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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
Tham khảo nha \(\)
1. Rút gọn:
a/ \(\left(x-3\right)\left(x^2+3x+9\right)+\left(54+x^3\right)\)
= \(x^3+3x^2+9x-3x^2-9x-27+54+x^3\)
= \(2x^3+27\)
b/ \(\left(3x+y\right)\left(9x^2-3xy+y^2\right)-\left(3x-y\right)\left(9x^2+3xy+y^2\right)\)
\(=27x^3-9x^2y+3xy^2+9x^2y-3xy^2+y^3-27x^3+9x^2y+3xy^2-9x^2y-3xy^2-y^3\)
\(=\left(27x^3-y^3\right)-\left(27x^3+y^3\right)\)
\(=27x^3-y^3-27x^3-y^3=-2y^3\)
2.Chứng minh rằng:
a/ \(a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)\)
Xét VP có:
\(=a^3+3a^2b+3ab^2+b^3-3a^2b-3ab^2\)
\(=a^3+b^3\)
=> VT=VP
=> \(a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)\)
b/ \(a^3-b^3=\left(a-b\right)^3+3ab\left(a-b\right)\)
Xét VP có:
\(=a^3-3a^2b+3ab^2-b^3+3a^2b-3ab^2\)
\(=a^3-b^3\)
=> VT=VP
=> \(a^3-b^3=\left(a-b\right)^3+3ab\left(a-b\right)\)
Chúc bạn học tốt ♥khong bt ai hay sao ma con tra loi gium nua cho hung du sao van cam on
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)\(a+b+c=0\) =>\(a^3+b^3+c^3+3a^2b+3ab^2+3b^2c+3bc^2+3c^2a+3a^2c+6abc=0\) =>\(a^3+b^3+c^3+3\left(a+b\right)\left(a+c\right)\left(b+c\right)=0\) =>\(a^3+b^3+c^3+3\left(-a\right)\left(-b\right)\left(-c\right)=0\) =>\(a^3+b^3+c^3=3abc\left(đpcm\right)\)
mik cần gấp
Bài 1:
\(a,\dfrac{1}{2}x^2y^2\left(2x+y\right)\left(x^2-xy+1\right)=\left(x^3y^2+\dfrac{1}{2}x^2y^3\right)\left(x^2-xy+1\right)=x^5y^2-x^4y^3+x^3y^2+\dfrac{1}{2}x^3y^3-\dfrac{1}{2}x^3y^4+\dfrac{1}{2}x^2y^3\)
\(b,\left(\dfrac{1}{2}x-1\right)\left(2x-3\right)=x^2-\dfrac{3}{2}x-2x+3=x^2-\dfrac{7}{2}x+3\)\(c,\left(x-7\right)\left(x-5\right)=x^2-5x-7x+35=x^2-12x+35\)\(f,\left(x-\dfrac{1}{2}\right)\left(x+\dfrac{1}{2}\right)\left(4x-1\right)=\left(x^2-\dfrac{1}{4}\right)\left(4x-1\right)=4x^3-x^2-x+\dfrac{1}{4}\)Bài 2 ,
\(\left(x-1\right)\left(x^2+x+1\right)=x^3+x^2+x-x^2-x-1=x^3-1\Rightarrowđpcm\)\(b,\left(x^3+x^2y+xy^2+y^3\right)\left(x-y\right)=x^4+x^3y+x^2y^2+y^3x+x^3y-x^2y^2-xy^3-y^4=x^4-y^4\)
Lời giải:
a)
$x^3+y^3+2x^2-2xy+2y^2=(x^3+y^3)+2(x^2-xy+y^2)$
$=(x+y)(x^2-xy+y^2)+2(x^2-xy+y^2)=(x^2-xy+y^2)(x+y+2)$
b)
$a^4+ab^3-a^3b-b^4=(a^4-a^3b)+(ab^3-b^4)$
$=a^3(a-b)+b^3(a-b)=(a-b)(a^3+b^3)=(a-b)(a+b)(a^2-ab+b^2)$
c)
\(a^3-b^3+3a^2+3ab+3b^2=(a^3-b^3)+3(a^2+ab+b^2)\)
\(=(a-b)(a^2+ab+b^2)+3(a^2+ab+b^2)=(a^2+ab+b^2)(a-b+3)\)
d)
\(x^4+x^3y-xy^3-y^4=x^3(x+y)-y^3(x+y)=(x+y)(x^3-y^3)=(x+y)(x-y)(x^2+xy+y^2)\)
\(\left(x+y\right)^3=x^3+3x^2y+3xy^2+y^3=\left(x^3-6x^2y+9xy^2\right)+\left(y^3-6xy^2+9x^2y\right)\)
\(=x\left(x^2-6xy+9y^2\right)+y\left(y^2-6xy+9x^2\right)=x\left(x-3y\right)^2+y\left(y-3x\right)^2\)
b/
\(\left(a+b\right)^3+\left(a-b\right)^3=a^3+3a^2b+3ab^2+b^3+a^3-3a^2b+3ab^2-b^3\)
\(=2a^3+6ab^2=2a\left(a^2+3b^2\right)\)
c/
\(\left(a+b\right)^3-\left(a-b\right)^3=a^3+3a^2b+3ab^2+b^3-\left(a^3-3a^2b+3ab^2-b^3\right)\)
\(=6a^2b+2b^3=2b\left(b^2+3a^2\right)\)
d/
\(a^3+b^3=a^3+3a^2b+3ab^2+b^3-\left(3a^2b+3ab^2\right)\)
\(=\left(a+b\right)^3-3ab\left(a+b\right)\)
e/
\(a^3-b^3=a^3-3a^2b+3ab^2-b^3+3a^2b-3ab^2\)
\(=\left(a-b\right)^3+3ab\left(a-b\right)\)