Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\dfrac{x^4-2x^2+1}{x^3+2x^2+x}=\dfrac{\left(x^2-1\right)^2}{x\left(x^2+2x+1\right)}=\dfrac{\left(x-1\right)^2\left(x+1\right)^2}{x\left(x+1\right)^2}=\dfrac{\left(x-1\right)^2}{x}\)
\(\dfrac{7x^3+14x^2+7x}{14x^2+14x}=\dfrac{7x\left(x^2+2x+1\right)}{14x\left(x+1\right)}=\dfrac{\left(x+1\right)^2}{2\left(x+1\right)}=\dfrac{x+1}{2}\)
`@` `\text {Ans}`
`\downarrow`
\((x+y)(x-y)+(xy^4-x^3y^2) \div (xy^2) \)
`= x(x-y) + y(x-y) + xy^4 \div xy^2 - x^3y^2 \div xy^2`
`= x^2 - xy + xy - y^2 + y^2 - x^2`
`= (x^2 - x^2) + (-xy + xy) + (-y^2 + y^2)`
`= 0`
a) \(\dfrac{x^3-1}{x^2+x+1}=\dfrac{\left(x-1\right)\left(x^2+x+1\right)}{x^2+x+1}=x-1\)
b) \(\dfrac{x^2+2xy+y^2}{2x^2+xy-y^2}\)
\(=\dfrac{\left(x+y\right)^2}{x^2+xy+x^2-y^2}=\dfrac{\left(x+y\right)^2}{x\left(x+y\right)+\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{\left(x+y\right)^2}{\left(2x-y\right)\left(x+y\right)}=\dfrac{x+y}{\left(2x-y\right)}\)
c) \(\dfrac{ax^4-a^4x}{a^2+ax+x^2}\)
\(=\dfrac{ax\left(x^3-a^3\right)}{a^2+ax+x^2}\)
\(=\dfrac{ax\left(x-a\right)\left(a^2+ax+x^2\right)}{a^2+ax+x^2}\)
\(=ax\left(x-a\right)\)
a) \(x^2+2x^2+x=x\left(x+2x+1\right)=x\left(x+1\right)^2\)
b) \(xy+y^2-x-y=\left(xy-x\right)+y^2-y=x\left(y-1\right)+y\left(y-1\right)=\left(y-1\right)\left(x+y\right)\)mấy câu sau bạn làm tương tự nhé, đặt biến x với x và y với y là được. có gì ib face cho mình
có gì sai xót mong m.n bỏ qua và nhắc nhở ạ
\(a,=2x\left(x+3\right)\\ b,=x^3\left(x+3\right)+\left(x+3\right)=\left(x^3+1\right)\left(x+3\right)\\ =\left(x+1\right)\left(x+3\right)\left(x^2-x+1\right)\\ c,=64-\left(x-y\right)^2=\left(8-x+y\right)\left(8+x-y\right)\\ A=x^2+6x+5+x^3-8-x^2-x+2\\ A=x^3+5x-1\)
a) 2x2+6x=2x(x+3)
b) x4+3x3+x+3=(x4+x)+(3x3+3)=x(x3+1)+3(x3+1)=(x+3)(x3+1)
c) 64-x2-y2+2xy=-(x2-2xy+y2)+82=8-(x+y)2=(8+x+y)(8-x-y)
A= (x+5)(x+1)+(x-2)(x2+2xx+4)-(x2+x-2)
A= x2+6x+5+x3-8-x2-x+2
A= x3+(x2-x2)+(6x-x)+(5-8+2)
A= x3+5x-1
\(3\left(x-y\right)^2-2\left(x+y\right)^2-\left(x+y\right)\left(x-y\right)\)
\(=3.\left(x^2-2xy+y^2\right)-2\left(x^2+2xy+y^2\right)-x^2+y^2\)
\(=3x^2-6xy+3y^2-2x^2-4xy-2y^2-x^2+y^2\)
\(=2y^2-10xy\)
\(=\dfrac{\left(x^2-y^2\right)\left(x^2+y^2\right)}{\left(y-x\right)\left(y^2+xy+x^2\right)}=\dfrac{-\left(y-x\right)\left(x+y\right)\left(x^2+y^2\right)}{\left(y-x\right)\left(y^2+xy+x^2\right)}=\dfrac{-\left(x+y\right)\left(x^2+y^2\right)}{x^2+xy+y^2}\)