Tính
a. \(^{-x^4+2x^3-2x^2+2x-1}\)
b. \(-2x^2-y^2+2xy+4x-40\)
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\(Bài1:\\ a,\left(4x-1\right)\left(2x^2-x-1\right)=4x\left(2x^2-x-1\right)-\left(2x^2-x-1\right)=8x^3-4x^2-4x-2x^2+x+1=8x^3-6x^2-3x+1\\ b,\left(4x^3+8x^2-2x\right):2x\\ =2x\left(2x^2+4x-1\right):2x\\ =2x^2+4x-1\)
\(Bài2:\\ a,2x^3-8x^2+8x=2x\left(x^2-4x+4\right)=2x\left(x-2\right)^2\\ b,2xy+2x+yz+z=2x\left(y+1\right)+z\left(y+1\right)=\left(y+1\right)\left(2x+z\right)\\ c,x^2+2x+1-y^2=\left(x+1\right)^2-y^2=\left(x-y+1\right)\left(x+y+1\right)\)
Bài 3:
A(x)⋮B(x)
=>\(3x^2+5x+m\) ⋮x-2
=>\(3x^2-6x+11x-22+m+22\) ⋮x-2
=>m+22=0
=>m=-22
Bài 2:
a: \(2x^3-8x^2+8x\)
\(=2x\left(x^2-4x+4\right)\)
\(=2x\left(x-2\right)^2\)
b: 2xy+2x+yz+z
=2x(y+1)+z(y+1)
=(y+1)(2x+z)
c: \(x^2+2x+1-y^2\)
\(=\left(x+1\right)^2-y^2\)
=(x+1-y)(x+1+y)
Câu 1:
a:\(\left(4x-1\right)\left(2x^2-x-1\right)\)
\(=8x^3-4x^2-4x-2x^2+x+1\)
\(=8x^3-6x^2-3x+1\)
b: \(\left(4x^3+8x^2-2x\right):2x\)
\(=\frac{4x^3}{2x}+\frac{8x^2}{2x}-\frac{2x}{2x}\)
\(=2x^2+4x-1\)
c: \(\left(6x^3-7x^2-16x+12\right):\left(2x+3\right)\)
\(=\left(6x^3+9x^2-16x^2-24x+8x+12\right):\left(2x+3\right)\)
\(=\left\lbrack3x^2\left(2x+3\right)-8x\left(2x+3\right)+4\left(2x+3\right)\right\rbrack:\left(2x+3\right)\)
\(=3x^2-8x+4\)
a: \(A=\left(a+b\right)^3+\left(a-b\right)^3-2a^3\)
\(=a^3+3a^2b+3ab^2+b^3+a^3-3a^2b+3ab^2-b^3-2a^3\)
\(=6ab^2\)
b: \(B=\left(2x-1\right)\left(4x^2+2x+1\right)-8\left(x+2\right)\left(x^2-2x+4\right)\)
\(=8x^3-1-8\left(x^3+8\right)\)
\(=8x^3-1-8x^3-64=-65\)
c: \(C=\left(x+2y\right)\left(x^2-2xy+4y^2\right)+\left(2x-y\right)\left(4x^2+2xy+y^2\right)-9x^3\)
\(=x^3+\left(2y\right)^3+\left(2x\right)^3-y^3-9x^3\)
\(=x^3+8y^3+8x^3-y^3-9x^3=7y^3\)
a) (x-y)(2x+3y)=2x2+3xy-2xy+3y2=2x2+xy+3y2
b) (2x-1)2-(2x-1)=0
<=> 2x-1=0 <=> x=\(\dfrac{1}{2}\)
a) Ta có: (x-y)(2x+3y)
\(=2x^2+3xy-2xy-3y^2\)
\(=2x^2+xy-3y^2\)
\(2x^3y-2xy^3-4xy^2-2xy\)
\(=2xy.\left(x^2-y^2-2y-1\right)\)
\(=2xy.[x^2-\left(y^2+2y+1\right)]\)
\(=2xy.[x^2-\left(y+1\right)^2]\)
\(=2xy.\left(x+y+1\right).\left(x-y-1\right)\)
Vậy chọn đáp án A
a,=(x+2).(x^2-2x+2^2)-18-x^3
=x^3 + 2^3 - 18 -x^3=(x^3-x^3)+(8-18) = -10
b, =(2x-y).((2x)2+2xy +y2) - (2x +y).((2x)^2-2xy +y^2)
=(2x)3-y3- (2x)3-y3= -2 .y3
2 ý này áp dụng HĐT : x3+y3=(x+y).(x2-xy+y2)
x3-y3=(x-y).(x2+xy+y2)
a) ( x + 2 )( x2 - 2x + 4 ) - ( 18 + x3 )
= x3 + 8 - 18 - x3 = -10
b) ( 2x - y )( 4x2 + 2xy + y2 ) - ( 2x + y )( 4x2 - 2xy + y2 )
= 8x3 - y3 - ( 8x3 + y3 )
= 8x3 - y3 - 8x3 - y3 = -2y3
c) ( x - 3 )( x + 3 ) - ( x + 5 )( x - 1 )
= x2 - 9 - ( x2 + 4x - 5 )
= x2 - 9 - x2 - 4x + 5 = -4x - 4
d) ( 3x - 2 )2 + ( x + 1 )2 + 2( 3x - 2 )( x + 1 )
= ( 3x - 2 + x + 1 )2
= ( 4x - 1 )2
a.\(-x^4+2x^3-2x^2+2x-1=-\left(x^4+2x^2+1\right)+\left(2x^3+2x\right)\)
=\(-\left(x^2+1\right)^2+2x\left(x^2+1\right)=\left(x^2+1\right)\left(-1+2x\right)\)
b.\(-2x^2-y^2+2xy+4x-10=-\left(x^2-4x+4\right)-\left(x^2-2xy+y^2\right)-6\)
=\(-\left(x-2\right)^2-\left(x-y\right)^2-6\)
a.=\(-\left(x^4+2x^2+\text{1}\right)+\left(2x^3+2x\right)\)=\(-\left(x^2+1\right)^2+2x\left(x^2+1\right)=\left(x^2+1\right)\left(-x^2+2x-1\right)\)
=\(-\left(x^2+1\right)\left(x-1\right)^2\)
.
b.=\(-\left(x^2-4x+4\right)-\left(x^2-2xy+y^2\right)-36\)=\(-\left(x-2\right)^2-\left(x-y\right)^2-36\)