Trang Chủ Sách bài tập lớp 8 SBT Toán 8

Bài 17, 18, 19 trang 28, 29 SBT Toán lớp 8 tập 1: Cộng các phân thức khác mẫu thức

CHIA SẺ
Bài 5 Phép cộng các phân thức đại số SBT Toán lớp 8. Giải bài 17, 18, 19 trang 28, 29 Sách bài tập Toán 8 tập 1. Câu 17: Cộng các phân thức cùng mẫu thức…

Câu 17: Cộng các phân thức cùng mẫu thức

a. \({{1 – 2x} \over {6{x^3}y}} + {{3 + 2y} \over {6{x^3}y}} + {{2x – 4} \over {6{x^3}y}}\)

b. \({{{x^2} – 2} \over {x{{\left( {x – 1} \right)}^2}}} + {{2 – x} \over {x{{\left( {x – 1} \right)}^2}}}\)

c. \({{3x + 1} \over {{x^2} – 3x + 1}} + {{{x^2} – 6x} \over {{x^2} – 3x + 1}}\)

d. \({{{x^2} + 38x + 4} \over {2{x^2} + 17x + 1}} + {{3{x^2} – 4x – 2} \over {2{x^2} + 17x + 1}}\)

a. \({{1 – 2x} \over {6{x^3}y}} + {{3 + 2y} \over {6{x^3}y}} + {{2x – 4} \over {6{x^3}y}}\) \( = {{1 – 2x + 3 + 2y + 2x – 4} \over {6{x^3}y}} = {{2y} \over {6{x^3}y}} = {1 \over {3{x^3}}}\)

b. \({{{x^2} – 2} \over {x{{\left( {x – 1} \right)}^2}}} + {{2 – x} \over {x{{\left( {x – 1} \right)}^2}}}\) \( = {{{x^2} – 2 + 2 – x} \over {x{{\left( {x – 1} \right)}^2}}} = {{x\left( {x – 1} \right)} \over {x{{\left( {x – 1} \right)}^2}}} = {1 \over {x – 1}}\)

c. \({{3x + 1} \over {{x^2} – 3x + 1}} + {{{x^2} – 6x} \over {{x^2} – 3x + 1}}\) \( = {{3x + 1 + {x^2} – 6x} \over {{x^2} – 3x + 1}} = {{{x^2} – 3x + 1} \over {{x^2} – 3x + 1}} = 1\)

d. \({{{x^2} + 38x + 4} \over {2{x^2} + 17x + 1}} + {{3{x^2} – 4x – 2} \over {2{x^2} + 17x + 1}}\) \( = {{{x^2} + 38x + 4 + 3{x^2} – 4x – 2} \over {2{x^2} + 17x + 1}} = {{4{x^2} + 34x + 2} \over {2{x^2} + 17x + 1}} = {{2\left( {2{x^2} + 17x + 1} \right)} \over {2{x^2} + 17x + 1}} = 2\)


Câu 18: Cộng các phân thức khác mẫu thức:

a. \({5 \over {6{x^2}y}} + {7 \over {12x{y^2}}} + {{11} \over {18xy}}\)

b. \({{4x + 2} \over {15{x^3}y}} + {{5y – 3} \over {9{x^2}y}} + {{x + 1} \over {5x{y^3}}}\)

c. \({3 \over {2x}} + {{3x – 3} \over {2x – 1}} + {{2{x^2} + 1} \over {4{x^2} – 2x}}\)

d. \({{{x^3} + 2x} \over {{x^3} + 1}} + {{2x} \over {{x^2} – x + 1}} + {1 \over {x + 1}}\)

a. \({5 \over {6{x^2}y}} + {7 \over {12x{y^2}}} + {{11} \over {18xy}}\)\( = {{30y} \over {36{x^2}{y^2}}} + {{21x} \over {36{x^2}{y^2}}} + {{22xy} \over {36{x^2}{y^2}}} = {{30y + 21x + 22xy} \over {36{x^2}{y^2}}}\)

b. \({{4x + 2} \over {15{x^3}y}} + {{5y – 3} \over {9{x^2}y}} + {{x + 1} \over {5x{y^3}}}\)\(\eqalign{  &  = {{3{y^2}\left( {4x + 2} \right)} \over {45{x^3}{y^3}}} + {{5x{y^2}\left( {5y – 3} \right)} \over {45{x^3}{y^3}}} + {{9{x^2}\left( {x + 1} \right)} \over {45{x^3}{y^3}}}  \cr  &  = {{12x{y^2} + 6{y^2} + 25x{y^3} – 15x{y^2} + 9{x^3} + 9{x^2}} \over {45{x^3}{y^3}}} = {{6{y^2} + 25x{y^3} – 3x{y^2} + 9{x^3} + 9{x^2}} \over {45{x^3}{y^3}}} \cr} \)

c. \({3 \over {2x}} + {{3x – 3} \over {2x – 1}} + {{2{x^2} + 1} \over {4{x^2} – 2x}}\)\( = {3 \over {2x}} + {{3x – 3} \over {2x – 1}} + {{2{x^2} + 1} \over {2x\left( {2x – 1} \right)}}\)

