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Quy đồng mẫu thức các phân thức sau :
a) 2514x2y;1421xy5
Mình ko ghi lại đề , bạn ghi ra xong rồi suy ra như mình nha .
1) \(=>A=\left(6x^2+3x-10x-5\right)-\left(6x^2+14x-9x-21\right)\)
\(=>A=-12x+16\)
2) \(=>B=8x^3+27-8x^3+2=29\)
3)\(=>C=[\left(x-1\right)-\left(x+1\right)]^3=\left(-2\right)^3=-8\)
4)\(=>D=[\left(2x+5\right)-\left(2x\right)]^3=5^3=125\)
5)\(=>E=\left(3x+1\right)^2-\left(3x+5\right)^2+12x+2\left(6x+3\right)\)
\(=>E=\left(3x+1+3x+5\right)\left(3x+1-3x-5\right)+12x+12x+6\)
\(=>E=\left(6x+6\right)\left(-4\right)+24x+6=-24x-24+24x+6=-18\)
6)\(=>F=\left(2x^2+3x-10x-15\right)-\left(2x^2-6x\right)+x+7=-8\)
k cho mik nha ,
Bài 2:
a: \(\dfrac{1}{2x^3y}=\dfrac{6yz^3}{12x^3y^2z^3}\)
\(\dfrac{2}{3xy^2z^3}=\dfrac{2\cdot4x^2}{12x^3y^2z^3}=\dfrac{8x^2}{12x^3y^2z^3}\)
\(a.\) Ta có:
\(MTC:\) \(\left(x+1\right)\left(x+2\right)\)
Do đó
\(\frac{3x}{x+1}=\frac{3x\left(x+2\right)}{\left(x+1\right)\left(x+2\right)}\)
\(\frac{x+4}{x+2}=\frac{\left(x+1\right)\left(x+4\right)}{\left(x+1\right)\left(x+2\right)}\)
\(b.\) Ta có:
\(x^2+x=x\left(x+1\right)\)
\(x^2-1=\left(x-1\right)\left(x+1\right)\)
nên \(MTC:\) \(x\left(x-1\right)\left(x+1\right)\)
Do đó:
\(\frac{5}{x^2+x}=\frac{5}{x\left(x+1\right)}=\frac{5\left(x-1\right)}{x\left(x-1\right)\left(x+1\right)}\)
\(\frac{6}{x^2-1}=\frac{6}{\left(x-1\right)\left(x+1\right)}=\frac{6x}{x\left(x-1\right)\left(x+1\right)}\)
\(c.\) Ta có:
\(x^2-5x+4=x^2-x-4x+4=x\left(x-1\right)-4\left(x-1\right)=\left(x-1\right)\left(x-4\right)\)
\(2x^2-8x=2x\left(x-4\right)\)
nên \(MTC:\) \(2x\left(x-1\right)\left(x-4\right)\)
Do đó:
\(\frac{4}{x^2-5x+4}=\frac{4}{\left(x-1\right)\left(x-4\right)}=\frac{8x}{2x\left(x-1\right)\left(x-4\right)}\)
\(\frac{x+1}{2x^2-8x}=\frac{x+1}{2x\left(x-4\right)}=\frac{\left(x-1\right)\left(x+1\right)}{2x\left(x-1\right)\left(x-4\right)}\)
Làm nốt d :P
\(\frac{x+3}{2x^2-15x-8};\frac{3}{x^2-8x}\)
Ta có : \(2x^2-15x-8=\left(2x+1\right)\left(x-8\right)\)
\(x^2-8x=x\left(x-8\right)\)
MTC : \(x\left(x-8\right)\left(2x+1\right)\)
\(\frac{x+3}{2x^2-15x-8}=\frac{x+3}{\left(2x+1\right)\left(x-8\right)}=\frac{x^2+3x}{x\left(x-8\right)\left(2x+1\right)}\)
\(\frac{3}{x^2-8x}=\frac{3}{x\left(x-8\right)}=\frac{6x+3}{x\left(x-8\right)\left(2x+1\right)}\)
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: \(\frac{10}{x+2}=\frac{10\cdot\left(-6\right)\cdot\left(x-2\right)}{\left(x+2\right)\cdot\left(-6\right)\cdot\left(x-2\right)}=\frac{-60\left(x-2\right)}{-6\left(x-2\right)\left(x+2\right)}=\frac{120-60x}{-6\left(x-2\right)\left(x+2\right)}\)
\(\frac{5}{2x-4}=\frac{5}{2\left(x-2\right)}=\frac{5\cdot\left(-3\right)\cdot\left(x+2\right)}{2\cdot\left(x-2\right)\cdot\left(-3\right)\cdot\left(x+2\right)}=\frac{-15\left(x+2\right)}{-6\left(x-2\right)\left(x+2\right)}=\frac{-15x-30}{-6\left(x-2\right)\left(x+2\right)}\)
\(\frac{1}{6-3x}=\frac{1}{3\left(2-x\right)}=\frac{-1}{3\left(x-2\right)}=\frac{-1\cdot\left(-2\right)\cdot\left(x+2\right)}{3\cdot\left(-2\right)\cdot\left(x-2\right)\cdot\left(x+2\right)}=\frac{2x+4}{-6\left(x-2\right)\left(x+2\right)}\)
\(\left(a\right)\frac{30x^3y^4}{12x^3y^4\left(x+3\right)}--\frac{7x+21}{12x^3y^4\left(x+3\right)}\)
\(\left(b\right)\frac{16x}{60x^4y^5}--\frac{55y^3}{60x^4y^5}\)
\(\left(c\right)\frac{5x-15}{2\left(x+3\right)\left(x-3\right)}--\frac{6}{2\left(x+3\right)\left(x-3\right)}\)
\(\left(d\right)\frac{6x}{3\left(x-4\right)^2}--\frac{x^2-4x}{3\left(x-4\right)^2}\)
\(\left(f\right)\frac{60}{6\left(x+2\right)}--\frac{-15}{6\left(x+2\right)}--\frac{-2}{6\left(x+2\right)}\)
\(\left(e\right)\frac{4x^2-3x+5}{x^3-1}--\frac{3x-2x^2-1}{x^3-1}--\frac{2-2x^3}{x^3-1}\)