a) Ta có:

x + y = 3

=> ( x + y)² = 9

=> x² + 2xy + y² = 9

=> 10 + 2xy = 9

=> 2xy = 9 - 10 = -1

=> xy = -1/2 

Ta có:

 x³ + y³ = (x + y)(x² - xy + y²)

 = 3.(10 + 1/2) = 63/2

b) Ta có: x + y = a

=> (x + y)² = a²

=> x² + 2xy + y² = a²

=> b + 2xy = a²

=> xy = (a² - b)/2

Ta có:  x³ + y³ = (x + y)(x² + xy + y²)

 = a[b + (a² - b )/2] = ab + (a³ - b)/2.

Làm b) công thức tổng quát luôn

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

Thay x^2+y^2=b  vào ta được:

b+2xy=a^2 => xy=(a^2-b)/2 

TA có x^3+y^3 =(x+y)(x^2+y^2 -xy)= a [b+(a^2-b)/2] =ab +(a^3-ab)/2=ab/2+a^3/2

a,

\(x^2+y^2=\left(x+y\right)^2-2xy=1^2-2\cdot\left(-6\right)=1-\left(-12\right)=13\)

\(x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)=1\cdot\left[13-\left(-6\right)\right]=19\)

\(x^5+y^5=\left(x+y\right)\left(x^2+y^2\right)^2-\left(2x^3y^2+xy^4+x^4y+2x^2y^3\right)=169-\left[2\left(xy\right)^2\left(x+y\right)+xy\left(x^3+y^3\right)\right]=169-\left[2\cdot36\cdot1-6\cdot19\right]=211\)

b,

\(x^2+y^2=\left(x-y\right)^2+2xy=1+12=13\)

\(x^3-y^3=\left(x-y\right)\left(x^2+xy+y^2\right)=1\cdot\left(13+6\right)=19\)

x - 4 = 2 => x = 2 + 4 => x = 6

x + y = 4 mà x = 6 => y = 4 - 6 => y = -2

=> xy = 6 \(\times\) (-2) = -12

x³ - y³ = 6³ - (-2)³ = 224

Ta có:\(x-4=2\Rightarrow x=6^{\left(1\right)}\)

Thay \(^{\left(1\right)}\) vào \(x+y=4\) ,ta được:

\(6+y=4\Rightarrow y=-2^{\left(2\right)}\)

Thay \(^{\left(1\right),\left(2\right)}\) vào xy ,ta được:

\(xy=6.\left(-2\right)=-12\)

thay \(^{\left(1\right),\left(2\right)}\) vào \(x^3-y^3\), ta được:

\(x^3-y^3=6^3-\left(-2\right)^3=216-\left(-8\right)=216+8=224\)

a) B= 2x²-3x+1

=(2x²-2x)-(x-1)

=2x(x-1)-(x-1)

=(2x-1)(x-1)

\(\left|x\right|=\frac{1}{2}\)nên ta có \(x=\frac{1}{2}\)hoặc\(x=\frac{-1}{2}\)

nếu \(x=\frac{1}{2}\)thì

B=(2*\(\frac{1}{2}\)-1)(\(\frac{1}{2}\)-1)

B=0

nếu x= -1/2

thì B= (2*(-1/2)-1)(-1/2-1)

B=(-2)*(-3/2)

B=3

giúp e câu b vs a Phong

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

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

\(=3-6xy-2-6xy=-12xy+1\)

c: \(=\left(x+y\right)^3-3\left(x^2+y^2+2xy\right)+3\left(x+y\right)+2012\)

\(=101^2-3\cdot101^2+3\cdot101+2012\)

=1002013

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

\(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=2^3-3\cdot\left(-3\right)\cdot2=8-\left(-18\right)=26\)

b,

\(x+y=a\\ \Rightarrow\left(x+y\right)^2=a^2\\ \Rightarrow x^2+2xy+y^2=a^2\\ \Rightarrow2xy=a^2-b\Rightarrow xy=\dfrac{a^2-b}{2}\)

\(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=a^3-3\cdot\dfrac{a^2-b}{2}\cdot a=a^3-\dfrac{3a\left(a^2-b\right)}{2}=a^3-\dfrac{3a^3-3ab}{2}=a^3-1,5a^3+3ab=\left(1-1,5\right)a^3+3ab=0,5a^3+3ab=0,5a\left(a^2+6b\right)\)

a: x-y-z=0

=>x=y+z; y=x-z; z=x-y

\(K=\dfrac{x-z}{x}\cdot\dfrac{y-x}{y}\cdot\dfrac{z+y}{z}=\dfrac{y\cdot\left(-z\right)\cdot x}{xyz}=-1\)

b: Tham khảo:

1,Thực hiện phép tính :

a, (x + 2)⁹ : (x + 2)⁶

=(x+2)^9-6

=(x+2)³

b, (x - y) ⁴ : (x - 2)³

=(x-y)^4-3

=x-y

c, ( x²+ 2x + 4)⁵ : (x² + 2x + 4)

=(x²+2x+4)^5-1

=(x²+2x+4)⁴

d, 2(x² + 1)³ : 1/3(x² + 1)

=(2÷1/3).[(x²+1)³÷(x²+1)]

=6(x²+1)²

e, 5 (x - y)⁵ : 5/6 (x - y)²

=(5÷5/6).[(x-y)⁵÷(x-y)²]

=6(x-y)⁾³