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\(a.\left(a-b\right)^3=-\left(b-a\right)^3\)
\(\Leftrightarrow\left(a-b\right)^3=\left(a-b\right)^3\)
Học tốt!
a) \(-\left(b-a\right)^3=-\left(b-a\right).\left(b-a\right)^2\)
\(=\left(a-b\right)\left(a-b\right)^2=\left(a-b\right)^3\)
b) \(\left(-a-b\right)^2=\left(-a-b\right)\left(-a-b\right)=\left(a+b\right)\left(a+b\right)=\left(a+b\right)^2\)
a) VP = -( b3 - 3b2a + 3ba2 - a3 ) = a3 - 3a2b + 3ab2 - b3 = ( a - b )3 = VT ( đpcm )
b) VT = ( -a )2 - 2(-a)b + b2 = a2 + 2ab + b2 = ( a + b )2 = VP ( đpcm )
a) (a-b)3=a3-3a2b+3ab2-b3 (1). -(b-a)3=-(b3-3b2a+3ba2-a3)=-b3+3ab2-3a2b+a3=a3-3a2b+3ab2-b3 (2). từ (1) và (2) => VT=VP => đpcm. b, (-a-b)2 =. (-a-b)2=[(-a)+(-b)]2=(-a)2+2.(-a).(-b)+(-b)2=a2+2ab+b2=(a+b)2 => VT=VP => đpcm
a) Ta có: \(\left(a^2+b^2\right)^2-4a^2b^2=\left(a^2+b^2\right)^2-\left(2ab\right)^2\)
\(=\left(a^2+b^2-2ab\right)\left(a^2+b^2+2ab\right)=\left(a-b\right)^2.\left(a+b\right)^2\)( đpcm )
b) Ta có: \(\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3-3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)
\(=\left(a-b+b-c\right)^3-3\left(a-b\right)\left(b-c\right)\left(a-b+b-c\right)+\left(c-a\right)^3\)
\(-3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)
\(=\left(a-c\right)^3-3\left(a-b\right)\left(b-c\right)\left(a-c\right)+\left(c-a\right)^3-3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)
\(=\left(a-c\right)^3+\left(c-a\right)^3-3\left(a-b\right)\left(b-c\right)\left(a-c\right)-3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)
\(=\left(a-c\right)^3-\left(a-c\right)^3+3\left(a-b\right)\left(b-c\right)\left(c-a\right)-3\left(a-b\right)\left(b-c\right)\left(c-a\right)=0\)
\(\Rightarrow\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3=3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)( đpcm )
1) Ta có: \(\left(a^2+b^2\right)^2-4a^2b^2\)
\(=a^4+2a^2b^2+b^4-4a^2b^2\)
\(=a^4-2a^2b^2+b^4\)
\(=\left(a^2-b^2\right)^2\)
\(=\left[\left(a-b\right)\left(a+b\right)\right]^2\)
\(=\left(a-b\right)^2\left(a+b\right)^2\)
2) Ta có: \(\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3\)
\(=\left(a-b+b-c\right)\left[\left(a-b\right)^2-\left(a-b\right)\left(b-c\right)+\left(b-c\right)^2\right]+\left(c-a\right)^3\)
\(=\left(a-c\right)\left(a^2-2ab+b^2-ab+ac+b^2-bc+b^2-2bc+c^2\right)+\left(c-a\right)^3\)
\(=-\left(c-a\right)\left(a^2+3b^2+c^2-3ab+ac-3bc\right)+\left(c-a\right)\left(c^2-2ca+a^2\right)\)
\(=\left(c-a\right)\left(c^2-2ca+a^2-a^2-3b^2-c^2+3ab-ac+3bc\right)\)
\(=\left(c-a\right)\left(3ab+3bc-3b^2-3ac\right)\)
\(=3\left(c-a\right)\left(ab-b^2-ac+bc\right)\)
\(=3\left(c-a\right)\left[b\left(a-b\right)-c\left(a-b\right)\right]\)
\(=3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)
3/ \(x^5+y^5\ge x^4y+xy^4\)
\(\Leftrightarrow x^4\left(x-y\right)-y^4\left(x-y\right)\ge0\)
\(\Leftrightarrow\left(x-y\right)\left(x^4-y^4\right)\ge0\)
\(\Leftrightarrow\left(x-y\right)^2\left(x+y\right)\left(x^2+y^2\right)\ge0\) (đúng)
bài 1
theo bài ra ta có
a + b + c = 0 => c = -[a+b] [ 1 ]
Thay (1) vao a^3+b^3+c^3 ta có:
a^3+b^3+[-(a+b)]^3=3ab[-(a+b)]
<=>a^3+b^3-(a+b)=-3ab(a+b)
<=> a3+ b3- a3 -3a2b- 3ab2- b3= -3a2b- 3ab2
<=> 0= 0
vậy ta có đpcm.
