Đề kiểm tra 15 phút - Đề số 5 - Bài 7 - Chương 2 - Đại số 8

Bài 1. Thực hiện phép tính : \(\left[ {{1 \over {{{\left( {2x - y} \right)}^2}}} + {2 \over {4{x^2} - {y^2}}} + {1 \over {{{\left( {2x + y} \right)}^2}}}} \right].{{4{x^2} + 4xy + {y^2}} \over {16x}}.\)

Bài 2. Cho \(x + {1 \over x} = a.\) Tính\({x^5} + {1 \over {{x^5}}}\) theo a.

Bài 1.

\(\left[ {{1 \over {{{\left( {2x - y} \right)}^2}}} + {2 \over {4{x^2} - {y^2}}} + {1 \over {{{\left( {2x + y} \right)}^2}}}} \right].{{4{x^2} + 4xy + {y^2}} \over {16x}}\)

\( = \left[ {{1 \over {{{\left( {2x - y} \right)}^2}}} + {2 \over {\left( {2x - y} \right)\left( {2x + y} \right)}} + {1 \over {{{\left( {2x + y} \right)}^2}}}} \right].{{4{x^2} + 4xy + {y^2}} \over {16x}}\)

\( = \left[ {{{{{\left( {2x + y} \right)}^2} + 2\left( {2x - y} \right)\left( {2x + y} \right) + {{\left( {2x - y} \right)}^2}} \over {{{\left( {2x + y} \right)}^2}{{\left( {2x - y} \right)}^2}}}} \right].{{{{\left( {2x + y} \right)}^2}} \over {16x}}\)

\( = {{{{\left( {2x + y + 2x - y} \right)}^2}} \over {{{\left( {2x - y} \right)}^2}{{\left( {2x + y} \right)}^2}}}.{{{{\left( {2x + y} \right)}^2}} \over {16x}} = {{{{\left( {4x} \right)}^2}} \over {16x{{\left( {2x - y} \right)}^2}}}\)

\(= {{16{x^2}} \over {16x{{\left( {2x - y} \right)}^2}}} = {x \over {{{\left( {2x - y} \right)}^2}}}.\)

Bài 2. Ta có :

\(x + {1 \over x} = a \)

\(\Rightarrow {a^2} = {x^2} + 2x.{1 \over x} + {1 \over {{x^2}}} = {x^2} + {1 \over {{x^2}}} + 2\)

\( \Rightarrow {a^2} - 2 = {x^2} + {1 \over {{x^2}}}\)

\({x^3} + {1 \over {{x^3}}} = {\left( {x + {1 \over x}} \right)^3} - 3\left( {x + {1 \over x}} \right)\)\(\; = {a^3} - 3a\)

Do đó :

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

\(= {x^5} + {{{x^2}} \over {{x^3}}} + {{{x^3}} \over {{x^2}}} + {1 \over {{x^5}}} \)

\(= {x^5} + {1 \over x} + x + {1 \over {{x^5}}}\)

 \( = {x^5} + {1 \over {{x^5}}} + x + {1 \over x}\)

\({x^5} + {1 \over {{x^5}}} \)

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

\(= \left( {{a^2} - 2} \right)\left( {{a^3} - 3a} \right) - a\)

\( = {a^5} - 3{a^3} - 2{a^3} + 6a - a = {a^5} - 5{a^3} + 5a.\)