1) 

Ta có: x+y=2

nên \(\left(x+y\right)^2=4\)

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

\(\Leftrightarrow2xy=2\)

hay xy=1

Ta có: \(x^3+y^3\)

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

\(=2^3-3\cdot1\cdot2\)

=2

2)\(x^2+y^2=\left(x+y\right)^2-2xy=8^2-2\cdot\left(-20\right)=104\)

\(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=8^3-3\cdot\left(-20\right)\cdot8=512+480=992\)

\(x^2+y^2+xy=\left(x+y\right)^2-xy=8^2-\left(-20\right)=64+20=84\)

D

\(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)\\ =2^3-3\cdot1\cdot2=8-6=2\left(D\right)\)

 Câu 30. Cho x + y = 2 và xy = 1 giá trị của biểu thức x³ + y³ bằng:

A . 8             B. 6                         C. 4                       D. 2

 Bài giải:

\(x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)=\left(x+y\right)\left((x+y)^2-3xy\right)=2\cdot\left(2^2-3\cdot1\right)=2\)

`x^3+y^3`

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

`=3[(x+y)^2-3xy]`

`=3(3^2-2.3)`

`=3(9-6)=3.3=9`

Lời giải:
a.

$x^3+y^3=(x+y)^3-3xy(x+y)=9^3-3.9.18=243$

$x^4+y^4=(x^2+y^2)^2-2x^2y^2=[(x+y)^2-2xy]^2-2x^2y^2$

$=[9^2-2.18]^2-2.18^2=1377$

Nếu $x\geq y$ thì:

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

$=|x-y|[(x+y)^2-xy]=\sqrt{(x+y)^2-4xy}[(x+y)^2-xy]$

$=\sqrt{9^2-4.18}(9^2-18)=189$

Nếu $x< y$ thì $x^3-y^3=-189$

b.

$A=(x+y)^2-6(x+y)+y-5$

$=(-9)^2-6(-9)+y-5=130+y$

Chưa đủ cơ sở để tính biểu thức.

a) \(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=9^3-3\cdot18\cdot9=243\)

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

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

\(=\left(9^2-2\cdot18\right)^2-2\cdot18^2\)

\(=45^2-2\cdot324\)

=1377

Bài 6:

M= 2.2 - 2.3+3.2.3

M= 4 - 6 + 18

M= 20

Bài 7: 

P= 1.2 - 5.-1.-2 + 8.-2.2

P = 2 -10 -32

P= -44

Bài 8:

A (thiếu dữ kiện bn ơi)

B= -1.2 . 3.2 + -1.3 +3.3 +-1.3

B= -2 . 6 + -3 + 9 +-3

B= -2 . 6 - 3 + 9 - 3

B= -12 - 3 + 9 - 3

B= -9

\(\left\{{}\begin{matrix}x-y=4\\xy=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=y+4\\y\left(y+4\right)=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=y+4\\y^2+4y-1=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=y+4\\\left[{}\begin{matrix}y=-2+\sqrt{5}\\y=-2-\sqrt{5}\end{matrix}\right.\end{matrix}\right.\)

Với \(y=-2+\sqrt{5}\Rightarrow x=2+\sqrt{5}\)

Với \(y=-2-\sqrt{5}\Rightarrow x=2-\sqrt{5}\)

\(\Rightarrow A=x^2+y^2=\left(-2+\sqrt{5}\right)^2+\left(2+\sqrt{5}\right)^2=\left(2-\sqrt{5}\right)^2+\left(-2-\sqrt{5}\right)^2=18\)

\(B=x^3+y^3\Rightarrow\left[{}\begin{matrix}B=\left(2+\sqrt{5}\right)^3+\left(-2+\sqrt{5}\right)^3=34\sqrt{5}\\B=\left(2-\sqrt{5}\right)^3+\left(-2-\sqrt{5}\right)^3=-34\sqrt{5}\end{matrix}\right.\)

\(\Rightarrow C=x^4+y^4=\left(-2+\sqrt{5}\right)^4+\left(2+\sqrt{5}\right)^4=\left(2-\sqrt{5}\right)^4+\left(-2-\sqrt{5}\right)^4=322\)

-Câu hỏi?

a. \(25-y^2=8\left(x-2009\right)\)

\(\Leftrightarrow25-y^2=8x-16072\)

\(\Leftrightarrow25-y^2+16072=8x\)

\(\Leftrightarrow16097-y^2=8x\)

\(\Leftrightarrow\dfrac{16097-y^2}{8}=x\)

\(\Leftrightarrow\dfrac{16096+1-y^2}{8}=x\)

\(\Leftrightarrow2012+\dfrac{1-y^2}{8}=x\)

-Vì \(x,y\) nguyên \(\Rightarrow\left(1-y^2\right)⋮8\)

\(\Leftrightarrow\left(1-y\right)\left(y+1\right)⋮8\)

\(\Leftrightarrow\left(1-y\right)⋮8\) hay \(\left(y+1\right)⋮8\)

\(\Leftrightarrow\left(1-y\right)\in\left\{1;2;4;8;-1;-2;-4;-8\right\}\) hay \(\left(y+1\right)\in\left\{1;2;4;8;-1;-2;-4;-8\right\}\)

\(\Leftrightarrow y\in\left\{0;-1;-3;-7;2;3;5;9\right\}\) hay \(y\in\left\{0;1;3;7;-2;-3;-5;-9\right\}\)

\(\Rightarrow y\in\left\{0;\pm1;\pm2;\pm3;\pm5;\pm7;\pm9\right\}\)

-Từ đó bạn tính các giá trị x tương ứng với mỗi giá trị của y.