Giải bài 2. 31 trang 51 SGK Toán 8 - Cùng khám phá — Không quảng cáo

Đề bài

Thực hiện các phép tính sau:

a)     $\frac{5}{{6x - 6}} + \frac{9}{{14x - 14}} + \frac{6}{{7x - 7}}$

b)    $\frac{2}{{y - 4}} + \frac{1}{y} - \frac{3}{{y - 3}}$

c)     $\frac{{8{a^2} + 18{b^2}}}{{4{a^2} - 9{b^2}}} - \frac{{2a + 3b}}{{2a - 3b}} + \frac{{2a - 3b}}{{2a + 3b}}$

d)    $\frac{{a - 4}}{{2a - 1}} + \frac{{5{a^2} + 9a + 14}}{{2{a^2} + 3a - 2}} - \frac{{3a - 5}}{{a + 2}}$

Lời giải chi tiết

a)

$\begin{array}{l}\frac{5}{{6x - 6}} + \frac{9}{{14x - 14}} + \frac{6}{{7x - 7}}\\ = \frac{5}{{6\left( {x - 1} \right)}} + \frac{9}{{14\left( {x - 1} \right)}} + \frac{6}{{7\left( {x - 1} \right)}}\\ = \frac{{5.14}}{{6.14.\left( {x - 1} \right)}} + \frac{{9.6}}{{6.14.\left( {x - 1} \right)}} + \frac{{6.12}}{{7.12.\left( {x - 1} \right)}}\\ = \frac{{70 + 54 + 72}}{{84\left( {x - 1} \right)}} = \frac{{196}}{{84\left( {x - 1} \right)}} = \frac{7}{{3\left( {x - 1} \right)}}\end{array}$

b)

$\begin{array}{l}\frac{2}{{y - 4}} + \frac{1}{y} - \frac{3}{{y - 3}}\\ = \frac{{2y\left( {y - 3} \right) + \left( {y - 4} \right)\left( {y - 3} \right) - 3y\left( {y - 4} \right)}}{{y\left( {y - 4} \right)\left( {y - 3} \right)}}\\ = \frac{{2{y^2} - 6y + {y^2} - y - 12 - 3{y^2} - 12y}}{{{y^3} - {y^2} - 12y}}\\ = \frac{{ - 19y - 12}}{{{y^3} - {y^2} - 12y}}\end{array}$

c)

$\begin{array}{l}\frac{{8{a^2} + 18{b^2}}}{{4{a^2} - 9{b^2}}} - \frac{{2a + 3b}}{{2a - 3b}} + \frac{{2a - 3b}}{{2a + 3b}}\\ = \frac{{8{a^2} + 18{b^2}}}{{\left( {2a - 3b} \right)\left( {2a + 3b} \right)}} - \frac{{2a + 3b}}{{2a - 3b}} + \frac{{2a - 3b}}{{2a + 3b}}\\ = \frac{{8{a^2} + 18{b^2} - \left( {2a + 3b} \right).\left( {2a + 3b} \right) + \left( {2a - 3b} \right)\left( {2a - 3b} \right)}}{{\left( {2a + 3b} \right)\left( {2a - 3b} \right)}}\\ =  = \frac{{8{a^2} + 18{b^2} - {{\left( {2a + 3b} \right)}^2} + {{\left( {2a - 3b} \right)}^2}}}{{\left( {2a + 3b} \right)\left( {2a - 3b} \right)}}\\ = \frac{{8{a^2} + 18{b^2} - 24ab}}{{4{a^2} - 9{b^2}}}\\ = \frac{{4{a^2} + 4{a^2} + 9{b^2} + 9{b^2} - 12ab - 12ab}}{{\left( {2a + 3b} \right)\left( {2a - 3b} \right)}}\\ = \frac{{\left( {4{a^2} - 12ab + 9{b^2}} \right) + \left( {4{a^2} - 12ab + 9{b^2}} \right)}}{{\left( {2a + 3b} \right)\left( {2a - 3b} \right)}}\\ = \frac{{{{\left( {2a - 3b} \right)}^2} + {{\left( {2a - 3b} \right)}^2}}}{{\left( {2a + 3b} \right)\left( {2a - 3b} \right)}}\\ = \frac{{2{{\left( {2a - 3b} \right)}^2}}}{{\left( {2a + 3b} \right)\left( {2a - 3b} \right)}}\\ = \frac{{2\left( {2a - 3b} \right)}}{{2a + 3b}}\end{array}$

d)

$\begin{array}{l}\frac{{a - 4}}{{2a - 1}} + \frac{{5{a^2} + 9a + 14}}{{2{a^2} + 3a - 2}} - \frac{{3a - 5}}{{a + 2}}\\ = \frac{{a - 4}}{{2a - 1}} + \frac{{5{a^2} + 9a + 14}}{{\left( {2a - 1} \right)\left( {a + 2} \right)}} - \frac{{3a - 5}}{{a + 2}}\\ = \frac{{\left( {a - 4} \right)\left( {a + 2} \right) + 5{a^2} + 9a + 14 - \left( {3a - 5} \right)\left( {2a - 1} \right)}}{{2{a^2} + 3a - 2}}\\ = \frac{{{a^2} - 2a - 8 + 5{a^2} + 9a + 14 - 6{a^2} + 13a - 5}}{{2{a^2} + 3a - 2}}\\ = \frac{{20a + 1}}{{2{a^2} + 3a - 2}}\end{array}$