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Rút gọn các phân thức:
a) \(\frac{\left(3x+2\right)^2-\left(x+2\right)^2}{x^3-x^2}=\frac{9x^2+12x+4-x^2-4x-4}{x^3-x^2}=\frac{8x^2+8x}{x^3-x^2}=\frac{8x\left(x+1\right)}{x^2\left(x-1\right)}=\frac{8\left(x+1\right)}{x-1}\)
b) \(\frac{x^4-1}{x^3+2x^2-x-2}=\frac{\left(x^2-1\right)\left(x^2+1\right)}{\left(x^3-x\right)+\left(2x^2-2\right)}=\frac{\left(x^2-1\right)\left(x^2+1\right)}{\left(x+2\right)\left(x^2-1\right)}=\frac{x^2+1}{x+2}\)
c) \(\frac{x^2+7x+12}{x^2+5x+6}=\frac{\left(x^2+3x\right)+\left(4x+12\right)}{\left(x^2+3x\right)+\left(2x+6\right)}=\frac{\left(x+3\right)\left(x+4\right)}{\left(x++3\right)\left(x+2\right)}=\frac{x+4}{x+2}\)
d) \(\frac{x^{10}-x^8+x^6-x^4+x^2-1}{x^4-1}=\frac{\left(x^{10}-x^8\right)+\left(x^6-x^4\right)+\left(x^2-1\right)}{\left(x^2-1\right)\left(x^2+1\right)}=\frac{\left(x^2-1\right)\left(x^8+x^4+1\right)}{\left(x^2-1\right)\left(x^2+1\right)}=\frac{x^8+x^4+1}{x^2+1}\)
Bài 1:
a, x2-3xy-10y2
=x2+2xy-5xy-10y2
=(x2+2xy)-(5xy+10y2)
=x(x+2y)-5y(x+2y)
=(x+2y)(x-5y)
b, 2x2-5x-7
=2x2+2x-7x-7
=(2x2+2x)-(7x+7)
=2x(x+1)-7(x+1)
=(x+1)(2x-7)
Bài 2:
a, x(x-2)-x+2=0
<=>x(x-2)-(x-2)=0
<=>(x-2)(x-1)=0
<=>\(\orbr{\begin{cases}x-2=0\\x-1=0\end{cases}}\)<=>\(\orbr{\begin{cases}x=2\\x=1\end{cases}}\)
b, x2(x2+1)-x2-1=0
<=>x2(x2+1)-(x2+1)=0
<=>(x2+1)(x2-1)=0
<=>x2+1=0 hoặc x2-1=0
1, x2+1=0 2, x2-1=0
<=>x2= -1(loại) <=>x2=1
<=>x=1 hoặc x= -1
c, 5x(x-3)2-5(x-1)3+15(x+2)(x-2)=5
<=>5x(x-3)2-5(x-1)3+15(x2-4)=5
<=>5x(x2-6x+9)-5(x3-3x2+3x-1)+15x2-60=5
<=>5x3-30x2+45x-5x3+15x2-15x+5+15x2-60=5
<=>30x-55=5
<=>30x=55+5
<=>30x=60
<=>x=2
d, (x+2)(3-4x)=x2+4x+4
<=>(x+2)(3-4x)=(x+2)2
<=>(x+2)(3-4x)-(x+2)2=0
<=>(x+2)(3-4x-x-2)=0
<=>(x+2)(1-5x)=0
<=>\(\orbr{\begin{cases}x+2=0\\1-5x=0\end{cases}}\)<=>\(\orbr{\begin{cases}x=-2\\-5x=-1\end{cases}}\)<=>\(\orbr{\begin{cases}x=-2\\x=\frac{-1}{-5}\end{cases}}\)<=>\(\orbr{\begin{cases}x=-2\\x=\frac{1}{5}\end{cases}}\)
Bài 3:
a, Sắp xếp lại: x3+4x2-5x-20
Thực hiện phép chia ta được kết quả là x2-5 dư 0
b, Sau khi thực hiện phép chia ta được :
Để đa thức x3-3x2+5x+a chia hết cho đa thức x-3 thì a+15=0
