(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

\(1.\)

\(a)\)

\(x^2+y^2\)

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

\(=a^2-2b\)

\(b)\)

\(x^3+y^3\)

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

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

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

\(=a^3-3ab\)

\(c)\)

\(x^4+y^4\)

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

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

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

\(d)\)

\(x^5+y^5\)

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

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

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

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

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

2)     => X/3 = Y/4

(2X^2 + Y^2)/(2.3^2 + 4^2) =  136/34 = 4

2X^2 = 4.18 = 72 => x  = 6

y^2 = 4.16 = 64 => y = 8

5)  (a+2b-3c)/(2+2.3 - 3.4) =  20/4 = 5

a = 10

2b = 30 => b = 15

3c = 60 => c = 20 

kho hieu qua 

18 voi 16 lay dau ra vay 

a.

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

Tại \(x=77;y=22\Rightarrow x\left(x+y+1\right)=77\left(77+22+1\right)=77.100=7700\)

b.

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

\(=\left(53-3\right)^2=50^2=2500\)

c.

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

\(=\left(2001+1999\right)\left(2001-1\right)=4000.2000=8000000\)

Bài 1:

Ta có:

[tex]\left\{\begin{matrix} xy^{2}+x+y+\frac{1}{y}=4 & \\ y^{2}+x+\frac{1}{y}=3 & \end{matrix}\right.(y\neq 0)[/tex]

Từ phương trình suy ra:

[tex]\left\{\begin{matrix} y(xy+1)+\frac{xy+1}{y}=4 & \\ y^{2}+\frac{xy+1}{y}=3 & \end{matrix}\right.[/tex]

Đặt [tex]xy+1=a,y=b(b\neq 0)[/tex] ta có:

[tex]\left\{\begin{matrix} b^{2}+\frac{a}{b}=3 & \\ ab+\frac{a}{b}=4 & \end{matrix}\right.[/tex]

[tex]\Rightarrow \left\{\begin{matrix} 3b-b^{3}=a & \\ ab^{2}+a=4b & \end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} 3b-b^{3}=a & \\ b\left ( 2b^{2}-b^{4}-1 \right )=0 & \end{matrix}\right.[/tex]

[tex]\Leftrightarrow \left\{\begin{matrix} b=0 & \\ a=0 & \end{matrix}\right.[/tex](Loại) hoặc [tex]\left\{\begin{matrix} b=1 & \\ a=2 & \end{matrix}\right.[/tex] hoặc [tex]\left\{\begin{matrix} b=-1 & \\ a=-2 & \end{matrix}\right.[/tex]

TH1: [tex]\left\{\begin{matrix} b=1 & \\ a=2 & \end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=1 & \\ y=1 & \end{matrix}\right.[/tex]

TH2: [tex]\left\{\begin{matrix} b=-1 & \\ a=-2 & \end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=3 & \\ y=-1 & \end{matrix}\right.[/tex]

Vậy hệ phương trình có hai nghiệm: [tex]\left\{\begin{matrix} x=1 & \\ y=1 & \end{matrix}\right.[/tex] hoặc [tex]\left\{\begin{matrix} x=3 & \\ y=-1 & \end{matrix}\right.[/tex]

Câu trả lời đầy đủ đây nhé:

Bài 2:

a: Ta có: \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)

\(\Leftrightarrow10x-16-12x+15=12x-16+11\)

\(\Leftrightarrow-14x=-4\)

hay \(x=\dfrac{2}{7}\)

b: Ta có: \(2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\)

\(\Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\)

\(\Leftrightarrow x^3=-8\)

hay x=-2

Bài 1: 

a: Ta có: \(I=x\left(y^2-xy^2\right)+y\left(x^2y-xy+x\right)\)

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

\(=xy\)

=1

b: Ta có: \(K=x^2\left(y^2+xy^2+1\right)-\left(x^3+x^2+1\right)\cdot y^2\)

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

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

\(=\dfrac{1}{4}-\dfrac{1}{4}=0\)

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

Với \(x=2-\sqrt{5};y=-2-\sqrt{5}\)

\(A=x^2+y^2=18\\ B=x^3-y^3=76\\ C=x^4+y^2=322\)

Với \(x=2+\sqrt{5};y=-2+\sqrt{5}\)

\(A=x^2+y^2=18\\ B=x^3-y^3=76\\ C=x^4+y^4=322\)

A=x^2+y^2

=(x-y)^2+2xy

=4^2+2=18

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

=4^3+3*1*4

=64+12=76

C=(x^2+y^2)^2-2x^2y^2

=18^2-2

=322

1) Ta có: \(\dfrac{1}{7}x^2y^3\cdot\left(-\dfrac{14}{3}xy^2\right)\cdot\left(-\dfrac{1}{2}xy\right)\left(x^2y^4\right)\)

\(=\left(-\dfrac{1}{7}\cdot\dfrac{14}{3}\cdot\dfrac{-1}{2}\right)\left(x^2y^3\cdot xy^2\cdot xy\cdot x^2y^4\right)\)

\(=\dfrac{1}{3}x^6y^{10}\)

2) Ta có: \(\left(3xy\right)^2\cdot\left(-\dfrac{1}{2}x^3y^2\right)\)

\(=9xy^2\cdot\dfrac{-1}{2}x^3y^2\)

\(=-\dfrac{9}{2}x^4y^4\)

3) Ta có: \(\left(-\dfrac{1}{4}x^2y\right)^2\cdot\left(\dfrac{2}{3}xy^4\right)^3\)

\(=\dfrac{1}{16}x^4y^2\cdot\dfrac{8}{27}x^3y^{12}\)

\(=\dfrac{1}{54}x^7y^{14}\)