Cho x + y = 10; xy = 4. Tính a) x^2 + y^2 b) x^3 + y^3 c) x^4 + y^4 d) x^5 + y^5 02/10/2021 Bởi Allison Cho x + y = 10; xy = 4. Tính a) x^2 + y^2 b) x^3 + y^3 c) x^4 + y^4 d) x^5 + y^5
a) ta có x+y= 10 => (x+y) ²=100 => x ²+y ² +2xy=100 => x ²+y ²=100-8=92 b)ta cos x+y=10 => (x+y) ³ =1000 => x ³+y ³ + 3xy(x+y) =1000 => x ³+y ³ = 1000-120 = 880 Bình luận
Đáp án: Giải thích các bước giải: ta có: \(\begin{array}{l}a)\,\,{x^2} + {y^2} = {x^2} + {y^2} + 2xy – 2xy\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\left( {x + y} \right)^2} – 2xy\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {10^2} – 2.4\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 92\end{array}\) \(\begin{array}{l}b)\,\,{x^3} + {y^3} = \left( {x + y} \right)\left( {{x^2} – xy + {y^2}} \right)\\ = \left( {x + y} \right)\left[ {{{\left( {x + y} \right)}^2} – 3xy} \right]\\ = 10.\left( {{{10}^2} – 3.4} \right)\\ = 10.88\\ = 880\end{array}\) \(\begin{array}{l}c)\,{x^4} + {y^4} = {\left( {{x^2}} \right)^2} + 2{x^2}{y^2} + {\left( {{y^2}} \right)^2} – 2{x^2}{y^2}\\ = {\left( {{x^2} + {y^2}} \right)^2} – 2xy.xy\\ = {92^2} – 2.4.4\\ = 8432\end{array}\) Bình luận
a) ta có x+y= 10
=> (x+y) ²=100
=> x ²+y ² +2xy=100
=> x ²+y ²=100-8=92
b)ta cos x+y=10
=> (x+y) ³ =1000
=> x ³+y ³ + 3xy(x+y) =1000
=> x ³+y ³ = 1000-120 = 880
Đáp án:
Giải thích các bước giải:
ta có:
\(\begin{array}{l}a)\,\,{x^2} + {y^2} = {x^2} + {y^2} + 2xy – 2xy\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\left( {x + y} \right)^2} – 2xy\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {10^2} – 2.4\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 92\end{array}\)
\(\begin{array}{l}b)\,\,{x^3} + {y^3} = \left( {x + y} \right)\left( {{x^2} – xy + {y^2}} \right)\\ = \left( {x + y} \right)\left[ {{{\left( {x + y} \right)}^2} – 3xy} \right]\\ = 10.\left( {{{10}^2} – 3.4} \right)\\ = 10.88\\ = 880\end{array}\)
\(\begin{array}{l}c)\,{x^4} + {y^4} = {\left( {{x^2}} \right)^2} + 2{x^2}{y^2} + {\left( {{y^2}} \right)^2} – 2{x^2}{y^2}\\ = {\left( {{x^2} + {y^2}} \right)^2} – 2xy.xy\\ = {92^2} – 2.4.4\\ = 8432\end{array}\)