\(\begin{array}{l}{(3 + \sqrt 2 )^5} - {(3 - \sqrt 2 )^5}\\ = {3^5} + {5.3^4}.\sqrt 2  + {10.3^3}{\left( {\sqrt 2 } \right)^2} + {10.3^2}{\left( {\sqrt 2 } \right)^3} + 5.3{\left( {\sqrt 2 } \right)^4} + {\sqrt 2 ^5}\\ - \left[ {{3^5} - {{5.3}^4}.\sqrt 2  + {{10.3}^3}{{\left( {\sqrt 2 } \right)}^2} - {{10.3}^2}{{\left( {\sqrt 2 } \right)}^3} + 5.3{{\left( {\sqrt 2 } \right)}^4} - {{\sqrt 2 }^5}} \right]\\ = 2\left( {{{5.3}^4}.\sqrt 2  + {{10.3}^2}{{\left( {\sqrt 2 } \right)}^3} + {{\sqrt 2 }^5}} \right)\\ = 810\sqrt 2  + 360\sqrt 2  + 8\sqrt 2 \\ = 1178\sqrt 2 \end{array}\)

a: \(r_6=3^{\text{1 , 414213 }}=4,7288\text{01466}\)

\(r_7=3^{\text{ 1 , 4142134}}=\text{4,728803544}\)

b: Khi \(n\rightarrow+\infty\) thì \(3^{r_n}\rightarrow3^{\sqrt{2}}\)

Tham khảo: Bài 4.8 trang 211 Sách bài tập Đại số và giải tích 11: Chứng minh rằng với |x| rất bé so với

Tham khảo cách giải:

Đặt \(x\left(y\right)=\sqrt{a^2+x}\) ta có:

\(y'\left(x\right)=\dfrac{\left(a^2+x\right)'}{2\sqrt{a^2+x}}=\dfrac{1}{2\sqrt{a^2+x}}\)

Từ đó:

\(\Delta y=y\left(x\right)-y\left(0\right)\approx y'\left(0\right)x\)

\(\Rightarrow\sqrt{a^2+x}-\sqrt{a^2+0}\approx\dfrac{1}{2\sqrt{a^2+0}}x\)

\(\Rightarrow\sqrt{a^2+x}-a\approx\dfrac{x}{2a}\)

\(\Rightarrow\sqrt{a^2+x}\approx a+\dfrac{x}{2a}\)

Áp dụng :

\(\sqrt{146}=\sqrt{12^2+2}\)

\(\approx12+\dfrac{2}{2.12}\approx12,0833\)

\(a,\sqrt{2^3}=2^{\dfrac{3}{2}}\\ b,\sqrt[5]{\dfrac{1}{27}}=\sqrt[5]{3^{-3}}=3^{-\dfrac{3}{5}}\\ c,\left(\sqrt[5]{a}\right)^4=\sqrt[5]{a^4}=a^{\dfrac{4}{5}}\)

a) Làm tròn đến hàng trăm

\(\begin{array}{l}1000\pi  = 3141,5926.... \approx 3100\,\\\, - 100\sqrt 2  =  - 141,4213... \approx  - 100\end{array}\)

b) Làm tròn đến hàng phần nghìn

\(\begin{array}{l} - \sqrt 5  \approx 2,23606... \approx 2,236;\,\,\\\,6,\left( {234} \right) \approx 6,234\end{array}\)

a: \(=3\cdot3^{\dfrac{1}{2}}\cdot3^{\dfrac{1}{.4}}\cdot3^{\dfrac{1}{8}}=3^{1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}}=3^{\dfrac{15}{16}}\)

b: \(=\sqrt{a\cdot\sqrt{a\cdot a^{\dfrac{1}{2}}}}\)

\(=\sqrt{a\cdot\sqrt{a^{\dfrac{3}{2}}}}=\sqrt{a\cdot a^{\dfrac{3}{4}}}=\sqrt{a^{\dfrac{7}{4}}}=a^{\dfrac{7}{4}\cdot\dfrac{1.}{2}}=a^{\dfrac{7}{8}}\)

c: \(=\dfrac{a^{\dfrac{1}{2}}\cdot a^{\dfrac{1}{3}}\cdot a^{\dfrac{1}{4}}}{\left(a^{\dfrac{1}{5}}\right)^3\cdot a^{\dfrac{2}{5}}}=\dfrac{a^{\dfrac{13}{12}}}{a}=a^{\dfrac{1}{12}}\)

\(\sqrt{14+\sqrt{40}+\sqrt{56}+\sqrt{140}}\)

\(=\sqrt{2+5+7+2\sqrt{2.5}+2\sqrt{2.7}+2\sqrt{5.7}}\)

\(=\sqrt{\left(\sqrt{2}+\sqrt{5}+\sqrt{7}\right)^2}=\sqrt{2}+\sqrt{5}+\sqrt{7}\)

\(\Rightarrow a+b+c=2+5+7=14\)

\(a,a^{\dfrac{3}{5}}\cdot a^{\dfrac{1}{2}}:a^{-\dfrac{2}{5}}=a^{\dfrac{3}{5}+\dfrac{1}{2}-\left(-\dfrac{2}{5}\right)}=a^{\dfrac{3}{2}}\\ b,\sqrt{a^{\dfrac{1}{2}}\sqrt{a^{\dfrac{1}{2}}\sqrt{a}}}\\ =\sqrt{a^{\dfrac{1}{2}}\sqrt{a^{\dfrac{1}{2}}\cdot a^{\dfrac{1}{2}}}}\\ =\sqrt{a^{\dfrac{1}{2}}\sqrt{a}}\\ =\sqrt{a^{\dfrac{1}{2}}\cdot a^{\dfrac{1}{2}}}\\ =\sqrt{a}\)