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Bài 1:
a) \(x^2-y^2+10x+25\)
\(=\left(x^2+10x+25\right)-y^2\)
\(=\left(x+5\right)^2-y^2\)
\(=\left(x+y+5\right)\left(x-y+5\right)\)
b) \(x^3-x^2-5x+125\)
\(=x^3+5x^2-6x^2-30x+25x+125\)
\(=x^2\left(x+5\right)-6x\left(x+5\right)+25\left(x+5\right)\)
\(=\left(x+5\right)\left(x^2-6x+25\right)\)
c) \(x^4+4y^4\)
\(=\left(x^2\right)^2+2x^22y^2+\left(2y^2\right)^2-2x^22y^2\)
\(=\left(x^2+2y^2\right)^2-\left(2xy\right)^2\)
\(=\left(x^2+2y^2-2xy\right)\left(x^2+2y^2+2xy\right)\)
d)Sửa đề \(a\left(b^2-c^2\right)+b\left(c^2-a^2\right)+c\left(a^2-b^2\right)\)
\(=a\left(b^2-c^2\right)-b\left[\left(b^2-c^2\right)+\left(a^2-b^2\right)\right]+c\left(a^2-b^2\right)\)
\(=a\left(b^2-c^2\right)-b\left(b^2-c^2\right)-b\left(a^2-b^2\right)+c\left(a^2-b^2\right)\)
\(=\left(a-b\right)\left(b^2-c^2\right)-\left(b-c\right)\left(a^2-b^2\right)\)
\(=\left(a-b\right)\left(b-c\right)\left(b+c\right)-\left(b-c\right)\left(a-b\right)\left(a+b\right)\)
\(=\left(a-b\right)\left(b-c\right)\left(b+c-a-b\right)\)
\(=\left(a-b\right)\left(b-c\right)\left(c-a\right)\)
e) \(7x^2-10xy+3y^2\)
\(=\left(\sqrt{7}x\right)^2-2.\sqrt{7}x.\sqrt{3}y+\left(\sqrt{3}y\right)^2\)
\(=\left(\sqrt{7}x-\sqrt{3}y\right)^2\)
f) Sửa đề \(a^3+b^3+c^3-3abc\)
\(=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+2ab-ac-bc+c^2\right)-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ac-bc+2ab-3ab\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ac-bc-ab\right)\)
h) \(xy\left(x+y\right)-yz\left(y+z\right)+xz\left(x-z\right)\)
\(=x^2y+xy^2-y^2z-yz^2+x^2z-xz^2\)
\(=\left(x^2y+x^2z\right)+\left(xy^2-xz^2\right)-yz\left(y+z\right)\)
\(=x^2\left(y+z\right)+x\left(y^2-z^2\right)-yz\left(y+z\right)\)
\(=x^2\left(y+z\right)+x\left(y+z\right)\left(y-z\right)-yz\left(y+z\right)\)
\(=\left(y+z\right)\left[x^2+x\left(y-z\right)-yz\right]\)
\(=\left(y+z\right)\left(x^2+xy-xz-yz\right)\)
\(=\left(y+z\right)\left[x\left(x+y\right)-z\left(x+y\right)\right]\)
\(=\left(y+z\right)\left(x+y\right)\left(x-z\right)\)
a) a3+a2c-abc+b2c+b3 =(a3+b3)+(a2c-abc+b2c)=(a+b)(a2-ab+b2)+c(a2-ab+b2)=(a2-ab+b2)(a+b-c)
b) x3-7x-6 = x3+x2-x2-x-6x-6=x2(x+1)-x(x+1)-6(x+1)=(x+1)(x2-x-6)=(x+1)(x-3)(x+2)
c) x3-x2-14x+24=x3-2x2+x2-2x-12x+24=x2(x-2)+x(x-2)-12(x-2)=(x-2)(x2+x-12)=(x-2)(x+4)(x-3)
\(x^2+y^2+z^2+3\ge2\left(x+y+z\right)\)
\(\Leftrightarrow\)\(x^2+y^2+z^2+3-2x-2y-2z\ge0\)
\(\Leftrightarrow\)\(\left(x^2-2x+1\right)+\left(y^2-2y+1\right)+\left(z^2-2z+1\right)\ge0\)
\(\Leftrightarrow\)\(\left(x-1\right)^2+\left(y-1\right)^2+\left(z-1\right)^2\ge0\)
Dáu "=" xảy ra \(\Leftrightarrow\) \(x=y=z=1\)
a,b,c,d > 0 ta có:
- a < b nên a.c < b.c
- c < d nên c.b < d.b
Áp dụng tính chất bắc cầu ta được: a.c < b.c < b.d hay a.c < b.d (đpcm)
a) a3+b3+a2c+b2c-abc
= (a+b)(a2-ab+b2)+c(a2+b2)-abc
=(a+b) [ (a+b)2-3ab]+c.[(a+b)2-2ab]-abc
=(a+b)(a+b)2-3ab(a+b)+c(a+b)2-3abc
=(a+b)2(a+b+c)-3ab(a+b+c)
=(a+b)2.0-3ab.0
=0
b) ax+ay+2x+2y+4
=a(x+y)+2(x+y)+4
=(x+y)(a+2)+4
=(a-2)(a+2)+4
=a2-4+4
=a2
c) A=1+x+x2+...+x49=>Ax=x+x2+x3+...+x50
- A=1+x+x2+...+x49
---> Ax-A=x50-1
d)(a+b)(a+c)+(c+a)(c+b)
=a2+ac+ab+bc+c2+bc+ac+ab
=a2+c2+2ac+2ab+2bc
=2b2+2bc+2ac+2ab
=2b(b+c)+2a(b+c)
=2b(b+c)(b+a)
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)
$(x+y)^3-(x-y)^3=x^3+3xy(x+y)+y^3-[x^3-3xy(x-y)-y^3]$
$=2y^3+3xy[(x+y)+(x-y)]=2y^3+6x^2y=2y(y^2+3x^2)$
\(\left(x+y\right)^3=x^3+3x^2y+3xy^2+y^3=\left(x^3-6x^2y+9xy^2\right)+\left(y^3-6xy^2+9x^2y\right)\)
\(=x\left(x^2-6xy+9y^2\right)+y\left(y^2-6xy+9x^2\right)=x\left(x-3y\right)^2+y\left(y-3x\right)^2\)
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/
\(a^3-b^3=a^3-3a^2b+3ab^2-b^3+3a^2b-3ab^2\)
\(=\left(a-b\right)^3+3ab\left(a-b\right)\)
ta có x+y=9, x.y=14=> x=9-y=14/y
nhân chéo ta có : 9y-y^2=14 <=> y^2+14-9y=0
<=> y^2-2y-7y+14=y(y-2)-7(y-2)=0
<=>(y-2)(y-7)=0
=> y=2 hoặc y=7
từ đó ta tính đc với y=2 thì x=7, với y=7 thì x=2
sau đó ta tính A, B, C dựa theo 2 trường hợp trên
Bài 3:
a: \(n\left(2n-3\right)-2n\left(n+1\right)\)
\(=2n^2-3n-2n^2-2n\)
=-5n chia hết cho 5
b: \(\left(n-1\right)\left(n+4\right)-\left(n-4\right)\left(n+1\right)\)
\(=n^2+4n-n-4-\left(n^2+n-4n-4\right)\)
\(=n^2+3n-4-\left(n^2-3n-4\right)\)
\(=6n⋮6\)