Bài 24. Phép nhân và phép chia phân thức đại số

\(\begin{array}{l}a)P.\frac{{x + 1}}{{2{\rm{x}} + 1}} = \frac{{{x^2} + x}}{{4{{\rm{x}}^2} - 1}}\\P = \frac{{{x^2} + x}}{{4{{\rm{x}}^2} - 1}}:\frac{{x + 1}}{{2{\rm{x}} + 1}}\\P = \frac{{{x^2} + x}}{{4{{\rm{x}}^2} - 1}}.\frac{{2{\rm{x}} + 1}}{{x + 1}}\\P = \frac{{x\left( {x + 1} \right).\left( {2{\rm{x}} + 1} \right)}}{{\left( {2{\rm{x}} - 1} \right)\left( {2{\rm{x}} + 1} \right)\left( {x + 1} \right)}}\\P = \frac{x}{{2{\rm{x}} - 1}}\end{array}\)

\(\begin{array}{l}b)Q:\frac{{{x^2}}}{{{x^2} + 4{\rm{x}} + 4}} = \frac{{\left( {x + 1} \right)\left( {x + 2} \right)}}{{{x^2} - 2{\rm{x}}}}\\Q = \frac{{\left( {x + 1} \right)\left( {x + 2} \right)}}{{{x^2} - 2{\rm{x}}}}.\frac{{{x^2}}}{{{x^2} + 4{\rm{x}} + 4}}\\Q = \frac{{\left( {x + 1} \right)\left( {x + 2} \right).{x^2}}}{{x\left( {x - 2} \right).{{\left( {x + 2} \right)}^2}}}\\Q = \frac{{x\left( {x + 1} \right)}}{{{x^2} - 4}}\end{array}\)

\(a,P=\dfrac{x^2+6x+9}{x^2+3x}\\ =\dfrac{x^2+2\cdot3\cdot x+3^2}{x\left(x+3\right)}\\ =\dfrac{\left(x+3\right)^2}{x\left(x+3\right)}\\ =\dfrac{x+3}{x}\\ Q=\dfrac{x^2+3x}{x^2-9}\\ =\dfrac{x\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}\\ =\dfrac{x}{x-3}\\ b,P\cdot Q=\dfrac{x+3}{x}\cdot\dfrac{x}{x-3}\\ =\dfrac{\left(x+3\right)\cdot x}{x\cdot\left(x-3\right)}\\ =\dfrac{x+3}{x-3}\\ P:Q=\dfrac{x+3}{x}:\dfrac{x}{x-3}\\ =\dfrac{x+3}{x}\cdot\dfrac{x-3}{x}\\ =\dfrac{x^2-9}{x^2}\)

Kết luận sau là sai vì đã là phép chia nó sẽ đổi dấu số thứ `2`

`(1/x: 1/x) : 1/x = 1 * x/1`

`1/x:(1/x:1/x)= 1/x : (1/x * x/1)=1/x *1`

\(a)\left( { - \frac{{3{\rm{x}}}}{{5{\rm{x}}{y^2}}}} \right):\left( { - \frac{{5{y^2}}}{{12{\rm{x}}y}}} \right) = \frac{{ - 3{\rm{x}}}}{{5{\rm{x}}{y^2}}}.\frac{{ - 12{\rm{x}}y}}{{5{y^2}}} = \frac{{36{{\rm{x}}^2}y}}{{25{\rm{x}}{y^4}}}\)

b) \(\frac{4{{\text{x}}^{2}}-1}{8{{\text{x}}^{3}}-1}:\frac{4{{\text{x}}^{2}}+4\text{x}+1}{4{{\text{x}}^{2}}+2\text{x}+1}=\frac{4{{\text{x}}^{2}}-1}{8{{\text{x}}^{3}}-1}.\frac{4{{\text{x}}^{2}}+2\text{x}+1}{4{{\text{x}}^{2}}+4\text{x}+1}\)

\(=\frac{\left( 2\text{x}-1 \right)\left( 2\text{x}+1 \right)\left( 4{{\text{x}}^{2}}+2\text{x}+1 \right)}{\left( 2\text{x}-1 \right)\left( 4{{\text{x}}^{2}}+2\text{x}+1 \right){{\left( 2\text{x}+1 \right)}^{2}}}=\frac{1}{2\text{x}+1}\).

