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1) \(\frac{1}{x-3}\) và \(\frac{5}{x^2-3x}\)
Ta có: \(1.\left(x^2-3x\right)=x^2-3x\)
\(\left(x-3\right).5=5x-15\)
\(\Rightarrow x^2-3x\ne5x-15\)
\(\Rightarrow1.\left(x^2-3x\right)\ne\left(x-3\right).5\)
Vậy: \(\frac{1}{x-3}\ne\frac{5}{x^2-3x}\)
2) \(\frac{x}{x^2+x}\) và \(\frac{2}{x-1}\) và \(\frac{x+2}{x^2-1}\)
Ta có: \(x.\left(x-1\right)=x^2-x\)
\(2.\left(x^2+x\right)=2x^2+2x\)
\(\Rightarrow x^2-x\ne2x^2+2x\)
\(\Rightarrow x.\left(x-1\right)\ne2.\left(x^2+x\right)\)
\(\Rightarrow\frac{1-3x}{2x}\ne\frac{2}{x-1}\) (1)
Ta lại có: \(2.\left(x^2-1\right)=2x^2-2\)
\(\left(x-1\right)\left(x+2\right)=x^2+2x-x-2\)
\(=x^2-x-2\)
\(\Rightarrow2x^2-2\ne x^2-x-2\)
\(\Rightarrow2.\left(x^2-1\right)\ne\left(x-1\right)\left(x+2\right)\)
\(\Rightarrow\frac{2}{x-1}\ne\frac{x+2}{x^2-1}\) (2)
Từ (1) và (2) => \(\frac{x}{x^2+x}\ne\frac{2}{x-1}\ne\frac{x+2}{x^2-1}\)
3) \(\frac{1-3x}{2x}\) và \(\frac{3x-2}{2x-1}\) và \(\frac{3x-2}{4x^2-2x}\)
Ta có:\(\left(1-3x\right)\left(2x-1\right)=2x-1-6x^2+3x\)
\(=5x-1-6x^2\)
\(2x.\left(3x-2\right)=6x^2-4x\)
\(\Rightarrow5x-1-6x^2\ne6x^2-4x\)
\(\Rightarrow\left(1-3x\right)\left(2x-1\right)\ne2x\left(3x-2\right)\)
\(\Rightarrow\frac{1-3x}{2x}\ne\frac{3x-2}{2x-1}\)(1)
Ta lại có: \(\left(3x-2\right)\left(4x^2-2x\right)=12x^2-6x^2-8x^2+4x\)
\(=12x^3-14x^2+4x\)
\(\left(2x-1\right)\left(3x-2\right)=6x^2-4x-3x+2\)
\(=6x^2-7x+2\)
\(\Rightarrow12x^3-14x^2+4x\ne6x^2-7x+2\)
\(\Rightarrow\left(3x-2\right)\left(4x^2-2x\right)\ne\left(2x-1\right)\left(3x-2\right)\)
\(\Rightarrow\frac{3x-2}{2x-1}\ne\frac{3x-2}{4x^2-2x}\) (2)
Từ (1) và (2) => \(\frac{1-3x}{2x}\ne\frac{3x-2}{2x-1}\ne\frac{3x-2}{4x^2-2x}\)
Bài 2: \(a,\frac{7x-1}{2x^2+6x}=\frac{7x-1}{2x\left(x+3\right)}=\frac{\left(7x-1\right)\left(x-3\right)}{2x\left(x+3\right)\left(x-3\right)}\)
\(\frac{5-3x}{x^2-9}=\frac{5-3x}{\left(x-3\right)\left(x+3\right)}=\frac{\left(5-3x\right)2x}{2x\left(x-3\right)\left(x+3\right)}\)
\(b,\frac{x+1}{x-x^2}=\frac{x+1}{x\left(1-x\right)}=-\frac{x+1}{x\left(x+1\right)}=-\frac{2\left(x-1\right)\left(x+1\right)}{2x\left(x-1\right)^2}\)
\(\frac{x+2}{2-4x+2x^2}=\frac{x+2}{2\left(x-1\right)^2}=\frac{2x\left(x+2\right)}{2x\left(x-1\right)^2}\)
\(c,\frac{4x^2-3x+5}{x^3-1}=\frac{4x^2-3x+5}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\frac{2x}{x^2+x+1}=\frac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\frac{6}{x-1}=\frac{6\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(d,\frac{7}{5x}=\frac{7.2\left(2y-x\right)\left(2y+x\right)}{2.5x\left(2y-x\right)\left(2y+x\right)}\)
\(\frac{4}{x-2y}=-\frac{4}{2y-x}=-\frac{4.2.5x\left(2x+x\right)}{2.5x\left(2y-x\right)\left(2y+x\right)}\)
\(\frac{x-y}{8y^2-2x^2}=\frac{x-y}{2\left(4y^2-x^2\right)}=\frac{x-y}{2\left(2y-x\right)\left(2y+x\right)}=\frac{5x\left(x-y\right)}{2.5x.\left(2y-x\right)\left(2y+x\right)}\)
mk ko biết làm
xin lỗi bn nhae
xin lỗi vì đã ko giúp được bn
chcus bn học gioi!
nhae@@@
a ) \(\frac{4}{x+2}+\frac{2}{x-2}+\frac{5x-6}{4-x^2}=\frac{4\left(x-2\right)+2\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{6-5x}{\left(x+2\right)\left(x-2\right)}=\frac{6x-4+6-5x}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{x+2}{\left(x+2\right)\left(x-2\right)}=\frac{1}{x+2}\)
b ) \(\frac{1-3x}{2x}+\frac{3x-2}{2x-1}+\frac{3x-2}{2x-4x^2}=\frac{\left(1-3x\right)\left(2x-1\right)+2x\left(3x-2\right)+2-3x}{2x\left(2x-1\right)}\)
\(=\frac{-6x^2+5x-1+6x^2-4x+2-3x}{2x\left(2x-1\right)}=\frac{-2x+1}{2x\left(2x-1\right)}=\frac{-1}{2x}\)
c ) \(\frac{1}{x^2+6x+9}+\frac{1}{6x-x^2-9}+\frac{x}{x^2-9}=\frac{1}{\left(x+3\right)^2}+\frac{1}{-\left(x-3\right)^2}+\frac{x}{\left(x+3\right)\left(x-3\right)}\)
\(=\frac{\left(x-3\right)^2-\left(x+3\right)^2+x\left(x+3\right)\left(x-3\right)}{\left(x+3\right)^2\left(x-3\right)^2}=\frac{-12x+x^3-9x}{\left(x+3\right)^2\left(x-3\right)^2}=\frac{x^3-21x}{x^4-18x^2+81}\)
d ) \(\frac{x^2+2}{x^3-1}+\frac{2}{x^2+x+1}+\frac{1}{1-x}=\frac{x^2+2+2\left(x-1\right)-\left(x^2+x+1\right)}{x^3-1}=\frac{x-1}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\frac{1}{x^2+x+1}\)
e ) \(\frac{x}{x-2y}+\frac{x}{x+2y}+\frac{4xy}{4y^2-x^2}=\frac{x\left(x+2y\right)+x\left(x-2y\right)-4xy}{\left(x-2y\right)\left(x+2y\right)}=\frac{2x\left(x-2y\right)}{\left(x-2y\right)\left(x+2y\right)}\)
\(=\frac{2x}{x+2y}\)