quy đồng mẫu thức các phân thức sau:
a)5/2x+6 và 7/12x^3y^4
b)4/15x^3y^5 và 11/12x^4y^2
c)5/2x+6 và 3/x^2-9
d)2x/x^2-8x+16 và x/3x^2-12x
e)4x^2-3x+5/x^3-1 và 1-2x/x^2+x+1 và -2
f)10/x+2 và 5/2x-4 và 1/6-3x
giúp mình với các bạn!!!mình tin các bạn có thể trả lời chính xác câu hỏi này!!
vì phân thức này ko có dấu gạch ngang nên mình dùng dấu /
xin trân trọng cảm ơn!!!
a) \(\frac{5}{2x+6}\) và \(\frac{7}{12x^3y^4}\)
Ta có: \(2x+6=2\left(x+3\right)\)
\(12x^3y^4=12x^3y^4\)
\(MSC=12x^3y^4\left(x+3\right)\)
Ta có: \(\frac{5}{2x+6}=\frac{5}{2\left(x+3\right)}=\frac{5\cdot6\cdot x^3y^4}{2\cdot6\cdot x^3y^4\cdot\left(x+3\right)}=\frac{30x^3y^4}{12x^3y^4\left(x+3\right)}\)
\(\frac{7}{12x^3y^4}=\frac{7\cdot\left(x+3\right)}{12x^3y^4\cdot\left(x+3\right)}=\frac{7x+21}{12x^3y\left(x+3\right)}\)
b)\(\frac{4}{15x^3y^5}\) và \(\frac{11}{12x^4y^2}\)
MSC=\(60x^4y^5\)
Ta có: \(\frac{4}{15x^3y^5}=\frac{4\cdot4\cdot x}{15x^3y^5\cdot4\cdot x}=\frac{16x}{60x^4y^5}\)
\(\frac{11}{12x^4y^2}=\frac{11\cdot5\cdot y^3}{12x^4y^2\cdot5\cdot y^3}=\frac{55y^3}{60x^4y^5}\)
c) \(\frac{5}{2x+6}\) và \(\frac{3}{x^2-9}\)
Ta có: \(2x+6=2\left(x+3\right)\)
\(x^2-9=\left(x-3\right)\left(x+3\right)\)
MSC=2(x+3)(x-3)
Ta có: \(\frac{5}{2x+6}=\frac{5}{2\left(x+3\right)}=\frac{5\cdot\left(x-3\right)}{2\left(x+3\right)\left(x-3\right)}=\frac{5x-15}{2\left(x+3\right)\left(x-3\right)}\)
\(\frac{3}{x^2-9}=\frac{3}{\left(x-3\right)\left(x+3\right)}=\frac{3\cdot2}{2\cdot\left(x-3\right)\cdot\left(x+3\right)}=\frac{6}{2\left(x-3\right)\left(x+3\right)}\)
d) \(\frac{2x}{x^2-8x+16}\) và \(\frac{x}{3x^2-12x}\)
Ta có: \(x^2-8x+16=\left(x-4\right)^2\)
\(3x^2-12x=3x\left(x-4\right)\)
MSC=\(3x\left(x-4\right)^2\)
Ta có: \(\frac{2x}{x^2-8x+16}=\frac{2x}{\left(x-4\right)^2}=\frac{2x\cdot3x}{3x\cdot\left(x-4\right)^2}=\frac{6x^2}{3x\left(x-4\right)^2}\)
\(\frac{x}{3x^2-12x}=\frac{x}{3x\left(x-4\right)}=\frac{x\left(x-4\right)}{3x\left(x-4\right)^2}=\frac{x^2-4x}{3x\left(x-4\right)^2}\)
e) \(\frac{4x^2-3x+5}{x^3-1}\); \(\frac{1-2x}{x^2+x+1}\) và -2
Ta có: \(x^3-1=\left(x-1\right)\left(x^2+x+1\right)\)
MSC=\(\left(x-1\right)\left(x^2+x+1\right)\)
Ta có: \(\frac{4x^2-3x+5}{x^3-1}=\frac{4x^2-2x+5}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\frac{1-2x}{x^2+x+1}=\frac{\left(1-2x\right)\left(x-1\right)}{\left(x^2+x+1\right)\left(x-1\right)}=\frac{3x-2x^2-1}{\left(x^2+x+1\right)\left(x-1\right)}\)
\(-2=\frac{-2\left(x^2+x+1\right)\left(x-1\right)}{\left(x^2+x+1\right)\left(x-1\right)}=\frac{-2\left(x^3-1\right)}{\left(x^2+x+1\right)\left(x-1\right)}=\frac{-2x^3+2}{\left(x^2+x+1\right)\left(x-1\right)}\)
f) \(\frac{10}{x+2}\) và \(\frac{5}{2x-4}\) và \(\frac{1}{6-3x}\)
Ta có: \(x+2=x+2\)
\(2x-4=2\left(x-2\right)\)
\(6-3x=3\left(2-x\right)=-3\left(x-2\right)\)
MSC=-6(x-2)(x+2)
Ta có: \(\frac{10}{x+2}=\frac{10\cdot\left(-6\right)\cdot\left(x-2\right)}{\left(x+2\right)\cdot\left(-6\right)\cdot\left(x-2\right)}=\frac{-60\left(x-2\right)}{-6\left(x-2\right)\left(x+2\right)}=\frac{120-60x}{-6\left(x-2\right)\left(x+2\right)}\)
\(\frac{5}{2x-4}=\frac{5}{2\left(x-2\right)}=\frac{5\cdot\left(-3\right)\cdot\left(x+2\right)}{2\cdot\left(x-2\right)\cdot\left(-3\right)\cdot\left(x+2\right)}=\frac{-15\left(x+2\right)}{-6\left(x-2\right)\left(x+2\right)}=\frac{-15x-30}{-6\left(x-2\right)\left(x+2\right)}\)
\(\frac{1}{6-3x}=\frac{1}{3\left(2-x\right)}=\frac{-1}{3\left(x-2\right)}=\frac{-1\cdot\left(-2\right)\cdot\left(x+2\right)}{3\cdot\left(-2\right)\cdot\left(x-2\right)\cdot\left(x+2\right)}=\frac{2x+4}{-6\left(x-2\right)\left(x+2\right)}\)
\(\left(a\right)\frac{30x^3y^4}{12x^3y^4\left(x+3\right)}--\frac{7x+21}{12x^3y^4\left(x+3\right)}\)
\(\left(b\right)\frac{16x}{60x^4y^5}--\frac{55y^3}{60x^4y^5}\)
\(\left(c\right)\frac{5x-15}{2\left(x+3\right)\left(x-3\right)}--\frac{6}{2\left(x+3\right)\left(x-3\right)}\)
\(\left(d\right)\frac{6x}{3\left(x-4\right)^2}--\frac{x^2-4x}{3\left(x-4\right)^2}\)
\(\left(f\right)\frac{60}{6\left(x+2\right)}--\frac{-15}{6\left(x+2\right)}--\frac{-2}{6\left(x+2\right)}\)
\(\left(e\right)\frac{4x^2-3x+5}{x^3-1}--\frac{3x-2x^2-1}{x^3-1}--\frac{2-2x^3}{x^3-1}\)
