Giải bài 2.31 trang 51 SGK Toán 8 – Cùng khám phá

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)   

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