Question

giúp mình làm 2 đề Tl này được không mng ơi đề cương gv dạy toán cho mai thi r cứu với

 

 

 

Answer 1

câu 13

a=x(x-1)

b=2xy(x-3)

Answer 2

\(\dfrac{x^2+3\text{x}y+2y^2}{\left(x^3+2\text{x}^2y\right)-\left(xy^2+2y^3\right)}\)

=\(\dfrac{x^2+3\text{x}y+2y^2}{x^2\left(x+2y\right)-y^2\left(x+2y\right)}\)=\(\dfrac{x^2+3\text{x}y+2y^2}{\left(x+2y\right)\left(x^2-y^2\right)}\)

câu 17

Answer 3

13 a.\(x\)²\(-x\)\(=x(x-1)\)

b.\(2x\)²\(y-6xy\)\(=2xy(x-3) \)

c.\(x\)²\(-3x+2\)=\(x\)²\(-4-3x+6\)

                    =\((x\)²\(-4)\)\(-(3x-6)\)

                    =\((x-2)(x+2)-3(x-2)\)

14 a. \(2x(x+2)-3(x+2)=0\)

      \(\Rightarrow\)\((x+2)(2x-3)=0\)

      \(\Rightarrow\)\(x+2=0\) \(hoặc \) \(2x-3=0\)

      \(\Rightarrow\) \(x=-2\) \(hoặc\) \(x=\)\(\dfrac{3}{2}\)

15 a. \(\dfrac{3}{2x+3}-\dfrac{x-3}{2x^2+3x}=\dfrac{3}{2x+3}-\dfrac{x-3}{x\left(2x+3\right)}\)

                                            \(=\dfrac{3.x-\left(x-3\right)}{x\left(2x+3\right)}=\dfrac{3x-x+3}{x\left(2x+3\right)}=\dfrac{2x+3}{x\left(2x+3\right)}=\dfrac{1}{x}\)

b. \(\dfrac{4x-24}{5x-5}\div\dfrac{x^2-36}{x^2-2x+1}=\dfrac{4\left(x-6\right)}{5\left(x-1\right)}\div\dfrac{\left(x-6\right)\left(x+6\right)}{\left(x-1\right)^2}=\dfrac{4\left(x-6\right)}{5\left(x-1\right)}\times\dfrac{\left(x-1\right)^2}{\left(x-6\right)\left(x+6\right)}=\dfrac{4\left(x-1\right)}{5\left(x+6\right)}\)