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\(Q=\frac{12x^2+20x+3}{6x^2+43x+7}=\frac{12x^2+18x+2x+3}{6x^2+42x+x+7}=\frac{6x\left(2x+3\right)+\left(2x+3\right)}{6x\left(x+7\right)+\left(x+7\right)}=\frac{\left(6x+1\right)\left(2x+3\right)}{\left(6x+1\right)\left(x+7\right)}=\frac{2x+3}{x+7}\)\(P=\frac{8x^2+36x+36}{x^2+10x+21}=\frac{4\left(2x^2+9x+9\right)}{x^2+3x+7x+21}=\frac{4\left(2x^2+3x+6x+9\right)}{x\left(x+3\right)+7\left(x+3\right)}=\frac{4\left[x\left(2x+3\right)+3\left(2x+3\right)\right]}{\left(x+3\right)\left(x+7\right)}=\frac{4\left(x+3\right)\left(2x+3\right)}{\left(x+3\right)\left(x+7\right)}=\frac{4\left(2x+3\right)}{x+7}\)
=> \(Q:P=\frac{2x+3}{x+7}:\frac{4\left(2x+3\right)}{x+7}=\frac{2x+3}{x+7}.\frac{x+7}{4\left(2x+3\right)}=\frac{1}{4}\)
=>\(Q=\frac{1}{4}P\)
\(\frac{Q\left(0\right)}{P\left(0\right)}=\frac{3.21}{7.36}=\frac{1}{4}\Rightarrow Q=\frac{1}{4}P\)
P không xác định khi x2+10x+21=x2+3x+7x+21=x(x+3)+7(x+3)=(x+7)(x+3)=0<=>x=-3 hoặc x=-7
\(Q=\frac{\left(6x+1\right)\left(2x+3\right)}{\left(6x+1\right)\left(x+7\right)}=\frac{2x+3}{x+7}\)
\(P=\frac{4\left(2x+3\right)\left(x+3\right)}{\left(x+7\right)\left(x+3\right)}=\frac{4\left(2x+3\right)}{x+7}=4Q\)
\(\frac{P}{Q}=4\)
Bài 1:
a) x≠2
Bài 2:
a) x≠0;x≠5
b) x2−10x+25x2−5x=(x−5)2x(x−5)=x−5x
c) Để phân thức có giá trị nguyên thì x−5x phải có giá trị nguyên.
=> x=−5
Bài 3:
a) (x+12x−2+3x2−1−x+32x+2)⋅(4x2−45)
=(x+12(x−1)+3(x−1)(x+1)−x+32(x+1))⋅2(2x2−2)5
=(x+1)2+6−(x−1)(x+3)2(x−1)(x+1)⋅2⋅2(x2−1)5
=(x+1)2+6−(x2+3x−x−3)(x−1)(x+1)⋅2(x−1)(x+1)5
=[(x+1)2+6−(x2+2x−3)]⋅25
=[(x+1)2+6−x2−2x+3]⋅25
=[(x+1)2+9−x2−2x]⋅25
=2(x+1)25+185−25x2−45x
=2(x2+2x+1)5+185−25x2−45x
=2x2+4x+25+185−25x2−45x
=2x2+4x+2+185−25x2−45x
=2x2+4x+205−25x2−45x
c) tự làm, đkxđ: x≠1;x≠−1
