Sửa đề: Các dấu bằng ở yêu cầu là dấu cộng.

1. Có: \(x+y=3\)

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

\(\Leftrightarrow x^2+2xy+y^2=9\)

\(\Leftrightarrow x^2+y^2=9-2\cdot1=7\) (do \(xy=1\))

\(------\)

Lại có: \(x+y=3\)

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

\(\Leftrightarrow x^3+y^3+3xy\left(x+y\right)=27\)

\(\Leftrightarrow x^3+y^3+3\cdot1\cdot3=27\) (do x + y = 3; xy = 1)

\(\Leftrightarrow x^3+y^3=18\)

Ta có: \(x^2+y^2=7\)

\(\Leftrightarrow\left(x^2+y^2\right)^2=7^2\)

\(\Leftrightarrow x^4+y^4+2\cdot\left(xy\right)^2=49\)

\(\Leftrightarrow x^4+y^4=49-2\cdot1=47\) (do xy = 1)

(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

\(1.\)

\(a)\)

\(x^2+y^2\)

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

\(=a^2-2b\)

\(b)\)

\(x^3+y^3\)

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

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

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

\(=a^3-3ab\)

\(c)\)

\(x^4+y^4\)

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

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

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

\(d)\)

\(x^5+y^5\)

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

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

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

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

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

câu 1:

ta có: \(x^2+y^2=4\Leftrightarrow\left(x^2+2xy+y^2\right)-2xy=4\Leftrightarrow\left(x+y\right)^2-2xy=4\Leftrightarrow9-2xy=4\Leftrightarrow-xy=-\frac{5}{2}\)

\(x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)=3.\left(4-xy\right)=3\left(4-\frac{5}{2}\right)=\frac{9}{2}\)

câu 2: tương tự ở trên tính xy rồi lắp vào hằng đẳng thức: \(x^3-y^3=\left(x-y\right)\left(x^2+y^2+xy\right)\)

6x = 24

  x = 24 : 6

  x = 4

Vậy x = 4

a, \(x^3+y^3+3xy=\left(x+y\right)\left(x^2-xy+y^2\right)+3xy=x^2-xy+y^2+3xy=x^2+2xy+y^2=\left(x+y\right)^2=1\)

b, tương tự a

c, Sửa đề Cho a+b=1. Tính giá trị của các biểu thứ :A= a³+b³+3ab(a²+b²)+ 6a²b²(a+b)

\(A=\left(a+b\right)\left(a^2-ab+b^2\right)+3ab\left[\left(a+b\right)^2-2ab\right]+6a^2b^2\left(a+b\right)\)

\(=\left(a+b\right)\left[\left(a+b\right)^2-3ab\right]+3ab\left[\left(a+b\right)^2-2ab\right]+6a^2b^2\left(a+b\right)\)

Thay a+b=1 vào A ta có:

\(A=1-3ab+3ab\left(1-2ab\right)+6a^2b^2\)

\(=1-3ab+3ab-6a^2b^2+6a^2b^2=1\)

d. \(B=x^2+2xy+y^2-4x-4y+1=\left(x+y\right)^2-4\left(x+y\right)+1=\left(x+y\right)\left(x+y-4\right)+1\)

Thay x+y=3 vào B ta có:

\(B=3\left(3-4\right)+1=3.\left(-1\right)+1=-3+1=-2\)

a. ta có : \(x^2+y^2=\left(x+y\right)^2-2xy=1^2-2\times\left(-6\right)=13\)

\(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=1^3-3\times\left(-6\right)\times1=19\)

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

\(=\left(x+y\right)\left[\left(x^2+y^2\right)^2-x^2y^2-xy\left(x^2+y^2\right)\right]=1.\left(13^2-\left(-6\right)^2-\left(-6\right).13\right)=211\)

b.\(x^2+y^2=\left(x-y\right)^2+2xy=1+2\times6=13\)

\(x^3-y^3=\left(x-y\right)^3+3xy\left(x-y\right)=1^3+6.3.1=19\)​​

​\(x^5-y^5=\left(x-y\right)\left[\left(x^4+x^3y+x^2y^2+xy^3+y^4\right)\right]\)​​

\(=\left(x-y\right)\left[\left(x^2+y^2\right)^2-x^2y^2+xy\left(x^2+y^2\right)\right]=1.\left(13^2-6^2+6.13\right)=211\)