\(\eqalign{  &  = {{3\left( {2x – 1} \right)} \over {2x\left( {2x – 1} \right)}} + {{2x\left( {3x – 3} \right)} \over {2x\left( {2x – 1} \right)}} + {{2{x^2} + 1} \over {2x\left( {2x – 1} \right)}} = {{6x – 3 + 6{x^2} – 6x + 2{x^2} + 1} \over {2x\left( {2x – 1} \right)}}  \cr  &  = {{8{x^2} – 2} \over {2x\left( {2x – 1} \right)}} = {{2\left( {4{x^2} – 1} \right)} \over {2x\left( {2x – 1} \right)}} = {{\left( {2x + 1} \right)\left( {2x – 1} \right)} \over {x\left( {2x – 1} \right)}} = {{2x + 1} \over x} \cr} \)

d. \({{{x^3} + 2x} \over {{x^3} + 1}} + {{2x} \over {{x^2} – x + 1}} + {1 \over {x + 1}}\)\( = {{{x^3} + 2x} \over {\left( {x + 1} \right)\left( {{x^2} – x + 1} \right)}} + {{2x} \over {{x^2} – x + 1}} + {1 \over {x + 1}}\)

\(\eqalign{  &  = {{{x^3} + 2x} \over {\left( {x + 1} \right)\left( {{x^2} – x + 1} \right)}} + {{2x\left( {x + 1} \right)} \over {\left( {x + 1} \right)\left( {{x^2} – x + 1} \right)}} + {{{x^2} – x + 1} \over {\left( {x + 1} \right)\left( {{x^2} – x + 1} \right)}}  \cr  &  = {{{x^3} + 2x + 2{x^2} + 2x + {x^2} – x + 1} \over {\left( {x + 1} \right)\left( {{x^2} – x + 1} \right)}} = {{{x^3} + 3{x^2} + 3x + 1} \over {\left( {x + 1} \right)\left( {{x^2} – x + 1} \right)}} = {{{{\left( {x + 1} \right)}^3}} \over {\left( {x + 1} \right)\left( {{x^2} – x + 1} \right)}}  \cr  &  = {{{{\left( {x + 1} \right)}^2}} \over {{x^2} – x + 1}} \cr} \)


Câu 19: Dùng quy tắc đổi dấu để tìm mẫu thức chung rồi thực hiện phép cộng:

a. \({4 \over {x + 2}} + {2 \over {x – 2}} + {{5x – 6} \over {4 – {x^2}}}\)

b. \({{1 – 3x} \over {2x}} + {{3x – 2} \over {2x – 1}} + {{3x – 2} \over {2x – 4{x^2}}}\)

c. \({1 \over {{x^2} + 6x + 9}} + {1 \over {6x – {x^2} – 9}} + {x \over {{x^2} – 9}}\)

d. \({{{x^2} + 2} \over {{x^3} – 1}} + {2 \over {{x^2} + x + 1}} + {1 \over {1 – x}}\)

e. \({x \over {x – 2y}} + {x \over {x + 2y}} + {{4xy} \over {4{y^2} – {x^2}}}\)

a. \({4 \over {x + 2}} + {2 \over {x – 2}} + {{5x – 6} \over {4 – {x^2}}}\) \( = {4 \over {x + 2}} + {2 \over {x – 2}} + {{6 – 5x} \over {\left( {x + 2} \right)\left( {x – 2} \right)}}\)

\(\eqalign{  &  = {{4\left( {x – 2} \right)} \over {\left( {x + 2} \right)\left( {x – 2} \right)}} + {{2\left( {x + 2} \right)} \over {\left( {x + 2} \right)\left( {x – 2} \right)}} + {{6 – 5x} \over {\left( {x + 2} \right)\left( {x – 2} \right)}} = {{4x – 8 + 2x + 4 + 6 – 5x} \over {\left( {x + 2} \right)\left( {x – 2} \right)}}  \cr  &  = {{x + 2} \over {\left( {x + 2} \right)\left( {x – 2} \right)}} = {1 \over {x – 2}} \cr} \)

b. \({{1 – 3x} \over {2x}} + {{3x – 2} \over {2x – 1}} + {{3x – 2} \over {2x – 4{x^2}}}\) \( = {{1 – 3x} \over {2x}} + {{3x – 2} \over {2x – 1}} + {{2 – 3x} \over {2x\left( {2x – 1} \right)}}\)