=>a= -15
a) \(x^3-4x^3+8x-8\)
\(=x^3-8+8x-4x^2\)
\(=\left(x-2\right)\left(x^2-2x+4\right)+4x\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2-2x+4+4x\right)=\left(x-2\right)\left(x^2+2x+4\right)\)
a)
\(x^5+x^3-x^2-1\)
\(=x^3\left(x^2+1\right)-\left(x^2+1\right)\)
\(=\left(x^2+1\right)\left(x^3-1\right)\)
\(=\left(x^2+1\right)\left(x-1\right)\left(x^2+x+1\right)\)
b)
\(x^2-3x^3-x+3\)
\(=x\left(x-1\right)-3\left(x^3-1\right)\)
\(=x\left(x-1\right)-3\left(x-1\right)\left(x^2+x+1\right)\)
\(=\left(x-1\right)\left(x-3x^2-3x-3\right)\)
\(=\left(x-1\right)\left(-3x^2-2x-3\right)\)
c)
\(x^2-6x+8\)
\(=x^2-2.x.3+9-1\)
\(=\left(x-3\right)^2-1\)
\(=\left(x-3-1\right)\left(x-3+1\right)\)
\(=\left(x-4\right)\left(x-2\right)\)
d)
\(4x^4-4x^2y^2-8y^4\)
\(=\left(2x^2\right)^2-2.\left(2x^2\right)y^2+y^2-9y^4\)
\(=\left(2x^2-y\right)^2-\left(3y^2\right)^2\)
\(=\left(2x^2-y-3y^2\right)\left(2x^2-y+3y^2\right)\)
a,nhóm x*5 với x*3,x*2 và 1: (x*5+ x*3) - (x*2+1) =x*3.(x*2+1)-(x*2+1) =....., câu b nhóm x*2 và -3x*3,x và 3, câu c bang (x-3)*2-1 =..., câu d đat 4 ra.
a) \(x^5+x^3-x^2-1=x^3\left(x^2+1\right)-\left(x^2+1\right)=\left(x^3-1\right)\left(x^2+1\right)\)
b) \(x^2-3x^3-x+3=x\left(x-1\right)-3\left(x^3-1\right)=x\left(x-1\right)-3\left(x-1\right)\left(x^2+x+1\right)=\left(x-1\right)\left[\left(x-3\right)-\left(x^2+x+1\right)\right]=\left(x-1\right)\left(-x^2-4\right)\)c) \(x^2-6x+8=x^2-6x+9-1=\left(x-3\right)^2-1=\left(x-2\right)\left(x-4\right)\)
d) \(4x^4+4x^2y^2-8y^4=4x^4+4x^2y^2+y^4-9y^4=\left(2x^2+y^2\right)^2-9y^4=\left(2x^2+4y^2\right)\left(2x^2-2y^2\right)=2\left(x^2+2y^2\right)2\left(x^2-y^2\right)=4\left(x^2+2y^2\right)\left(x+y\right)\left(x-y\right)\)
\(x^8+x+1\)
\(=x^8+x^7+x^6-x^7-x^6-x^5+x^5+x^4+x^3-x^4-x^3-x^2+x^2+x+1\)
\(=x^6\left(x^2+x+1\right)-x^5\left(x^2+x+1\right)+x^3\left(x^2+x+1\right)-x^2\left(x^2+x+1\right)+x^2+x+1\)
\(=\left(x^2+x+1\right)\left(x^6-x^5+x^3-x^2+1\right)\)
a: \(=\dfrac{1}{x^2+x+1}+\dfrac{x^3-1}{x^2+x+1}=\dfrac{x^3}{x^2+x+1}\)
b: \(=\dfrac{-x^4}{x-1}+\dfrac{\left(x-1\right)\left(x^3+x^2+x+1\right)}{x-1}\)
\(=\dfrac{-x^4+x^4-1}{x-1}=\dfrac{-1}{x-1}\)