(a+b)3=(a+b)(a+b)(a+b)
=a(a+b)(a+b)+b(a+b)(a+b)
=(a2+ab)(a+b)+(ab+b2)(a+b)
=(a3+a2b+a2b+ab2)+(a2b+ab2+ab2+b3)
=a3+a2b+a2b+ab2+a2b+ab2+ab2+b3
=a3+a2b+a2b+a2b+ab2+ab2+ab2+b3
=a3+3a2b+3ab2+b3
vậy (a+b)3 = a3 +3a2b +3ab2 + b3 =>dpcm
\(VT=\left(a-b\right)\left(a^2+ab+b^2\right)+ab\left(a-b\right)\)
\(=\left(a-b\right)\left(a^2+ab+b^2+ab\right)\)
\(=\left(a-b\right)\left(a^2+2ab+b^2\right)\)
\(=\left(a-b\right)\left(a+b\right)^2\)
\(=VP\left(đpcm\right)\)
Ta có: \(a^3-b^3+ab\left(a-b\right)=\left(a-b\right)\left(a^2+ab+b^2\right)+ab\left(a-b\right)\)
\(=\left(a-b\right)\left(a^2+ab+b^2+ab\right)=\left(a-b\right)\left(a^2+2ab+b^2\right)\)
\(=\left(a-b\right)\left(a+b\right)^2\)( đpcm )
Lời giải:
a)
$(a-b)^3=(a-b)^2.(a-b)=(b-a)^2.-(b-a)=-(b-a)^3$
b)
$(-a-b)^2=[-(a+b)]^2=(-1)^2(a+b)^2=(a+b)^2$
c)
$(x+y)^3=x^3+3x^2y+3xy^2+y^3$
$=x^3-6x^2y+9x^2y-6xy^2+9xy^2+y^3$
$=(x^3-6x^2y+9xy^2)+(y^3-6xy^2+9x^2y)$
$=x(x^2-6xy+9y^2)+y(y^2-6xy+9x^2)$
$=x(x-3y)^2+y(y-3x)^2$
d)
$(x+y)^3-(x-y)^3=x^3+3xy(x+y)+y^3-[x^3-3xy(x-y)-y^3]$
$=2y^3+3xy[(x+y)+(x-y)]=2y^3+6x^2y=2y(y^2+3x^2)$
Lời giải:
a)
$(a-b)^3=(a-b)^2.(a-b)=(b-a)^2.-(b-a)=-(b-a)^3$
b)
$(-a-b)^2=[-(a+b)]^2=(-1)^2(a+b)^2=(a+b)^2$
c)
$(x+y)^3=x^3+3x^2y+3xy^2+y^3$
$=x^3-6x^2y+9x^2y-6xy^2+9xy^2+y^3$
$=(x^3-6x^2y+9xy^2)+(y^3-6xy^2+9x^2y)$
$=x(x^2-6xy+9y^2)+y(y^2-6xy+9x^2)$
$=x(x-3y)^2+y(y-3x)^2$
d)
$(x+y)^3-(x-y)^3=x^3+3xy(x+y)+y^3-[x^3-3xy(x-y)-y^3]$
$=2y^3+3xy[(x+y)+(x-y)]=2y^3+6x^2y=2y(y^2+3x^2)$
a) ta có: \(\left(a-b\right)^3=a^3-3a^2b+3ab^2-b^3\)(1)
\(-\left(b-a\right)^3=-\left(b^3-3b^2a+3ba^2-a^3\right)\)
\(=a^3-3a^2b+3ab^2-b^3\)(2)
từ (1) và (2) \(\Rightarrow\left(a-b\right)^3=-\left(b-a\right)^3\)
b) ta có: \(\left(a+b\right)^2=a^2+2ab+b^2\)(3)
\(\left(-a-b^2\right)=\left(-a\right)^2-2\left(-a\right)\cdot b+\left(-b\right)^2\)
\(=a^2+2ab+b^2\)(4)
từ (3) và (4) \(\Rightarrow\left(-a-b\right)^2=\left(a+b\right)^2\)