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Lời giải:
a. $a^4+a^3+a^2+a=(a^4+a^3)+(a^2+a)$
$=a^3(a+1)+a(a+1)=(a+1)(a^3+a)=a(a+1)(a^2+1)$
b. $3xy^2+5y-3x^2y+(-5x)=(3xy^2-3x^2y)+(5y-5x)$
$=3xy(y-x)+5(y-x)=(y-x)(3xy+5)$
c. $xy-z+y-xz=(xy+y)-(z+xz)=y(x+1)-z(x+1)=(x+1)(y-z)$
d.
$x^2-bx+ax-ab=(a^2+ax)-(bx+ab)=a(a+x)-b(a+x)=(a+x)(a-b)$
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\(=\dfrac{2\left(x+y\right)}{\left(a+b\right)^2}.\dfrac{a\left(x-y\right)+b\left(x-y\right)}{2\left(x^2-y^2\right)}\)
\(=\dfrac{2\left(x+y\right)}{\left(a+b\right)^2}.\dfrac{\left(x-y\right)\left(a+b\right)}{2\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{1}{a+b}\)
\(b,\dfrac{a+b-c}{a^2+2ab+b^2-c^2}.\dfrac{a^2+2ab+b^2+ac+bc}{a^2-b^2}\)
\(=\dfrac{a+b-c}{\left(a+b\right)^2-c^2}.\dfrac{\left(a+b\right)^2+c\left(a+b\right)}{\left(a-b\right)\left(a+b\right)}\)
\(=\dfrac{a+b-c}{\left(a+b-c\right)\left(a+b+c\right)}.\dfrac{\left(a+b\right)\left(a+b+c\right)}{\left(a-b\right)\left(a+b\right)}\)
\(=\dfrac{1}{a-b}\)
\(c,\dfrac{x^3+1}{x^2+2x+1}.\dfrac{x^2-1}{2x^2-2x+2}\)
\(=\dfrac{\left(x+1\right)\left(x^2-x+1\right)}{\left(x+1\right)^2}.\dfrac{\left(x-1\right)\left(x+1\right)}{2\left(x^2-x+1\right)}\) \(=\dfrac{x-1}{2}\) \(d,\dfrac{x^8-1}{x+1}.\dfrac{1}{\left(x^2+1\right)\left(x^4+1\right)}\) \(=\dfrac{\left(x^4\right)^2-1}{x+1}.\dfrac{1}{\left(x^2+1\right)\left(x^4+1\right)}\) \(=\dfrac{\left(x^4-1\right)\left(x^4+1\right)}{x+1}.\dfrac{1}{\left(x^2+1\right)\left(x^4+1\right)}\) \(=\dfrac{\left(x^2+1\right)\left(x^2-1\right)}{x+1}.\dfrac{1}{x^2+1}\) \(=\dfrac{\left(x-1\right)\left(x+1\right)}{x+1}\) \(=x-1\) \(e,\dfrac{x-y}{xy+y^2}-\dfrac{3x+y}{x^2-xy}.\dfrac{y-x}{x+y}\) \(=\dfrac{x-y}{y\left(x+y\right)}-\dfrac{3x+y}{x\left(x-y\right)}.\dfrac{-\left(x-y\right)}{x+y}\) \(=\dfrac{x-y}{y\left(x+y\right)}-\dfrac{3x+y}{x}.\dfrac{-1}{x+y}\) \(=\dfrac{x-y}{y\left(x+y\right)}-\dfrac{-3x-y}{x\left(x+y\right)}\) \(=\dfrac{x\left(x-y\right)+y\left(3x+y\right)}{xy\left(x+y\right)}\) \(=\dfrac{x^2-xy+3xy+y^2}{xy\left(x+y\right)}\) \(=\dfrac{x^2+2xy+y^2}{xy\left(x+y\right)}\) \(=\dfrac{\left(x+y\right)^2}{xy\left(x+y\right)}=\dfrac{x+y}{xy}\)tìm giá trị của m để pt 2x-m=1-x nhận giá trị x=-2 là nghiệm
giải hộ e với :)
![](https://rs.olm.vn/images/avt/0.png?1311)
Thực hiện phép nhân đa thức với đa thức ở vế trái.
=> VT = VP (đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a^2+b^2=\left(a+b\right)^2-2ab=\left(-3\right)^2-2\cdot\left(-2\right)=9+4=13\)
\(a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)\)
\(=\left(-3\right)^3-3\cdot\left(-2\right)\cdot\left(-3\right)\)
\(=-27-18=-45\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a^2+b^2=a^3+b^3=a^4+b^4\)
\(\Rightarrow\left(a^3+b^3\right)^2=\left(a^2+b^2\right)\left(a^4+b^4\right)\)
\(\Rightarrow a^6+b^6+2a^3b^3=a^6+b^6+a^2b^4+a^4b^2\)
\(\Rightarrow2a^3b^3=a^2b^2\left(a^2+b^2\right)\)
\(\Rightarrow2ab=a^2+b^2\)
\(\Rightarrow\left(a-b\right)^2=0\)
\(\Rightarrow a=b\)
Thế vào \(a^2+b^2=a^3+b^3\)
\(\Rightarrow a^2+a^2=a^3+a^3\Rightarrow2a^3=2a^2\Rightarrow a=b=1\)
\(\Rightarrow a+b=2\)