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

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

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

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

\(=x^4+x^2y^2+x^3y+xy^3-x^3y-xy^3-x^2y^2-y^4\)

\(=x^4-y^4\)

c)\(\left(a+b+c\right)\left(ab+bc+ac\right)-abc\)

\(=a^2b+abc+a^2c+ab^2+b^2c+abc+abc+bc^2+ac^2-abc\)

\(=2abc+a^2b+a^2c+ab^2+b^2c+bc^2+ac^2\left(1\right)\)

\(\left(a+b\right)\left(a+c\right)\left(b+c\right)\)

\(=a^2+ac+ab+bc\left(b+c\right)\)

\(=a^2b+abc+ab^2+b^2c+a^2c+ac^2+abc+bc^2\)

\(=2abc+a^2b+ab^2+b^2c+a^2c+ac^2+bc^2\left(2\right)\)

Từ (1)(2) => đpcm

đẽ thu gọn vế vd a) ta có vt: ( x-y) .(x+y)=x^2 -y^2

                                                                 =vp

                                                               ->dpcm

b) (x-y) . (x^3 +xy^2 +x^2y+y^3)

  =(x-y ).(x^3 + y^3) 

= x.x^3 -y.y^3

=x^4 - y^4 =vp

->dpcm

c) (a +b+ c) (ab +bc +ac) -abc 

=nhân vô rút gọn 

=(a^2b +2abc +c^b) +(a^2c+c^2a) + (ab^2+b^2c )

=b(a+c)^2 +ac(a+c) +b^2 (a+c) 

=(a+c).[b(a+c)+b^2 +ac+b^2]

=(a+c)(ab+b^2+bc+ac)

=(a+c) [b(a+b)+c(a+b)]

=(a+b)(a+c)(b+c)=vp 

->dpcm

a/VT=x⁵+x^4.y+x^3.y^2+x^2.y^4+x.y^4-x^4.y-x^3.y^2-x^2.y^3-x.y^4-y^5

=x^5-y^5=VP

=>dpcm

Các câu khác tương tự

\(x-y=1\Rightarrow x^2-2xy+y^2=1\Rightarrow x^2+xy+y^2=19\Rightarrow x^3-y^3=\left(x-y\right)\left(x^2+xy+y^2\right)=1.19=19\)

\(2,a^2+b^2+c^2=ab+bc+ca\Leftrightarrow2\left(a^2+b^2+c^2\right)=2ab+2bc+2ca\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ac+a^2\right)=0\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0ma:\left\{{}\begin{matrix}\left(a-b\right)^2\ge0\\\left(b-c\right)^2\ge0\\\left(c-a\right)^2\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a-b=0\\b-c=0\\c-a=0\end{matrix}\right.\Leftrightarrow a=b=c\)

\(a+b+c=0\Leftrightarrow\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ca=0\Leftrightarrow a^2+b^2+c^2=-2\left(ab+bc+ca\right)\Rightarrow a^4+b^4+c^4+2a^2b^2+2b^2c^2+2c^2a^2=4a^2b^2+4b^2c^2+4c^2a^2+4abc\left(a+b+c\right)=4a^2b^2+4c^2a^2+4b^2c^2\Rightarrow a^4+b^4+c^4=2a^2b^2+2b^2c^2+2c^2a^2\Leftrightarrow2\left(a^4+b^4+c^4\right)=a^4+b^4+c^4+2a^2b^2+2b^2c^2+2c^2a^2=\left(a^2+b^2+c^2\right)^2\left(dpcm\right)\)

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

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

\(=x^2+2xy+y^2-2zx-2yz+z^2\)

\(=x^2+y^2+z^2+2xy-2yz-2zx\)

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

\(=x^4+x^3y+x^2y^2+xy^3-x^3y-x^2y^2-xy^3-y^4\)

\(=x^4-y^4\)

c) \(\left(x+y\right)\left(x^4-x^3y+x^2y^2-xy^3+y^4\right)\)

\(=x^5-x^4y+x^3y^2-x^2y^3+xy^4+x^4y-x^3y^2+x^2y^3-xy^4+y^5\)

\(=x^5+y^5\)

nhìn zậy thoy chứ dễ lắm mik làm vd 2 bài còn lại bn làm có gì bí thì hỏi mik

a) biến đổi vế trái ta có : \(\left(x+y\right)^2-y^2=\left(x+y-y\right)\left(x+y+y\right)=x\left(x+2y\right)\)( = vế phải )

b) BĐVT ta có : \(\left(x^2+y^2\right)^2-\left(2xy\right)^2=\left(x^2+y^2-2xy\right)\left(x^2+y^2+2xy\right)=\left(x-y\right)^2\left(x+y\right)^2\)= VP

Ta có:\(x+y=a\)

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

=>\(x^2+y^2=a^2-2xy=a^2-2b\left(đpcm\right)\)

Ta lại có:\(x^3+3x^2y+3xy^2+y^3=a^3\)

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

=>\(x^3+y^3=a^3-3xy\left(x+y\right)=a^3-3ab\left(đpcm\right)\)

b)\(a+b+c=0\) =>\(a^3+b^3+c^3+3a^2b+3ab^2+3b^2c+3bc^2+3c^2a+3a^2c+6abc=0\) =>\(a^3+b^3+c^3+3\left(a+b\right)\left(a+c\right)\left(b+c\right)=0\) =>\(a^3+b^3+c^3+3\left(-a\right)\left(-b\right)\left(-c\right)=0\) =>\(a^3+b^3+c^3=3abc\left(đpcm\right)\)