Giải bài 1 trang 47 SBT Toán 10 Chân trời sáng tạo tập 2 – CTST

Phương pháp giải

Khai triển \({\left( {a + b} \right)^5} = C_5^0{a^5} + C_5^1{a^4}{b^1} + C_5^2{a^3}{b^2} + C_5^3{a^2}{b^3} + C_5^4{b^1}{a^4} + C_5^5{a^5}\)

Khai triển \({\left( {a + b} \right)^4} = C_4^0{a^4} + C_4^1{a^3}{b^1} + C_4^2{a^2}{b^2} + C_4^3{a^1}{b^3} + C_4^4{b^4}\)

Lời giải chi tiết

a) \({\left( {x + 3y} \right)^4} = C_4^0{x^4}{\left( {3y} \right)^0} + C_4^1{x^3}{\left( {3y} \right)^1} + C_4^2{x^2}{\left( {3y} \right)^2} + C_4^3{x^1}{\left( {3y} \right)^3} + C_4^4{x^0}{\left( {3y} \right)^4}\)

\({x^4} + 12{x^3}y + 54{x^2}{y^2} + 108{x^1}{y^3} + 81{y^4}\)

b) \(\begin{array}{l}{\left( {3 – 2x} \right)^5} = C_5^0{3^5}{\left( { – 2x} \right)^0} + C_5^1{3^4}{\left( { – 2x} \right)^1} + C_5^2{3^3}{\left( { – 2x} \right)^2} + C_5^3{3^2}{\left( { – 2x} \right)^3} + C_5^4{3^1}{\left( { – 2x} \right)^4} + C_5^5{3^0}{\left( { – 2x} \right)^5}\\ = 243 – 810{x^1} + 1080{x^2} – 720{x^3} + 240{x^4} – 32{x^5}\end{array}\)

c) \(\begin{array}{l}{\left( {x – \frac{2}{x}} \right)^5} = C_5^0{x^5}{\left( { – \frac{2}{x}} \right)^0} + C_5^1{x^4}{\left( { – \frac{2}{x}} \right)^1} + C_5^2{x^3}{\left( { – \frac{2}{x}} \right)^2} + C_5^3{x^2}{\left( { – \frac{2}{x}} \right)^3} + C_5^4{x^1}{\left( { – \frac{2}{x}} \right)^4} + C_5^5{x^0}{\left( { – \frac{2}{x}} \right)^5}\\ = {x^5} – 10{x^3} + 40x – \frac{{80}}{x} + \frac{{80}}{{{x^3}}} – \frac{{32}}{{{x^5}}}\end{array}\)

d)

\(\begin{array}{l}{\left( {3\sqrt x  – \frac{1}{{\sqrt x }}} \right)^4} = C_4^0{\left( {3\sqrt x } \right)^4}{\left( { – \frac{1}{{\sqrt x }}} \right)^0} + C_4^1{\left( {3\sqrt x } \right)^3}{\left( { – \frac{1}{{\sqrt x }}} \right)^1} + C_4^2{\left( {3\sqrt x } \right)^2}{\left( { – \frac{1}{{\sqrt x }}} \right)^2}\\ + C_4^3{\left( {3\sqrt x } \right)^1}{\left( { – \frac{1}{{\sqrt x }}} \right)^3} + C_4^4{\left( {3\sqrt x } \right)^0}{\left( { – \frac{1}{{\sqrt x }}} \right)^4}\\ = 81{x^2} – 108x + \frac{2}{3} – \frac{{12}}{x} + \frac{1}{{{x^2}}}\end{array}\)

— *****