Tìm phần dư của phép chia
\(x^{19}+x^5-x^{1995}:\left(x^2-1\right)\)
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(u-1)^2 + (v+3/2)^2 + 11,75 \(\ge\)11,75
''='' <=> u = 1, v = -3/2
=> Min = 11,75 <=> u = 1, v = -3/2
a(b^2-a^2+a^2-c^2) + b(c^2-a^2) + c(a^2-b^2)
= a(b^2-a^2) + a(a^2-c^2) + b(c^2-a^2) + c(a^2-b^2)
= (a^2-b^2)(c-a) + (c^2-a^2)(b-a)
= (a-b)(a+b)(c-a) + (c-a)(c+a)(b-a)
= (a-b)(c-a)(a+b-c-a) = (a-b)(b-c)(c-a)
A = 8x^3 - 12x^2 + 18x - 12x^2 + 18x - 27 + 24x^2 - 16x
= 8x^3 + 20x - 27
Đặt \(x^2+y^2=a;y^2+z^2=b\)
\(\Rightarrow z^2-x^2=\left(y^2+z^2\right)-\left(x^2+y^2\right)=b-a\)
\(\Rightarrow A=a^3+\left(b-a\right)^3-b^3\)
\(=\left(a-b\right)\left(a^2+ab+b^2\right)-\left(a-b\right)^3\)
\(=\left(a-b\right)\left[a^2+ab+b^2-a^2+2ab-b^2\right]\)
\(=3ab\left(a-b\right)=3\left(x^2+y^2\right)\left(y^2+z^2\right)\left(x^2-z^2\right)\)
\(=3\left(x^2+y^2\right)\left(y^2+z^2\right)\left(x-z\right)\left(x+z\right)\)
\(B=\left(x+y+z\right)^3-x^3-y^3-z^3\)
\(B=x^3+y^3+z^3+3.\left(x+y\right)\left(y+z\right)\left(z+x\right)-x^3-y^3-z^3\)
\(B=3.\left(x+y\right)\left(y+z\right)\left(z+x\right)\)
Đây là hằng đẳng thức:
\(\left(a+b+c\right)^3=a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
a) 5xy ( x - y ) - 2x + 2y
= 5xy ( x - y ) - 2 ( x - y )
= ( x - y ) ( 5xy - 2 )
b) 6x-2y-x(y-3x)
= 2 ( y - 3x ) - x ( y - 3x )
= ( y - 3x ( ( 2 - x )
c) x2 + 4x - xy-4y
= x ( x + 4 ) - y ( x + 4 )
( x + 4 ) ( x - y )
d) 3xy + 2z - 6y - xz
= ( 3xy - 6y ) + ( 2z - xz )
= 3y ( x - 2 ) + z ( x - 2 )
= ( x - 2 ) ( 3y + z )
a,5xy(x-y)-2x+2y=5xy(x-y)-2(x-y)=(x-y)(5xy-2)
b,6x-2y-x(y-3x)=-2(y-3x)-x(y-3x)=(y-3x)(-2-x)
c,x^2+4x-xy-4y=x(x+4)-y(x+4)=(x+4)(x-y)
d,3xy+2z-6y-xz=(3xy-6y)+(2z-xz)=3y(x-2)+z(2-x)=3y(x-2)-z(x-2)=(x-2)(3y-z)
11)
a,4-9x^2=0
(2-3x)(2+3x)=0
2-3x=0=>x=2/3 hoặc 2+3x=0=>x=-2/3
b,x^2 +x+1/4=0
(x+1/2)^2 =0
x+1/2=0
x=-1/2
c,2x(x-3)+(x-3)=0
(x-3)(2x+1)=0
x-3=0=>x=3 hoặc 2x+1=0=>x=-1/2
d,3x(x-4)-x+4=0
3x(x-4)-(x-4)=0
(x-4)(3x-1)=0
x-4=0=>x=4 hoặc 3x-1=0=>x=1/3
e,x^3-1/9x=0
x(x^2-1/9)=0
x(x+1/3)(x-1/3)=0
x=0 hoặc x+1/3=0=>x=-1/3 hoặc x-1/3=0=>x=1/3
f,(3x-y)^2-(x-y)^2 =0
(3x-y-x+y)(3x-y+x-y)=0
2x(4x-2y)=0
4x(2x-y)=0
x=0hoặc 2x-y=0=>x=y/2
\(a,\left(a+b\right)^3-\left(a-b\right)^3-6a^2b\)
\(=a^3+3a^2b+3ab^2+b^3-a^3+3a^2b-3ab^2+b^3-6a^2b\)
\(=2b^3\)
\(b,\left(a+b\right)^3+\left(a-b\right)^3-6ab^2\)
\(=a^3+3a^2b+3ab^2+b^3+a^3-3a^2b+3ab^2-b^3-6ab^2\)
\(=2a^3\)