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\(\dfrac{7x^3+14x^2+7x}{14x^2+14x}=\dfrac{7x\left(x^2+2x+1\right)}{14x\left(x+1\right)}=\dfrac{\left(x+1\right)^2}{2\left(x+1\right)}=\dfrac{x+1}{2}\)
\(\dfrac{x^4-2x^2+1}{x^3+2x^2+x}=\dfrac{\left(x^2-1\right)^2}{x\left(x^2+2x+1\right)}=\dfrac{\left(x-1\right)^2\left(x+1\right)^2}{x\left(x+1\right)^2}=\dfrac{\left(x-1\right)^2}{x}\)
\(\left(2x-3\right)\left(4x^2+6x+9\right)-4x\left(2x^2-1\right)\)
\(=8x^3-27-8x^3+4x\\ =8x^3-8x^3+4x-27\\ =4x-27\)
\(=\dfrac{\left(x^2-y^2\right)\left(x^2+y^2\right)}{\left(y-x\right)\left(y^2+xy+x^2\right)}=\dfrac{-\left(y-x\right)\left(x+y\right)\left(x^2+y^2\right)}{\left(y-x\right)\left(y^2+xy+x^2\right)}=\dfrac{-\left(x+y\right)\left(x^2+y^2\right)}{x^2+xy+y^2}\)
a: \(=\dfrac{\left(3x-1\right)^3}{3x-1}\cdot\dfrac{8xy}{12x^3}=\dfrac{2y\left(3x-1\right)^2}{3x^2}\)
b: \(=\dfrac{\left(x-5\right)\left(x+5\right)}{x\left(x-5\right)}=\dfrac{x+5}{x}\)
Nhận xét: \(b^3c-cb^3=0;b^2c-cb^2=0.\).Nên phân thức trở thành:
\(\frac{a^3b-ab^3+c^3a-ca^3}{a^2b-ab^2+c^2a-ca^2}=\frac{a^3\left(b-c\right)-a\left(b^3-c^3\right)}{a^2\left(b-c\right)-a\left(b^2-c^2\right)}\)
\(=\frac{a\left(b-c\right)\left\{a^2-\left(b^2-bc+c^2\right)\right\}}{a\left(b-c\right)\left\{a-\left(b+c\right)\right\}}\)
\(=\frac{a^2-\left(b^2-bc+c^2\right)}{a-\left(b+c\right)}=\frac{a^2-\left(b+c\right)^2+3bc}{a-\left(b+c\right)}\)
\(=a+b+c+\frac{3bc}{a-b-c}\).
Bài 2:
\(\Leftrightarrow\left(x-1\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\)
\(a,=\left(x+8-x+2\right)^2=10^2=100\\ b,=x^2\left(x^2-16\right)-\left(x^4-1\right)=x^4-16x^2-x^4+1=1-16x^2\\ c,=x^3+1-x^3+1=2\)
\(\dfrac{a+2b-ab-2b^2}{1-b}=\dfrac{\left(a+2b\right)-b\left(a+2b\right)}{1-b}=\dfrac{\left(1-b\right)\left(a+2b\right)}{1-b}=a+2b\)
Cám ơn bạn nhiều ạ