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Bài 1
a) (6x4y2 - 3x3y3) : 3x3y2 = 6x4y2 : 3x3y2 - 3x3y3 : 3x3y2 = 2x - y
b) (2x - 1)(x2 - x + 3) = 2x3 - 2x2 + 6x - x2 + x - 3 = 2x3 - 3x2 + 7x - 3
Bài 2
1) (x - 2)2 - (x - 3)2 = (x - 2 - x + 3)(x - 2 + x - 3) = 2x - 5>
2) 4x2 - 4xy + 2y2 + 1 = (4x2 - 4xy + y2) + y2 + 1 = (2x - y)2 + y2 + 1 > 0
vì \(\hept{\begin{cases}\left(2x-y\right)^2\ge0\\y^2\ge0\end{cases}}\)
Bài 2:
a: \(A=\left(2x-y\right)^2=\left(12-2\right)^2=100\)
b: \(=\left(x-3\right)^3=100^3=1000000\)
c: \(=\left(x-y\right)^2-9z^2\)
\(=\left(x-y-3z\right)\left(x-y+3z\right)\)
\(=\left(6+4-90\right)\left(6+4+90\right)=-80\cdot100=-8000\)
giải
a.(2x-3)(4x^2+6x+9)-2x(4x^2-1)
=8x^3+12x^2+18x-12x^2-18x-27-8x^3+2x
=2x-27
bài 1
b.(x+y)2+2(x+y)(x-y)+(x-y)2
= [(x+y)+(x-y)]2
= (x+y-x+y)2
= (2y)2
= 4y2
bài 2
a. 2x2y+4xy+2y
=2y(x2+2x+1)
=2y(x+1)2
b.9x2+6xy-4z2+y2
= (9x2+6xy+y2)-4z2
= (3x+y)2-(2z)2
= (3x+y-2z)(3x+y+2z)
Bài 3:
a) \(\left(4x^2+4xy+y^2\right):\left(2x+y\right)=\left(2x+y\right)^2:\left(2x+y\right)=2x+y\)
b) \(\left(27x^3+1\right):\left(3x+1\right)=\left(3x+1\right)\left(9x^2-9x+1\right):\left(3x+1\right)=9x^2-9x+1\)
c) \(\left(x^2-6xy+9y^2\right):\left(3y-x\right)=\left(x-3y\right)^2:\left(3y-x\right)=\left(3y-x\right)^2:\left(3y-x\right)=3y-x\)
d) \(\left(8x^3-1\right):\left(4x^2+2x+1\right)=\left(2x-1\right)\left(4x^2+2x+1\right):\left(4x^2+2x+1\right)=2x-1\)
Bài 4: Tương tự bài 3 '-'
Bài 4 :
a ) ( 4x4 - 9 ) : ( 2x2 - 3 )
= ( 2x2 + 3 )( 2x2 - 3 ) : ( 2x2 - 3 )
= 2x2 + 3
b ) ( 8x3 - 27 ) : ( 4x2 + 6x + 9 )
= ( 2x - 3 )( 4x2 + 6x + 9 ) : ( 4x2 + 6x + 9 )
= 2x - 3
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a,Để phép chia thực hiện đc<=>x^n<=x^5=>n<=5=>n=(0;1;2;3;4;5)
y^n<=y=>n<=1=>n=(1;0)
Từ hai ý trên=>n=(0;1)
b,,Để phép chia thực hiện đc<=>x^n+1<=x^2=>n+1<=2=>n=(0;1)
y^n=1<=y^2=>n+1<=2=>n=(0;1)
Từ hai ý trên =>n=(0;1)
Bài 1:
\(B=\dfrac{1}{9}x^2-2x+9\)
\(=\left(\dfrac{1}{3}x\right)^2-2\cdot\dfrac{1}{3}x\cdot3+3^2=\left(\dfrac{1}{2}x-3\right)^2\)
\(C=x^3-9x^2+27x-27=\left(x-3\right)^3\)
\(D=27x^3+27x^2+9x+1=\left(3x+1\right)^3\)
\(E=\left(x-2y\right)^3\)
a ) \(\left(x+1\right)\left(x-2\right)=x^2-2x+x-2=x^2-x-2\)
b ) \(\left(4x^4y^4-12x^2y^2\right):4x^2y^2=x^2y^2-3\)
c ) \(\frac{3x^2-1}{2x}+\frac{x^2+1}{2x}=\frac{3x^2-1+x^2+1}{2x}=\frac{4x^2}{2x}=2x\)
d ) \(\frac{x^2}{x-1}+\frac{2x}{1-x}+\frac{1}{x-1}=\left(\frac{x^2}{x-1}+\frac{1}{x-1}\right)+\frac{2x}{1-x}\)
\(=\frac{x^2+1}{x-1}+\frac{2x}{1-x}=\frac{x^2+1}{x-1}+\frac{-2x}{x-1}=\frac{x^2+1-2x}{x-1}=\frac{\left(x-1\right)^2}{x-1}=x-1\)
a) .......=x2-x-2
b) .........=x2y2-3
c) .......=(3x2-1+x2+1)/2x=4x2/2x=2x
d) x2 /(x-1)+(-2x)/(x-1)+1/(x-1)=(x2-2x+1)/(x-1)=(x-1)2/(x-1)=x-1
e)...
x-y=4
=> x2-2xy+y2=16
<=> 106-2xy =16 (vì x2+y2 =106)
=>xy=(106-16)/2=45
ta có x3 -y3 =(x-y)(x2+xy+y2 )
=4(106+45)=604
\(A=\left(\dfrac{3}{4}x^4y^2-\dfrac{9}{2}x^3y^2+9x^2y^2\right):\dfrac{3}{4}xy^2\)
\(=x^3-6x^2+12x\)
\(=1^3-6\cdot1^2+12\cdot1=1+12-6=13-6=7\)