Question

[Nguyễn Thị Ngọc Ánh]

cho hai phân thức Q=\(\frac{12x^2+20x+3}{6x^2+43x+7}\)và P=\(\frac{8x^2+36x+36}{x^2+10x+21}\) với các giá trị của x để P và Q xác định thì Q=...P

Answer 1

[Trà My]

\(Q=\frac{12x^2+20x+3}{6x^2+43x+7}=\frac{12x^2+18x+2x+3}{6x^2+42x+x+7}=\frac{6x\left(2x+3\right)+\left(2x+3\right)}{6x\left(x+7\right)+\left(x+7\right)}=\frac{\left(6x+1\right)\left(2x+3\right)}{\left(6x+1\right)\left(x+7\right)}=\frac{2x+3}{x+7}\)\(P=\frac{8x^2+36x+36}{x^2+10x+21}=\frac{4\left(2x^2+9x+9\right)}{x^2+3x+7x+21}=\frac{4\left(2x^2+3x+6x+9\right)}{x\left(x+3\right)+7\left(x+3\right)}=\frac{4\left[x\left(2x+3\right)+3\left(2x+3\right)\right]}{\left(x+3\right)\left(x+7\right)}=\frac{4\left(x+3\right)\left(2x+3\right)}{\left(x+3\right)\left(x+7\right)}=\frac{4\left(2x+3\right)}{x+7}\)

=> \(Q:P=\frac{2x+3}{x+7}:\frac{4\left(2x+3\right)}{x+7}=\frac{2x+3}{x+7}.\frac{x+7}{4\left(2x+3\right)}=\frac{1}{4}\)

=>\(Q=\frac{1}{4}P\)

Answer 2

[ngonhuminh]

\(\frac{Q\left(0\right)}{P\left(0\right)}=\frac{3.21}{7.36}=\frac{1}{4}\Rightarrow Q=\frac{1}{4}P\)