\(A=x^3+y^3+3xy=\left(x+y\right)^3-3xy\left(x+y\right)+3xy=1+0=1\)

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

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

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

\(x+y=2;x^2+y^2=10\text{ do đó:}xy=-3\text{ nên }\left(x-y\right)^2=16\text{ do đó: }x-y=4\text{ hoặc }x-y=-4\)

\(\text{giải ra được:}x=3;y=-1\text{ hoặc ngược lại nên }x^3+y^3=-26\text{ hoặc }26\)

A = x³ + y³ + 3xy

= x³ + 3x²y + 3xy² + y³ - 3x²y - 3xy² + 3xy

= ( x³ + 3x² + 3xy² + y³ ) - ( 3x²y + 3xy - 3xy )

= ( x + y )³ - 3xy( x + y - 1 )

= 1³ - 3xy( 1 - 1 )

= 1³ - 3xy.0

= 1 - 0 = 1

Vậy A = 1

b) B = x³ - y³ - 3xy

= x³ - 3x²y + 3xy² - y³ + 3x²y - 3xy² - 3xy

= ( x³ - 3x²y + 3xy² - y³ ) + ( 3x²y - 3xy² - 3xy )

= ( x - y )³ + 3xy( x - y - 1 )

= 1³ + 3xy( 1 - 1 )

= 1 + 3xy.0

= 1 + 0 = 1

Vậy B = 1

M = a³ + b³ + 3ab( a² + b² ) + 6a²b²( a + b )

= ( a + b )( a² - ab + b² ) + 3ab[ ( a + b )² - 2ab ] + 6a²b²( a + b )

= ( a + b )[ ( a + b )² - 3ab ] + 3ab[ ( a + b )² - 2ab ] + 6a²b²( a + b )

= 1.( 1 - 3ab ) + 3ab( 1 - 2ab ) + 6a²b².1

= 1 - 3ab + 3ab - 6a²b² + 6a²b²

= 1

Vậy M = 1

d) x + y = 2

⇔ ( x + y )² = 4

⇔ x² + 2xy + y² = 4

⇔ 10 + 2xy = 4 ( gt x² + y² = 10 )

⇔ 2xy = -6

⇔ xy = -3

x³ + y³ = x³ + 3x²y + 3xy² + y³ - 3x²y - 3xy²

            = ( x³ + 3x²y + 3xy² + y³ ) - ( 3x²y + 3xy² )

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

            = 2³ - 3.(-3).(2)

            = 8 + 18 = 26

d đâu bạn

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

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

a) Vì \(x-y=1\)

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

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

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

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

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

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

\(=4x^2+4xy+4y^2-3x^2-6xy-3y^2\)

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

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

\(=4\)

Bài 8: Cho a+b= 1 nha ( mk thiếu đề)

Bài 1:

Theo bài ra ta có:

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

\(=\left(5-y\right)^2-2\times2+\left(5-x\right)^2\)

\(=5^2-2\times5y+y^2-4+5^2-2\times5x+x^2\)

\(=25-10y+y^2+25-10x+x^2-4\)

\(=\left(25+25\right)-\left(10x+10y\right)+x^2+y^2-4\)

\(=50-10\left(x+y\right)+x^2+2xy+y^2-2xy-4\)

\(=50-10\times5+\left(x+y\right)^2-2\times2-4\)

\(=50-50+5^2-4-4\)

\(=25-8=17\)

Vậy giá trị của \(\left(x-y\right)^2\)là 17

ta có : M=2.(a^3  +b^3) -3.(a^2 + b^2)

       <=>M=2.(a+b)(a^2  -ab  +b^2)  - 3(a^2  +3b^2)

      <=>M=2(a^2  -ab  +b^2)  -3(a^2 +b^2)               vì a+b=1(gt)

      <=>M=-(a^2 +b^2 +2ab)

      <=>M=-(a+b)^2

      <=>M=-1  (vì a+b=1)

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