Áp dụng BĐT Bunhiacôpxki:

\(123^2=\left(m\sqrt{123-n^2}+n\sqrt{123-m^2}\right)^2\)

\(\Rightarrow123^2\le\left(m^2+n^2\right)\left(123-n^2+123-m^2\right)\)

\(\Leftrightarrow123^2\le\left(m^2+n^2\right)\left(2.123-m^2-n^2\right)\)

Đặt \(m^2+n^2=x\)

\(\Rightarrow123^2\le x\left(2.123-x\right)\)

\(\Leftrightarrow x^2-2.x.123+123^2\le0\)

\(\Leftrightarrow\left(x-123\right)^2\le0\)

\(\Leftrightarrow x-123=0\Rightarrow x=123\)

Lời giải:

Áp dụng BĐT Bunhiacopxky và Cauchy ngược dấu ta có:

\((m\sqrt{123-n^2}+n\sqrt{123-m^2})^2\leq (m^2+n^2)(123-n^2+123-m^2)\leq \left(\frac{m^2+n^2+123-n^2+123-m^2}{2}\right)^2\)

\(\Leftrightarrow (m\sqrt{123-n^2}+n\sqrt{123-m^2})^2\leq 123^2\)

\(\Rightarrow m\sqrt{123-n^2}+n\sqrt{123-m^2}\leq 123\)

Dấu "=" xảy ra khi \(\left\{\begin{matrix} \frac{m}{\sqrt{123-n^2}}=\frac{n}{\sqrt{123-m^2}}\\ m^2+n^2=123-n^2+123-m^2(1)\end{matrix}\right.\)

Từ (1) \(\Rightarrow m^2+n^2=123\)

\(x^2+y^2=\left(x+y\right)^2-2xy=m^2-2n\\ x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=m^3-3mn\\ \Rightarrow x^5+y^5=\left(x^3+y^3\right)\left(x^2+y^2\right)-x^2y^2\left(x+y\right)=\left(m^3-3mn\right)\left(m^2-2n\right)-n^2m\\ \Rightarrow x^7+y^7=\left(x^2+y^2\right)\left(x^5+y^5\right)-x^2y^2\left(x^3+y^3\right)=.....\)

tick mik nha

Ta có \(\frac{2.3.4+2.3.4.2018+2.3.4.2019-2.3.4.2020}{5.6.7+5.6.7.2018+5.6.7.2019-5.6.7.2020}\)

\(=\frac{2.3.4\left(1+2018+2019-2020\right)}{5.6.7.\left(1+2018+2019-2020\right)}=\frac{2.3.4}{5.6.7}=\frac{4}{35}\)