Bài 35 trang 50 SGK Toán 8 tập 1

Chủ đề

Bạn Đang Xem: Bài 35 trang 50 SGK Toán 8 tập 1

Thực hiện phép tính:

Xem Thêm : Theo em, Thạch Sanh thuộc kiểu nhân vật nào

a) \(\dfrac{{x + 1}}{{x – 3}} – \dfrac{{1 – x}}{{x + 3}} – \dfrac{{2x ) \left({1 – x} \right)}}{{9 – {x^2}}}\)

b) \(\dfrac{{3x + 1}}{{{\left( {x – 1} \right)}^2}}} – \dfrac{1}{{ x + 1}} + \dfrac{{x + 3}}{{1 – {x^2}}}\)

Giải thích chi tiết

Xem Thêm : Bài thơ về tiểu đội xe không kính – Ngữ văn 9 Tập 1

\(\eqalign{ & a)\;{{x + 1} \over {x – 3}} – {{1 – x} \over {x + 3}} – { {2x\left( {1 – x} \right)} \over {9 – {x^2}}} (x \ne 3;x \ne -3 )\cr & = { {x + 1} \over {x – 3}} + {{ – \left( {1 – x} \right)} \over {x + 3}} + {{2x\left ( { 1 – x} \right)} \over { – \left( {9 – {x^2}} \right)}} \cr & = {{x + 1} \over {x – 3}} + {{x – 1} \over {x + 3}} + {{2x\left({1 – x} \right)} \over {{x^2} – 9} } \cr & = {{x + 1} \over {x – 3}} + {{x – 1} \over {x + 3}} + {{2x – 2{x^2}} \over {\left({x – 3}\right)\left({x + 3}\right)}} \cr & = {{(x + 1)(x+3) } \vượt {(x – 3)(x+3)}} + {{(x – 1)(x-3)} \vượt {(x + 3)(x-3)}} + {{ 2x – 2{x^2}} \over {\left( {x – 3} \right)\left( {x + 3} \right)}} \cr & = {{ left( {x + 1} \right)\left( {x + 3} \right) + \left( {x – 1} \right)\left( {x – 3} \ phải) + 2x – 2{x^2}} \over {\left( {x – 3} \right)\left( {x + 3} \right)}} \ cr & = {{{x^2} + 3x + x + 3 + {x^2} – 3x – x + 3 + 2x – 2{x^2}} \over {\left( {x – 3} right)\left( {x + 3} \right)}} \cr & = {{2x + 6} \over {\left( {x – 3} \right)\ left( {x + 3} \right)}}\cr& = {{2\left( {x + 3} \right)} \over {\left( {x – 3} \ phải)\left( {x + 3} \right)}} = {2 \over {x – 3}} \cr} \)

\(\eqalign{ & b)\,\,{{3x + 1} \ qua {{{\left( {x – 1} \right)}^2}} } – {1 \over {x + 1}} + {{x + 3} \over {1 – {x^2}}} (x \ne 1; x \ne -1 )\cr & = {{3x + 1} \over {{{\left( {x – 1} \right)}^2}}} + {{ – 1} \over {x + 1}} + {{ – \left( {x + 3} \right)} \over { – \left( {1 – {x^2}} \right)}} \cr & = {{3x + 1} \over {{{\left({x – 1} \right)}^2}}} + {{ – 1} \over {x + 1}} + {{ – \left ( {x + 3} \right)} \over {{x^2} – 1}} \cr & = {{3x + 1} \over {{{\left( {x – 1 } \right)}^2}}} + {{ – 1} \over {x + 1}} + {{ – \left( {x + 3} \right)} \over {\ left( {x – 1} \right)\left( {x + 1} \right)}} \cr & = {{(3x + 1)(x+1)} \over {{ {\left( {x – 1} \right)}^2(x+1)}}} + {{ – (x -1)^2} \over {(x + 1)(x-1 )^2}} + {{ – \left( {x + 3} \right)(x-1)} \ qua {\left( {x – 1} \right)^2\left ( {x + 1} \right)}} \cr & = {{\left( {3x + 1} \right)\left( {x + 1} \right) – {{ left( {x – 1} \right)}^2} – \left( {x + 3} \right)\left( {x – 1} \right)} \over {{{\left( {x – 1} \right)}^2} \left( {x + 1} \right)}} \cr & = {{3{x^2} + 4x + 1 – \left( {{x^2} – 2x + 1} \right) – \left( {{x^2} + 2x – 3} right)} \over {{{\left( {x – 1} \right)}^2}\left( {x + 1} \right)}} \cr & = {{ 3{x^2} + 4x + 1 – {x^2} + 2x – 1 – {x^2} – 2x + 3} \ qua {{{\left( {x – 1} right) }^2}\left( {x + 1} \right)}} \cr & = {{{x^2} + 4x + 3} \over {{{\left( {x – 1} \right)}^2}\left( {x + 1} \right)} } = {{{x^2} + x + 3x + 3} \over {{{\left( {x – 1} \right)}^2}\left( {x + 1} \right)}} \cr & = {{\left( {{x^2} + x} right) + \left( {3x + 3} \right)} \over {{{\left( {x – 1} \right)}^2}\left( {x + 1} \right)}} \cr & = {{x\left( {x + 1} \right) + 3\left( {x + 1} \right)} \over {{{ \left({x – 1} \right)}^2}\left({x + 1} \right)}} \cr & = {{\left({x + 1} right)\left( {x + 3} \right)} \ qua {{{\left( {x – 1} \right)} ^2}\left( {x + 1} right)}} = {{x + 3} \over {{{\left( {x – 1} \right )}^2}}} \cr} \)

loigiaihay.com