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Câu hỏi của jihoon oppa! I'm May - Toán lớp 8 - Học toán với OnlineMath

zZz Cool Kid zZz cái đó cô chi nhầm nên bị ngược dấu rồi nhé! Mình đang nghĩ cách:(

1) Ta có: \(2020^2=\left(2019+1\right)^2=2019^2+2.2019+1.\)

\(\Rightarrow1+2019^2=2020^2-2.2019\)

\(\Rightarrow M=\sqrt{1+2019^2+\frac{2019^2}{2020^2}}+\frac{2019}{2020}=\sqrt{2020^2-2.2019+\frac{2019^2}{2020^2}}+\frac{2019}{2020}\)

\(=\sqrt{2020^2-2.2020.\frac{2019}{2020}+\left(\frac{2019}{2020}\right)^2}+\frac{2019}{2020}\)

\(=\sqrt{\left(2020-\frac{2019}{2020}\right)^2}+\frac{2019}{2020}=2020-\frac{2019}{2020}+\frac{2019}{2020}\)

\(=2020\)

Vậy M=2020.

2) Xét  : \(k\in N;k\ge2\)ta có:

\(\left(1+\frac{1}{k-1}-\frac{1}{k}\right)^2=1+\frac{1}{\left(k-1\right)^2}+\frac{1}{k^2}+\frac{2}{k-1}-\frac{2}{\left(k-1\right)k}-\frac{2}{k}\)

                                          \(=1+\frac{1}{\left(k-1\right)^2}+\frac{1}{k^2}+\frac{2}{k-1}-\frac{2}{k-1}+\frac{2}{k}-\frac{2}{k}\)

\(\Rightarrow\left(1+\frac{1}{k-1}-\frac{1}{k}\right)^2=1+\frac{1}{\left(k-1\right)^2}+\frac{1}{k^2}\)

\(\Rightarrow\sqrt{1+\frac{1}{\left(k-1\right)^2}+\frac{1}{k^2}}=1+\frac{1}{k-1}+\frac{1}{k}\)

Cho \(k=3,4,...,2020.\)Ta có:

\(N=\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+\sqrt{1+\frac{1}{3^2}+\frac{1}{4^2}}+...+\sqrt{1+\frac{1}{2019^2}+\frac{1}{2020^2}}\)

\(=\left(1+\frac{1}{2}-\frac{1}{3}\right)+\left(1+\frac{1}{3}-\frac{1}{4}\right)+...+\left(1+\frac{1}{2018}-\frac{1}{2019}\right)+\left(1+\frac{1}{2019}-\frac{1}{2020}\right)\)

\(=2018+\frac{1}{2}-\frac{1}{2020}=2018\frac{1009}{2020}\)

Vậy \(N=2018\frac{1009}{2020}.\)

\(P=2-\sqrt{x^2-x}\)

để P max thì \(2-\sqrt{x^2-x}\)max hay \(\sqrt{x^2-x}\)min 

Mà \(\sqrt{x^2-x}\ge0\)

Dấu " = " xảy ra \(\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)

nên P max = 2 \(\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)

\(\left(5\sqrt{3}+3\sqrt{5}\right):15\)

\(=\sqrt{5}.\sqrt{3}\left(\sqrt{5}+\sqrt{3}\right):15\)

\(=\sqrt{15}\left(\sqrt{5}+\sqrt{3}\right):15=\frac{\sqrt{5}+\sqrt{3}}{\sqrt{15}}\)

a) \(\sqrt{75}-\sqrt{5\frac{1}{3}}+\frac{9}{2}\sqrt{2\frac{2}{3}}+2\sqrt{27}\)

\(=\sqrt{75}-\sqrt{\frac{16}{3}}+\frac{9}{2}\sqrt{\frac{8}{3}}+2\sqrt{27}\)

\(=5\sqrt{3}-\frac{4}{\sqrt{3}}+3\sqrt{6}+6\sqrt{3}\)

\(=-\frac{4}{\sqrt{3}}+5\sqrt{3}+3\sqrt{6}+6\sqrt{3}\)

\(=-\frac{4}{\sqrt{3}}+11\sqrt{3}+3\sqrt{6}\)

\(=-\frac{4\sqrt{3}}{3}+11\sqrt{3}+3\sqrt{6}\)

b) \(\sqrt{48}-\sqrt{5\frac{1}{3}}+2\sqrt{75}-5\sqrt{1\frac{1}{3}}\)

\(=\sqrt{48}-\sqrt{\frac{16}{3}}+2\sqrt{75}-5\sqrt{\frac{4}{3}}\)

\(=4\sqrt{3}-\frac{4}{\sqrt{3}}+10\sqrt{3}-\frac{10}{\sqrt{3}}\)

\(=-\frac{4}{\sqrt{3}}-\frac{10}{\sqrt{3}}+4\sqrt{3}+10\sqrt{3}\)

\(=-\frac{14\sqrt{3}}{3}+4\sqrt{3}+10\sqrt{3}\)

\(=-\frac{14\sqrt{3}}{3}+14\sqrt{3}\)

c)\(\left(\sqrt{15}+2\sqrt{3}\right)^2+12\sqrt{5}\)

\(=27+12\sqrt{5}+12\sqrt{5}\)

\(=27+24\sqrt{5}\)

d)\(\left(\sqrt{6}+2\right)\left(\sqrt{3}-\sqrt{2}\right)\)

\(=\sqrt{6}+2-\sqrt{3}-\sqrt{2}\)

e) \(\left(\sqrt{3}+1\right)^2-2\sqrt{3}+4\)

\(=4+2\sqrt{3}-2\sqrt{3}+4\)

= 8

f) \(\frac{1}{7+4\sqrt{3}}+\frac{1}{7-4\sqrt{3}}\)

\(=\frac{7-4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}+\frac{7+4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}\)

\(=\frac{7-4\sqrt{3}+7+4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}\)

\(=\frac{14}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}\)

= 14

a) \(2\sqrt{2}.\left(\sqrt{3}-2\right)+\left(1+2\sqrt{2}\right)^2-2\sqrt{6}=9\)

\(=2\sqrt{2}.\left(\sqrt{3}-2\right)+9+4\sqrt{2}-2\sqrt{6}\)

\(=2\sqrt{6}-4\sqrt{2}+9+4\sqrt{2}-2\sqrt{6}\)

= 9 (đpcm)

b) \(\sqrt{\sqrt{2}+1}-\sqrt{\sqrt{2}-1}=\sqrt{2\left(\sqrt{2}-1\right)}\)

\(=\sqrt{\sqrt{2}+1}-\sqrt{\sqrt{2}-1}=\sqrt{2^{\frac{1}{2}}\left(\sqrt{2}-1\right)}\)

\(=\sqrt{2\left(\sqrt{2}-1\right)}\) (đpcm)

Tìm GTNN của P=a^7+b^7+c^7 biết a^3b^3+b^3c^3+c^3a^3>=1 - Sasu ka