2}\end{cases}}}\)( với k là hằng số )
\(\Rightarrow y\left(y-2\right)=\frac{k+2}{2}\cdot\frac{k-2}{2}\)
\(\Rightarrow\frac{\left(k+2\right)\left(k-2\right)}{4}=3\)
\(\Rightarrow k^2-4=12\)
\(\Rightarrow k^2=16\)
\(\Rightarrow k=4;-4\)
- Nếu k = 4 thì:
\(\Rightarrow\hept{\begin{cases}y=\frac{k+2}{2}=3\\x=2-y=-1\end{cases}\Rightarrow x^3+y^3=-1+27=26}\)
- Nếu k = -4 thì:
\(\Rightarrow\hept{\begin{cases}y=\frac{k+2}{2}=-1\\x=2-y=3\end{cases}\Rightarrow x^3+y^3=27+-1=26}\)
Vậy x³ + y³ = 26

a, \(x+y=2\Rightarrow\left(x+y\right)^2=4\Rightarrow x^2+2xy+y^2=4\Rightarrow10+2xy=4\Rightarrow xy=-3\)

\(\Rightarrow x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)=2.13=26\)

vậy............

b, \(x+y=a\Rightarrow\left(x+y\right)^2=a^2\)

\(\Rightarrow x^2+2xy+y^2=a^2\)

\(\Rightarrow xy=\frac{a^2-b}{2}\)

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

Vậy....

1. Cho các số tự nhiên a,b,c thỏa mãn a²+b²+c²=ab+bc+ca và a+b+c=3. Tính M=a²⁰¹⁶+2015b²⁰¹⁵+2020c

a²+b²+c²=ab+bc+ca

<=> 2( a²+b²+c² ) =2( ab+bc+ca )

<=> 2a² + 2b² + 2c² = 2ab + 2bc + 2ca

<=> 2a² + 2b² + 2c² - 2ab - 2bc - 2ca = 0

<=> ( a² - 2ab + b² ) + ( b² - 2bc + c² ) + ( c² - 2ca + a² ) = 0

<=> ( a - b )² + ( b - c )² + ( c - a )² = 0

Dễ chứng minh VT ≥ 0 ∀ a,b,c. Dấu "=" xảy ra <=> a=b=c

Lại có a+b+c=3 => a=b=c=1

từ đây bạn thế vào tính M nhé :))

2.Cho x>y>0. Chứng minh \(\frac{x-y}{x+y}< \frac{x^2-y^2}{x^2+y^2}\)

Ta có : \(\frac{x^2-y^2}{x^2+y^2}>\frac{x-y}{x+y}\)

<=> \(\frac{x^2-y^2}{x^2+y^2}-\frac{x-y}{x+y}>0\)

<=> \(\frac{\left(x^2-y^2\right)\left(x+y\right)}{\left(x^2+y^2\right)\left(x+y\right)}-\frac{\left(x^2+y^2\right)\left(x-y\right)}{\left(x^2+y^2\right)\left(x+y\right)}>0\)

<=> \(\frac{x^3+x^2y-xy^2-y^3}{\left(x^2+y^2\right)\left(x+y\right)}-\frac{x^3-x^2y+xy^2-y^3}{\left(x^2+y^2\right)\left(x+y\right)}>0\)

<=> \(\frac{x^3+x^2y-xy^2-y^3-x^3+x^2y-xy^2+y^3}{\left(x^2+y^2\right)\left(x+y\right)}>0\)

<=> \(\frac{2x^2y-2xy^2}{\left(x^2+y^2\right)\left(x+y\right)}>0\)

<=> \(\frac{2xy\left(x-y\right)}{\left(x^2+y^2\right)\left(x+y\right)}>0\)( đúng vì x > y > 0 )

=> đpcm 

a , Ta có: \(a+b=10\Rightarrow a=10-b\)

\(a+b=10\Rightarrow\left(a+b\right)^2=100\)

\(\Leftrightarrow a^2+2ab+b^2=100\)

\(\Leftrightarrow2ab=100-\left(a^2+b^2\right)=100-52=48\Rightarrow ab=24\)\(\Leftrightarrow\left(10-b\right)b=24\Leftrightarrow10b+b^2-24=0\)

\(\Leftrightarrow b^2+10b+25-49=0\)

\(\Leftrightarrow\left(b+5\right)^2=49\Rightarrow\left[{}\begin{matrix}b+5=7\\b+5=-7\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}b=2\\b=-12\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}a=8\\a=-2\end{matrix}\right.\)b, ta có:

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

\(\Leftrightarrow x^3+3x^2y+3xy^2+y^3=1\)

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

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

c, \(a+b=13\Rightarrow\left(a+b\right)^2=169\)

\(\Leftrightarrow a^2+2ab+b^2=169\Rightarrow a^2+b^2=169-2ab=169-2.9=151\)\(\Rightarrow a^3+b^3=\left(a+b\right)\left(a^2+b^2+ab\right)=13.\left(151+9\right)=2080\)

\(d,x+y=7\Rightarrow\left(x+y\right)^2=49\)

\(\Leftrightarrow x^2+2xy+y^2=49\Rightarrow2xy=49-\left(x^2+y^2\right)=16\Rightarrow xy=8\)

\(\Rightarrow x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)=7.\left(33-8\right)=175\)

\(1,A=\frac{1}{x^2+y^2}+\frac{1}{xy}=\frac{1}{x^2+y^2}+\frac{1}{2xy}+\frac{1}{2xy}\)

                                             \(\ge\frac{4}{\left(x+y^2\right)}+\frac{1}{\frac{\left(x+y\right)^2}{2}}\ge\frac{4}{1}+\frac{2}{1}=6\)

Dấu "=" <=> x= y = 1/2

\(2,A=\frac{x^2+y^2}{xy}=\frac{x}{y}+\frac{y}{x}=\left(\frac{x}{9y}+\frac{y}{x}\right)+\frac{8x}{9y}\ge2\sqrt{\frac{x}{9y}.\frac{y}{x}}+\frac{8.3y}{9y}\)

                                                                                                  \(=2\sqrt{\frac{1}{9}}+\frac{8.3}{9}=\frac{10}{3}\)

Dấu "=" <=> x = 3y

bài 3 min hay max ?

làm họ mình đi

huhu co to chua day