a/\(\left(x-1\right)\left(x^2+x+1\right)=x^3+x^2+x-x^2-x-1=x^3-1\left(đpcm\right)\)

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

c/ \(\left(x+y+z\right)^2=\left(x+y+z\right)\left(x+y+z\right)=x^2+xy+xz+y^2+xy+yz+z^2+zx+yz=x^2+y^2+z^2+2xy+2yz+2zx\left(đpcm\right)\)

d/ \(\left(x+y+z\right)^3=\left(x+y\right)^3+3\left(x+y\right)^2z+3z^2\left(x+y\right)+z^3\)

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

\(=x^3+3x^2y+3xy^2+y^3+3x^2z+6xyz+3y^2z+3z^2x+3yz^2+z^3\)

\(=x^3+y^3+z^3+3xyz+3x^2y+3xy^2+3x^2z+3y^2z+3y^2x+3yz^2+3xyz\)

\(=x^3+y^3+z^3+\left(x+z\right)\left(3xy+3xz+3y^2+3yz\right)\)

\(=x^3+y^3+z^3+\left(x+z\right)\left[3x\left(y+z\right)+3y\left(y+z\right)\right]\)

\(=x^3+y^3+z^3+\left(x+z\right)\left(y+z\right)\left(3x+3y\right)\)

\(=x^3+y^3+z^3+3\left(x+y\right)\left(y+z\right)\left(z+x\right)\) (đpcm)

a, Xét vế trái ta có:

(x-1)(x^2+ x+1)=x^3+ x^2+ x- x^2- x-1

=x^3+ (x^2- x^2)+(x-x)-1

=x^3-1

Vậy...

b,Xét vế trái ta có:(x^3+ x^2y+ xy^2+ y^3)(x-y)

=x^4- x^3y+ x^3y- x^2- y^2+ x^2y^2- xy^3+ xy^3- y^4

=x^4-y^4

Vậy ........

c, Xét vế trái ta có:

(x+y+z)^2=(x+y+z)(x+y+z)

=x^2+ xy+ xz+ yx+y^2+ yz+ zx+ zy+ z^2

=x^2+ y^2+ z^2+ 2xy+ 2xz+ 2yz

Vậy...............

d, Xé vế trái ta có:

(x+y+x)^3=(x+y+z)(x+y+z)(x+y+z)(x+y+z)

=(x^2+y^2+z^2+2xy+2xz+2yz)(x+y+z)

=x^3+ xy^2+ xz^2+ 2x^2y+ 2xyz+ 2x^2z+ x^2y+ y^3+ yz^2+2xy^2+ 2y^2z+z^3+ 2xyz+ x^2z+ y^2z+2xyz+ 2yz^2+ 2xz^2

=x^3+ 3xy^2+ 6xy+ 3x^2y+3xz^2+ 3x^2z+ 3yz^2+ y^3z^3 (1)

Xét vế phải ta có:x^3+ y^3+ z^3+ 3(x+y)(x+y)(y+z)

=x^3+ y^3+ z^3+ 3(xy+ xz+ y^2+ yz)(z+x)

=x^3+ y^3+ z^3+ 3(xyz+ xz^2+ y^2z+ yz^2+ x^2y+ x^2z+ xy^2+xyz)

=x^2+ y^3+ z^3 +3(2xyz+ xz^2+ y^2z+ yz^2+x^2y+x^2z+ xy^2)

=x^3+ y^3+ z^3+6xyz+ 3xz^2+ 3y^2z+3yz^2+ 3x^2y+3x^2z+3xy^2(2)

Từ (1) và (2)=>.......

Hướng dẫn :\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\Rightarrow\frac{xy+yz+zx}{xyz}=0\Rightarrow xy+yz+zx=0\)

Thay vào:\(x^2+2yz=x^2+yz+yz=x^2+yz-xy-zx=x\left(x-y\right)-z\left(x-y\right)=\left(x-y\right)\left(x-z\right)\)

Tương tự thay vào mà quy đồng

Hầy mình không nghĩ lớp 7 đã phải làm những bài biến đổi như thế này. Cái này phù hợp với lớp 8-9 hơn.

1.

Đặt $x^2-y^2=a; y^2-z^2=b; z^2-x^2=c$. 

Khi đó: $a+b+c=0\Rightarrow a+b=-c$

$\text{VT}=a^3+b^3+c^3=(a+b)^3-3ab(a+b)+c^3$

$=(-c)^3-3ab(-c)+c^3=3abc$

$=3(x^2-y^2)(y^2-z^2)(z^2-x^2)$

$=3(x-y)(x+y)(y-z)(y+z)(z-x)(z+x)$

$=3(x-y)(y-z)(z-x)(x+y)(y+z)(x+z)$

$=3.4(x-y)(y-z)(z-x)=12(x-y)(y-z)(z-x)$

Ta có đpcm.

Bài 2:

Áp dụng kết quả của bài 1:

Mẫu:

$(x^2-y^2)^3+(y^2-z^2)^3+(z^2-x^2)^3=3(x-y)(y-z)(z-x)(x+y)(y+z)(z+x)=3(x-y)(y-z)(z-x)(1)$

Tử: 

Đặt $x-y=a; y-z=b; z-x=c$ thì $a+b+c=0$

$(x-y)^3+(y-z)^3+(z-x)^3=a^3+b^3+c^3$

$=(a+b)^3-3ab(a+b)+c^3=(-c)^3-3ab(-c)+c^3=3abc$

$=3(x-y)(y-z)(z-x)(2)$

Từ $(1);(2)$ suy ra \(\frac{(x-y)^3+(y-z)^3+(z-x)^3}{(x^2-y^2)^3+(y^2-z^2)^3+(z^2-x^2)^3}=1\)

Bài 3:

\(ab+bc+ac=\frac{(a+b+c)^2-(a^2+b^2+c^2)}{2}=\frac{2^2-2}{2}=1\)

Do đó:

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{ab+bc+ac}{abc}=\frac{1}{abc}\)

Ta có đpcm.

1: \(=\dfrac{\left(x^2+2xy+y^2\right)-1}{\left(x^2+2x+1\right)-y^2}\)

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

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

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

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

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

\(=\dfrac{x+y+z}{2}\)