​x+31​−x−32​=(x−3)⋅(x+3)3⋅(x+5)​+x+31​−x−32​
x2−93x+15​+x+31​−x−32​=(x−3)⋅(x+3)3⋅(x+5)​+x+31​−x−32​=(x−3)⋅(x+3)3⋅(x+5)​+(x+3)⋅(x−3)x−3​−(x−3)⋅(x+3)2⋅(x+3)​\(=\frac{3\cdot\left(\right.x+5\left.\right)+x-3-2\cdot\left(\right.x+3\left.\right)}{\left(\right.x-3\left.\right)\cdot\left(\right.x+3\left.\right)}=\frac{3x+15+x-3-2x-6}{\left(\right.x-3\left.\right)\cdot\left(\right.x+3\left.\right)}=\frac{2x+6}{\left(\right.x+3\left.\right)\cdot\left(\right.x-3\left.\right)}=\frac{2\cdot\left(\right.x+3\left.\right)}{\left(\right.x+3\left.\right)\cdot\left(\right.x-3\left.\right)}=\frac{2}{x-3}\)

câu b

để \(\frac{2}{x-3}=\frac{2}{3}\) thì \(x-3=3\)

\(\Rightarrow x=3+3=6\)

vậy  \(x=6\) thì \(x=\frac{2}{3}\)

​x+31​−x−32​=(x−3)⋅(x+3)3⋅(x+5)​+x+31​−x−32​
x2−93x+15​+x+31​−x−32​=(x−3)⋅(x+3)3⋅(x+5)​+x+31​−x−32​=(x−3)⋅(x+3)3⋅(x+5)​+(x+3)⋅(x−3)x−3​−(x−3)⋅(x+3)2⋅(x+3)​\(=\frac{3\cdot\left(\right.x+5\left.\right)+x-3-2\cdot\left(\right.x+3\left.\right)}{\left(\right.x-3\left.\right)\cdot\left(\right.x+3\left.\right)}=\frac{3x+15+x-3-2x-6}{\left(\right.x-3\left.\right)\cdot\left(\right.x+3\left.\right)}=\frac{2x+6}{\left(\right.x+3\left.\right)\cdot\left(\right.x-3\left.\right)}=\frac{2\cdot\left(\right.x+3\left.\right)}{\left(\right.x+3\left.\right)\cdot\left(\right.x-3\left.\right)}=\frac{2}{x-3}\)

câu b

để \(\frac{2}{x-3}=\frac{2}{3}\) thì \(x-3=3\)

\(\Rightarrow x=3+3=6\)

vậy  \(x=6\) thì \(x=\frac{2}{3}\)

​x+31​−x−32​=(x−3)⋅(x+3)3⋅(x+5)​+x+31​−x−32​
x2−93x+15​+x+31​−x−32​=(x−3)⋅(x+3)3⋅(x+5)​+x+31​−x−32​=(x−3)⋅(x+3)3⋅(x+5)​+(x+3)⋅(x−3)x−3​−(x−3)⋅(x+3)2⋅(x+3)​\(=\frac{3\cdot\left(\right.x+5\left.\right)+x-3-2\cdot\left(\right.x+3\left.\right)}{\left(\right.x-3\left.\right)\cdot\left(\right.x+3\left.\right)}=\frac{3x+15+x-3-2x-6}{\left(\right.x-3\left.\right)\cdot\left(\right.x+3\left.\right)}=\frac{2x+6}{\left(\right.x+3\left.\right)\cdot\left(\right.x-3\left.\right)}=\frac{2\cdot\left(\right.x+3\left.\right)}{\left(\right.x+3\left.\right)\cdot\left(\right.x-3\left.\right)}=\frac{2}{x-3}\)

câu b

để \(\frac{2}{x-3}=\frac{2}{3}\) thì \(x-3=3\)

\(\Rightarrow x=3+3=6\)

vậy  \(x=6\) thì \(x=\frac{2}{3}\)