a) Số tiền x (triệu đồng) mà bác Châu phải trả mỗi tháng là: 

\(\begin{array}{l}x = \frac{{1200}}{y} + \left( {1200.\frac{r}{{12}}} \right)\\ \Rightarrow x = \frac{{1200}}{y} + 100{\rm{r}}\\ \Rightarrow r = \frac{{xy - 1200}}{{100y}}\end{array}\)

b) Thay x = 30, y = 48, ta có: r = 0.05

Lãi suất năm của khoản vay khi mỗi tháng trả góp 30 triệu đồng trong vòng 4 năm là:

\(r = \frac{{30.48 - 1200}}{{100.48}} = 0,05 = 5(\% )\)

\(\begin{array}{l}\frac{{3{\rm{x}}}}{{2{y^2}}}:\left( {\frac{{ - 5{{\rm{x}}^2}}}{{12{y^3}}}} \right)\\ = \frac{{3{\rm{x}}}}{{2{y^2}}}.\frac{{12{y^3}}}{{ - 5{{\rm{x}}^2}}}\\ = \frac{{3{\rm{x}}.12{y^3}}}{{2{y^2}.\left( { - 5{{\rm{x}}^2}} \right)}} = \frac{{36{\rm{x}}{y^3}}}{{ - 10{{\rm{x}}^2}{y^2}}} = \frac{{ - 18y}}{{5y}}\end{array}\)

Ta có: \(\frac{{2{\rm{x}}}}{{x + 1}}.\frac{{x - 1}}{x} = \frac{{2{\rm{x}}\left( {x - 1} \right)}}{{x\left( {x + 1} \right)}} = \frac{{2\left( {x - 1} \right)}}{{x + 1}}\)

Ta có: \(\frac{{2{\rm{x}}}}{{x + 1}}.\frac{{x - 1}}{x} = \frac{{2{\rm{x}}.\left( {x - 1} \right)}}{{\left( {x + 1} \right).x}}\)

Vậy nhận định của cả Pi và tròn đều đúng

\(a)\frac{x}{{x + y}}.\frac{{2{\rm{x}} + 2y}}{{3{\rm{x}}y}}\)

\(\begin{array}{l} = \frac{{2{{\rm{x}}^2} + 2{\rm{x}}y}}{{3{\rm{x}}y(x + y)}}\\ = \frac{{2{\rm{x}}(x + y)}}{{3{\rm{x}}y(x + y)}} = \frac{{2{\rm{x}}}}{{3{\rm{x}}y}}\end{array}\)

\(b)\frac{{3{\rm{x}}}}{{4{{\rm{x}}^2} - 1}}.\frac{{ - 2{\rm{x}} + 1}}{{2{{\rm{x}}^2}}}\)

\(\begin{array}{l} = \frac{{3{\rm{x}}( - 2{\rm{x}} + 1)}}{{2{{\rm{x}}^2}(4{{\rm{x}}^2} - 1)}}\\ = \frac{{ - 3{\rm{x}}}}{{2{{\rm{x}}^2}(2{\rm{x}} + 1)}}\end{array}\)

\(a)\left( { - \frac{{3{\rm{x}}}}{{5{\rm{x}}{y^2}}}} \right).\left( { - \frac{{5{y^2}}}{{12{\rm{x}}y}}} \right) = \frac{{\left( { - 3{\rm{x}}} \right).\left( { - 5{y^2}} \right)}}{{5{\rm{x}}{y^2}.12{\rm{x}}y}} = \frac{1}{{4{\rm{x}}y}}\)

\(b)\frac{{{x^2} - x}}{{2{\rm{x}} + 1}}.\frac{{4{{\rm{x}}^2} - 1}}{{{x^3} - 1}} = \frac{{x\left( {x - 1} \right).\left( {2{\rm{x}} - 1} \right)\left( {2{\rm{x}} + 1} \right)}}{{\left( {2{\rm{x}} + 1} \right).\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} = \frac{{x\left( {2{\rm{x}} - 1} \right)}}{{{x^2} + x + 1}}\)