\(\eqalign{  &  = {{\left( {1 – 3x} \right)\left( {2x – 1} \right)} \over {2x\left( {2x – 1} \right)}} + {{\left( {3x – 2} \right).2x} \over {2x\left( {2x – 1} \right)}} + {{2 – 3x} \over {2x\left( {2x – 1} \right)}}  \cr  &  = {{2x – 1 – 6{x^2} + 3x + 6{x^2} – 4x + 2 – 3x} \over {2x\left( {2x – 1} \right)}} = {{1 – 2x} \over {2x\left( {2x – 1} \right)}} = {{ – \left( {2x – 1} \right)} \over {2x\left( {2x – 1} \right)}} = {{ – 1} \over {2x}} \cr} \)

c. \({1 \over {{x^2} + 6x + 9}} + {1 \over {6x – {x^2} – 9}} + {x \over {{x^2} – 9}}\)\( = {1 \over {{{\left( {x + 3} \right)}^2}}} + {{ – 1} \over {{{\left( {x – 3} \right)}^2}}} + {x \over {\left( {x + 3} \right)\left( {x – 3} \right)}}\)

\(\eqalign{  &  = {{{{\left( {x – 3} \right)}^2}} \over {{{\left( {x + 3} \right)}^2}{{\left( {x – 3} \right)}^2}}} + {{ – {{\left( {x + 3} \right)}^2}} \over {{{\left( {x + 3} \right)}^2}{{\left( {x – 3} \right)}^2}}} + {{x\left( {x + 3} \right)\left( {x – 3} \right)} \over {{{\left( {x + 3} \right)}^2}{{\left( {x – 3} \right)}^2}}}  \cr  &  = {{{x^2} – 6x + 9 – {x^2} – 6x – 9 + {x^3} – 9x} \over {{{\left( {x + 3} \right)}^2}{{\left( {x – 3} \right)}^2}}} = {{{x^3} – 21x} \over {{{\left( {x + 3} \right)}^2}{{\left( {x – 3} \right)}^2}}} \cr} \)

d. \({{{x^2} + 2} \over {{x^3} – 1}} + {2 \over {{x^2} + x + 1}} + {1 \over {1 – x}}\)\( = {{{x^2} + 2} \over {\left( {x – 1} \right)\left( {{x^2} + x + 1} \right)}} + {2 \over {{x^2} + x + 1}} + {{ – 1} \over {x – 1}}\)

\(\eqalign{  &  = {{{x^2} + 2} \over {\left( {x – 1} \right)\left( {{x^2} + x + 1} \right)}} + {{2\left( {x – 1} \right)} \over {\left( {x – 1} \right)\left( {{x^2} + x + 1} \right)}} + {{ – \left( {{x^2} + x + 1} \right)} \over {\left( {x – 1} \right)\left( {{x^2} + x + 1} \right)}}  \cr  &  = {{{x^2} + 2 + 2x – 2 – {x^2} – x – 1} \over {\left( {x – 1} \right)\left( {{x^2} + x + 1} \right)}} = {{x – 1} \over {\left( {x – 1} \right)\left( {{x^2} + x + 1} \right)}} = {1 \over {{x^2} + x + 1}} \cr} \)

e. \({x \over {x – 2y}} + {x \over {x + 2y}} + {{4xy} \over {4{y^2} – {x^2}}}\)\( = {x \over {x – 2y}} + {x \over {x + 2y}} + {{ – 4xy} \over {\left( {x + 2y} \right)\left( {x – 2y} \right)}}\)

\(\eqalign{  &  = {{x\left( {x + 2y} \right)} \over {\left( {x – 2y} \right)\left( {x + 2y} \right)}} + {{x\left( {x – 2y} \right)} \over {\left( {x – 2y} \right)\left( {x + 2y} \right)}} + {{ – 4xy} \over {\left( {x – 2y} \right)\left( {x + 2y} \right)}}  \cr  &  = {{{x^2} + 2xy + {x^2} – 2xy – 4xy} \over {\left( {x – 2y} \right)\left( {x + 2y} \right)}} = {{2{x^2} – 4xy} \over {\left( {x – 2y} \right)\left( {x + 2y} \right)}} = {{2x\left( {x – 2y} \right)} \over {\left( {x – 2y} \right)\left( {x + 2y} \right)}}  \cr  &  = {{2x} \over {x + 2y}} \cr} \)