​x+31​−x−32​=(x−3)⋅(x+3)3⋅(x+5)​+x+31​−x−32​
x2−93x+15​+x+31​−x−32​=(x−3)⋅(x+3)3⋅(x+5)​+x+31​−x−32​=(x−3)⋅(x+3)3⋅(x+5)​+(x+3)⋅(x−3)x−3​−(x−3)⋅(x+3)2⋅(x+3)​\(=\frac{3\cdot\left(\right.x+5\left.\right)+x-3-2\cdot\left(\right.x+3\left.\right)}{\left(\right.x-3\left.\right)\cdot\left(\right.x+3\left.\right)}=\frac{3x+15+x-3-2x-6}{\left(\right.x-3\left.\right)\cdot\left(\right.x+3\left.\right)}=\frac{2x+6}{\left(\right.x+3\left.\right)\cdot\left(\right.x-3\left.\right)}=\frac{2\cdot\left(\right.x+3\left.\right)}{\left(\right.x+3\left.\right)\cdot\left(\right.x-3\left.\right)}=\frac{2}{x-3}\)

câu b

để \(\frac{2}{x-3}=\frac{2}{3}\) thì \(x-3=3\)

\(\Rightarrow x=3+3=6\)

vậy  \(x=6\) thì \(x=\frac{2}{3}\)

​x+31​−x−32​=(x−3)⋅(x+3)3⋅(x+5)​+x+31​−x−32​
x2−93x+15​+x+31​−x−32​=(x−3)⋅(x+3)3⋅(x+5)​+x+31​−x−32​=(x−3)⋅(x+3)3⋅(x+5)​+(x+3)⋅(x−3)x−3​−(x−3)⋅(x+3)2⋅(x+3)​\(=\frac{3\cdot\left(\right.x+5\left.\right)+x-3-2\cdot\left(\right.x+3\left.\right)}{\left(\right.x-3\left.\right)\cdot\left(\right.x+3\left.\right)}=\frac{3x+15+x-3-2x-6}{\left(\right.x-3\left.\right)\cdot\left(\right.x+3\left.\right)}=\frac{2x+6}{\left(\right.x+3\left.\right)\cdot\left(\right.x-3\left.\right)}=\frac{2\cdot\left(\right.x+3\left.\right)}{\left(\right.x+3\left.\right)\cdot\left(\right.x-3\left.\right)}=\frac{2}{x-3}\)

câu b

để \(\frac{2}{x-3}=\frac{2}{3}\) thì \(x-3=3\)

\(\Rightarrow x=3+3=6\)

vậy  \(x=6\) thì \(x=\frac{2}{3}\)

​x+31​−x−32​=(x−3)⋅(x+3)3⋅(x+5)​+x+31​−x−32​
x2−93x+15​+x+31​−x−32​=(x−3)⋅(x+3)3⋅(x+5)​+x+31​−x−32​=(x−3)⋅(x+3)3⋅(x+5)​+(x+3)⋅(x−3)x−3​−(x−3)⋅(x+3)2⋅(x+3)​\(=\frac{3\cdot\left(\right.x+5\left.\right)+x-3-2\cdot\left(\right.x+3\left.\right)}{\left(\right.x-3\left.\right)\cdot\left(\right.x+3\left.\right)}=\frac{3x+15+x-3-2x-6}{\left(\right.x-3\left.\right)\cdot\left(\right.x+3\left.\right)}=\frac{2x+6}{\left(\right.x+3\left.\right)\cdot\left(\right.x-3\left.\right)}=\frac{2\cdot\left(\right.x+3\left.\right)}{\left(\right.x+3\left.\right)\cdot\left(\right.x-3\left.\right)}=\frac{2}{x-3}\)

câu b

để \(\frac{2}{x-3}=\frac{2}{3}\) thì \(x-3=3\)

\(\Rightarrow x=3+3=6\)

vậy  \(x=6\) thì \(x=\frac{2}{3}\)

​x+31​−x−32​=(x−3)⋅(x+3)3⋅(x+5)​+x+31​−x−32​
x2−93x+15​+x+31​−x−32​=(x−3)⋅(x+3)3⋅(x+5)​+x+31​−x−32​=(x−3)⋅(x+3)3⋅(x+5)​+(x+3)⋅(x−3)x−3​−(x−3)⋅(x+3)2⋅(x+3)​\(=\frac{3\cdot\left(\right.x+5\left.\right)+x-3-2\cdot\left(\right.x+3\left.\right)}{\left(\right.x-3\left.\right)\cdot\left(\right.x+3\left.\right)}=\frac{3x+15+x-3-2x-6}{\left(\right.x-3\left.\right)\cdot\left(\right.x+3\left.\right)}=\frac{2x+6}{\left(\right.x+3\left.\right)\cdot\left(\right.x-3\left.\right)}=\frac{2\cdot\left(\right.x+3\left.\right)}{\left(\right.x+3\left.\right)\cdot\left(\right.x-3\left.\right)}=\frac{2}{x-3}\)

câu b

để \(\frac{2}{x-3}=\frac{2}{3}\) thì \(x-3=3\)