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: \(A=x\left(x+2\right)+y\left(y-2\right)-2xy+37\)

\(=x^2+2x+y^2-2y-2xy+37\)

\(=\left(x^2-2xy+y^2\right)+\left(2x-2y\right)+37\)

\(=\left(x-y\right)^2+2\left(x-y\right)+37\)

\(=\left(x-y\right)\left(x-y+2\right)+37\)(1)

Thay x-y=7 vào biểu thức (1), ta được:

\(A=7\cdot\left(7+2\right)+37=7\cdot9+37=100\)

Vậy: Khi x-y=7 thì A=100

b) Ta có: \(x+y=2\)

\(\Leftrightarrow\left(x+y\right)^2=4\)

\(\Leftrightarrow x^2+y^2+2xy=4\)

\(\Leftrightarrow2xy+10=4\)

\(\Leftrightarrow2xy=-6\)

\(\Leftrightarrow xy=-3\)

Ta có: \(A=x^3+y^3\)

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

Thay x+y=2; \(x^2+y^2=10\) và xy=-3 vào biểu thức (2), ta được:

\(A=2\cdot\left(10+3\right)=2\cdot13=26\)

Vậy: Khi x+y=2 và \(x^2+y^2=10\) thì A=26

\(\Rightarrow A=x^2+2x+y^2-2y-2xy+37=x^2-2xy+y^2+2\left(x-y\right)+37=\left(x-y\right)^2+2\left(x-y\right)+37=7^2+2\cdot7+37=100\)

\(\Rightarrow A=x^3+y^3=\left(x+y\right)\left(x^2+y^2-xy\right)=\left(x+y\right)\left[x^2+y^2-\dfrac{\left(x+y\right)^2-\left(x^2+y^2\right)}{2}\right]=2\cdot\left[10+3\right]=2\cdot13=26\) \(\Rightarrow\left\{{}\begin{matrix}x+y=-z\\x+z=-y\\y+z=-x\end{matrix}\right.\) \(\Rightarrow P=\left(\dfrac{x+y}{y}\right)\left(\dfrac{y+z}{z}\right)\left(\dfrac{x+z}{x}\right)=-\dfrac{z}{y}\cdot\dfrac{-x}{z}\cdot-\dfrac{y}{x}=-1\)

\(B=8x^2+2x-8x^3-8x^2+8x^3-2x+3=3\)

\(C=x^3-3x^2+3x-1+x^3+3x^2+3x+1+2x^3-8x=4x^3-2x\)

\(D=\left(x+y-5\right)^2-2\left(x+y-5\right)\left(x+3\right)+\left(x+3\right)^2=\left(x+y-5-x-3\right)^2=\left(y-8\right)^2\)

câu 2. ta có 

a.\(\left(x-y\right)^2=\left(x+y\right)^2-4xy=7^2-4\times12=1\)

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

Các bài này đưa về dạng Hằng đẳng thức là được . Làm ra dài lắm bạn ạ !

Ns nghe dễ lắm, lm thử đee =))

A. \(x+y=1\Rightarrow\left(x+y\right)^3=1\Rightarrow x^3+3x^2y+3xy^2+y^3=1\)

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

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

B. \(x-y=1\Rightarrow\left(x-y\right)^3=1\Rightarrow x^3-3x^2y+3xy^2-y^3=1\)

\(\Rightarrow x^3-3xy\cdot\left(x-y\right)-y^3=1\)

\(\Rightarrow x^3-3xy-y^3=1\)

C. \(M=a^3+b^3+3ab\left[\left(a+b\right)^2-2ab\right]+6a^2b^2\left(a+b\right)\)

\(M=a^3+b^3+3ab\left(1-2ab\right)+6a^2b^2=a^3+b^3+3ab-6a^2b^2+6a^2b^2\)

\(M=a^3+b^3+3ab=1\)(Theo hệ quả câu A).

D. Từ \(x+y=2\Rightarrow\left(x+y\right)^2=4\Rightarrow x^2+y^2+2xy=4\Rightarrow10+2xy=4\Rightarrow xy=-3\)

Mà, \(\left(x+y\right)^3=x^3+y^3+3xy\left(x+y\right)\)

\(\Leftrightarrow2^3=x^3+y^3+3\left(-2\right)\cdot2\Leftrightarrow x^3+y^3=8+12=20\)