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

1. Ta có: \(\dfrac{x^7+x^6+x^5+x^4+x^3+x^2+x+1}{x^2-1}\)

\(=\dfrac{x^6\left(x+1\right)+x^4\left(x+1\right)+x^2\left(x+1\right)+\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}\)

\(=\dfrac{\left(x+1\right)\left(x^6+x^4+x^2+1\right)}{\left(x+1\right)\left(x-1\right)}\)

\(=\dfrac{\left(x^6+x^4+x^2+1\right)}{\left(x-1\right)}\)

\(=\dfrac{x^4\left(x^2+1\right)+x^2+1}{x-1}\)

\(=\dfrac{\left(x^2+1\right)\left(x^4+1\right)}{x-1}\)

2.Ta có: \(\dfrac{x^2+y^2+z^2-2xy+2xz-2xz}{x^2-2xy+y^2-z^2}\)

\(=\dfrac{\left(x-y+z\right)^2}{\left(x-y\right)^2-z^2}=\dfrac{\left(x-y+z\right)\left(x-y+z\right)}{\left(x-y-z\right)\left(x-y+z\right)}=\dfrac{x-y+z}{x-y-z}\)

_Chúc bạn học tốt_

\(\text{1) }\dfrac{x^7+x^6+x^5+x^4+x^3+x^2+x+1}{x^2-1}\\ =\dfrac{\left(x^7+x^6\right)+\left(x^5+x^4\right)+\left(x^3+x^2\right)+\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{x^6\left(x+1\right)+x^4\left(x+1\right)+x^2\left(x+1\right)+\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{\left(x+1\right)\left(x^6+x^4+x^2+1\right)}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{x^6+x^4+x^2+1}{\left(x-1\right)}\\ \)

\(\text{2) }\dfrac{x^2+y^2+z^2-2xy+2xz-2yz}{x^2-2xy+y^2-z^2}\\ =\dfrac{\left(x^2-2xy+y^2\right)+\left(2xz-2yz\right)+z^2}{\left(x^2-2xy+y^2\right)-z^2}\\ =\dfrac{\left(x-y\right)^2+2z\left(x-y\right)+z^2}{\left(x-y\right)^2-z^2}\\ =\dfrac{\left(x-y+z\right)^2}{\left(x-y+z\right)\left(x-y-z\right)}\\ =\dfrac{x-y+z}{x-y-z}\)

\(\text{1) }\dfrac{x^7+x^6+x^5+x^4+x^3+x^2+x+1}{x^2-1}\\ =\dfrac{\left(x^7+x^6\right)+\left(x^5+x^4\right)+\left(x^3+x^2\right)+\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{x^6\left(x+1\right)+x^4\left(x+1\right)+x^2\left(x+1\right)+\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{\left(x^6+x^4+x^2+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{x^6+x^4+x^2+1}{x-1}\)

\(\text{3) }\dfrac{x^2+y^2+z^2-2xy+2xz-2yz}{x^2-2xy+y^2-z^2}\\ =\dfrac{\left(x^2-2xy+y^2\right)+\left(2xz-2yz\right)+z^2}{\left(x^2-2xy+y^2\right)-z^2}\\ =\dfrac{\left(x-y\right)^2+2\left(x-y\right)z+z^2}{\left(x-y\right)^2-z^2}\\ =\dfrac{\left(x-y+z\right)^2}{\left(x-y+z\right)\left(x-y-z\right)}\\ =\dfrac{x-y+z}{x-y-z}\)

Câu 1 :

Ta có :

\(\dfrac{x^7+x^6+x^5+x^4+x^3+x^2+x+1}{x^2-1}\)

\(=\dfrac{x^6\left(x+1\right)+x^4\left(x+1\right)+x^2\left(x+1\right)+\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{\left(x+1\right)\left(x^6+x^4+x^2+1\right)}{\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{\left(x^6+x^4+x^2+1\right)}{\left(x-1\right)}\)

\(=\dfrac{x^4\left(x^2+1\right)+\left(x^2+1\right)}{\left(x-1\right)}\)

\(=\dfrac{\left(x^2+1\right)\left(x^4+1\right)}{x-1}\)

Câu 2 : Ta có VT :

\(\dfrac{2x^2+3xy+y^2}{2x^3+x^2y-2xy^2-y^3}\)

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

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

\(=\dfrac{\left(x+y\right)\left(2x+y\right)}{\left(x+y\right)\left(x-y\right)\left(2x+y\right)}\)

\(=\dfrac{1}{x-y}=VP\left(đpcm\right)\)

Câu 3 :

\(\)Ta có :

\(\dfrac{x^2+y^2+z^2-2xy+2xz-2yz}{x^2-2xy+y^2-z^2}\)

\(=\dfrac{\left(x-y+z\right)^2}{\left(x-y\right)^2-z^2}=\dfrac{\left(x-y+z\right)\left(x-y+z\right)}{\left(x-y-z\right)\left(x-y+z\right)}\)

\(=\dfrac{\left(x-y+z\right)}{\left(x-y-z\right)}\)

\(\)

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

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

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

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

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

BBieesn đổi hằng đẳng thức 

x²+4x+4

=x²+2.2x+2²

=(x+2)²

Ta có:

\(x^2+4x+4\)

\(=x^2+2.2x+2^2\)

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

Ta có : x² + 4x + 4 = (x + 2)²

\(P=\left(x-y\right)\left(x^2+y^2\right)\left(x^4+y^4\right)-x^8+y^8+1\)

\(\Leftrightarrow P=\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)\left(x^4+y^4\right)-x^8+y^8+1\) (Vì: \(x-y=1\))

\(\Leftrightarrow P=\left(x^2-y^2\right)\left(x^2+y^2\right)\left(x^4+y^4\right)-x^8+y^8+1\)

\(\Leftrightarrow P=\left(x^4-y^4\right)\left(x^4+y^4\right)-x^8+y^8+1\)

\(\Leftrightarrow P=x^8-y^8-x^8+y^8+1\)

\(\Leftrightarrow P=1\)

bài bạn làm hơi sai

\(x^2+10x+26+y^2+2y\)

\(=\left(x^2+10x+25\right)+\left(y^2+2y+1\right)\)

\(=\left(x+5\right)^2+\left(y+1\right)^2\)

\(\left(x+y+4\right)\left(x+y-4\right)\)

\(=\left(x+y\right)^2-16\)

\(=x^2+y^2+2xy-16\)

a, =(x^2 +10x+25) +(y^2 +2y+1)

    = (x+5)^2 +(y+1)^2

b, =(x+y)^2 -4^2

    = x^2 + 2xy+ y^2 -16

x²+10x+26+y²+2y=(x+5)²+(y+1)²

(x+y+4)(x+y-4)=(x+y)²-16

Lời giải:

a) (x2 + 2xy + y2) : (x + y)

= (x + y)2 : (x + y)

= x + y

b) (125x3 + 1) : (5x + 1)

= [(5x)3 + 1] : (5x + 1)

= (5x + 1)[(5x)2 – 5x + 1]] : (5x + 1)

= (5x)2 – 5x + 1

= 25x2 – 5x + 1

c) (x2 – 2xy + y2) : (y – x)

= (x – y)2 : [-(x – y)]

= -(x – y)

= y – x

Hoặc (x2 – 2xy + y2) : (y – x)

= (y2 – 2yx + x2) : (y – x)

= (y – x)2 : (y – x)

= y – x

\(\text{a) (x^2 + 2xy + y^2) : (x + y)}\\ \left(x+y\right)^2:\left(x+y\right)=x+y\)

a)x+y

b)25x²-5x+1

c)-x+y