\(\Rightarrow x=3+3=6\)

vậy  \(x=6\) thì \(x=\frac{2}{3}\)

​x+31​−x−32​=(x−3)⋅(x+3)3⋅(x+5)​+x+31​−x−32​
x2−93x+15​+x+31​−x−32​=(x−3)⋅(x+3)3⋅(x+5)​+x+31​−x−32​=(x−3)⋅(x+3)3⋅(x+5)​+(x+3)⋅(x−3)x−3​−(x−3)⋅(x+3)2⋅(x+3)​\(=\frac{3\cdot\left(\right.x+5\left.\right)+x-3-2\cdot\left(\right.x+3\left.\right)}{\left(\right.x-3\left.\right)\cdot\left(\right.x+3\left.\right)}=\frac{3x+15+x-3-2x-6}{\left(\right.x-3\left.\right)\cdot\left(\right.x+3\left.\right)}=\frac{2x+6}{\left(\right.x+3\left.\right)\cdot\left(\right.x-3\left.\right)}=\frac{2\cdot\left(\right.x+3\left.\right)}{\left(\right.x+3\left.\right)\cdot\left(\right.x-3\left.\right)}=\frac{2}{x-3}\)

câu b

để \(\frac{2}{x-3}=\frac{2}{3}\) thì \(x-3=3\)

\(\Rightarrow x=3+3=6\)

vậy  \(x=6\) thì \(x=\frac{2}{3}\)

​x+31​−x−32​=(x−3)⋅(x+3)3⋅(x+5)​+x+31​−x−32​
x2−93x+15​+x+31​−x−32​=(x−3)⋅(x+3)3⋅(x+5)​+x+31​−x−32​=(x−3)⋅(x+3)3⋅(x+5)​+(x+3)⋅(x−3)x−3​−(x−3)⋅(x+3)2⋅(x+3)​\(=\frac{3\cdot\left(\right.x+5\left.\right)+x-3-2\cdot\left(\right.x+3\left.\right)}{\left(\right.x-3\left.\right)\cdot\left(\right.x+3\left.\right)}=\frac{3x+15+x-3-2x-6}{\left(\right.x-3\left.\right)\cdot\left(\right.x+3\left.\right)}=\frac{2x+6}{\left(\right.x+3\left.\right)\cdot\left(\right.x-3\left.\right)}=\frac{2\cdot\left(\right.x+3\left.\right)}{\left(\right.x+3\left.\right)\cdot\left(\right.x-3\left.\right)}=\frac{2}{x-3}\)

câu b

để \(\frac{2}{x-3}=\frac{2}{3}\) thì \(x-3=3\)

\(\Rightarrow x=3+3=6\)

vậy  \(x=6\) thì \(x=\frac{2}{3}\)

​x+31​−x−32​=(x−3)⋅(x+3)3⋅(x+5)​+x+31​−x−32​
x2−93x+15​+x+31​−x−32​=(x−3)⋅(x+3)3⋅(x+5)​+x+31​−x−32​=(x−3)⋅(x+3)3⋅(x+5)​+(x+3)⋅(x−3)x−3​−(x−3)⋅(x+3)2⋅(x+3)​\(=\frac{3\cdot\left(\right.x+5\left.\right)+x-3-2\cdot\left(\right.x+3\left.\right)}{\left(\right.x-3\left.\right)\cdot\left(\right.x+3\left.\right)}=\frac{3x+15+x-3-2x-6}{\left(\right.x-3\left.\right)\cdot\left(\right.x+3\left.\right)}=\frac{2x+6}{\left(\right.x+3\left.\right)\cdot\left(\right.x-3\left.\right)}=\frac{2\cdot\left(\right.x+3\left.\right)}{\left(\right.x+3\left.\right)\cdot\left(\right.x-3\left.\right)}=\frac{2}{x-3}\)

câu b

để \(\frac{2}{x-3}=\frac{2}{3}\) thì \(x-3=3\)

\(\Rightarrow x=3+3=6\)

vậy  \(x=6\) thì \(x=\frac{2}{3}\)