Nội dung bài giảng
Rút gọn biểu thức:
a. \({{{x^4} + 15x + 7} \over {2{x^3} + 2}}.{x \over {14{x^2} + 1}}.{{4{x^3} + 4} \over {{x^4} + 15x + 7}}\)
b. \({{{x^7} + 3{x^2} + 2} \over {{x^3} - 1}}.{{3x} \over {x + 1}}.{{{x^2} + x + 1} \over {{x^7} + 3{x^2} + 2}}\)
Giải:
a. \({{{x^4} + 15x + 7} \over {2{x^3} + 2}}.{x \over {14{x^2} + 1}}.{{4{x^3} + 4} \over {{x^4} + 15x + 7}}\)
\( = {{\left( {{x^4} + 15x + 7} \right).x.\left( {4{x^3} + 4} \right)} \over {\left( {2{x^3} + 2} \right).\left( {14{x^2} + 1} \right).\left( {{x^4} + 15x + 7} \right)}} = {{4x\left( {{x^3} + 1} \right)} \over {2\left( {{x^3} + 1} \right)\left( {14{x^2} + 1} \right)}} = {{2x} \over {14{x^2} + 1}}\)
b. \({{{x^7} + 3{x^2} + 2} \over {{x^3} - 1}}.{{3x} \over {x + 1}}.{{{x^2} + x + 1} \over {{x^7} + 3{x^2} + 2}}\)\( = {{\left( {{x^7} + 3{x^2} + 2} \right).3x.\left( {{x^2} + x + 1} \right)} \over {\left( {{x^3} - 1} \right)\left( {x + 1} \right)\left( {{x^7} + 3{x^2} + 2} \right)}}\)
\( = {{3x\left( {{x^2} + x + 1} \right)} \over {\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)\left( {x + 1} \right)}} = {{3x} \over {\left( {x - 1} \right)\left( {x + 1} \right)}}\)