98)\(\)\(a^n=1\Rightarrow n=0\)

Hoặc với \(a=1\) thì với mọi \(n\)

99)\(3^2+4^2=9+16=25=5^2\) là số chính phương

\(5^2+12^2=25+144=169=13^2\)là số chính phương

\(\)Bạn viết thiếu 1 vế đúng không?

\(\dfrac{x-69}{30}+\dfrac{x-67}{32}+\dfrac{x-65}{34}=\dfrac{x-63}{36}+\dfrac{x-61}{38}+\dfrac{x-59}{40}\)\(\Rightarrow\left(\dfrac{x-69}{30}-1\right)+\left(\dfrac{x-67}{32}-1\right)+\left(\dfrac{x-65}{34}-1\right)=\left(\dfrac{x-63}{36}-1\right)+\left(\dfrac{x-61}{38}-1\right)+\left(\dfrac{x-59}{40}-1\right)\)

\(\Rightarrow\dfrac{x-99}{30}+\dfrac{x-99}{32}+\dfrac{x-99}{34}=\dfrac{x-99}{36}+\dfrac{x-99}{38}+\dfrac{x-99}{40}\)

\(\Rightarrow\dfrac{x-99}{30}+\dfrac{x-99}{32}+\dfrac{x-99}{34}-\dfrac{x-99}{36}-\dfrac{x-99}{38}-\dfrac{x-99}{40}=0\)\(\Rightarrow\left(x-99\right)\left(\dfrac{1}{30}+\dfrac{1}{32}+\dfrac{1}{34}-\dfrac{1}{36}-\dfrac{1}{38}-\dfrac{1}{40}\right)=0\)

Vì \(\dfrac{1}{30}+\dfrac{1}{32}+\dfrac{1}{34}-\dfrac{1}{36}-\dfrac{1}{38}-\dfrac{1}{40}\ne0\)

Nên:

\(x-99=0\Rightarrow x=99\)

1)

a)\(2=\sqrt{4}< \sqrt{5}\)

b) \(5=\sqrt{25}>\sqrt{23}\)

c) \(\sqrt{83}>\sqrt{81}=9\)

\(\left\{{}\begin{matrix}\sqrt{23}< \sqrt{25}=5\\\sqrt{13}< \sqrt{16}=4\end{matrix}\right.\)

\(\sqrt{23}+\sqrt{13}< 4+5=9\)

Vậy \(\sqrt{23}+\sqrt{13}< \sqrt{83}\)

2) Ta có:

\(\sqrt{x}=5\Rightarrow x=25\)

\(3\sqrt{x}=6\Rightarrow\sqrt{x}=2\Rightarrow x=4\)

\(3-\sqrt{3+1}=1\)

Nên:

\(3-2=1\)(luôn đúng)

\(A=\dfrac{2^{10}.3^{31}+2^{40}.3^6}{2^{11}.3^{31}+2^{41}.3^6}\)

\(A=\dfrac{2^{10}.3^6.3^{25}+2^{10}.2^{30}.3^6}{2^{11}.3^{25}.3^6+2^{11}.2^{30}.3^6}\)

\(A=\dfrac{2^{10}.3^6\left(3^{25}+2^{30}\right)}{2^{10}.3^6\left(3^{25}+2^{30}\right)}=1\)

\(\)\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{n^2}\)

\(A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{\left(n-1\right)n}\)

\(A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\)

\(A< 1-\dfrac{1}{n}< 1\)

\(B=\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{2n^2}\)

\(B=\dfrac{1}{2}\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2n^2}\right)\)

\(B=\dfrac{1}{4}+\dfrac{1}{2}\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2n^2}\right)\)

\(B< \dfrac{1}{4}+\dfrac{1}{2}\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+.....+\dfrac{1}{\left(n-1\right)n}\right)\)

\(\)\(\left|x-1,5\right|+\left|2,5-x\right|=0\)

Với mọi \(x\in R\) thì:

\(\left\{{}\begin{matrix}\left|x-1,5\right|\ge0\\\left|2,5-x\right|\ge0\end{matrix}\right.\) \(\Rightarrow\left|x-1,5\right|+\left|2,5-x\right|\ge0\)

Dấu "=" xảy ra khi:

\(\left\{{}\begin{matrix} \left|x-1,5\right|=0\\ \left|2,5-x\right|=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1,5\\x=2,5\end{matrix}\right.\)

Khi đó không tồn tại giá trị x

\(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{6}\)

\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\sqrt{\dfrac{1}{6}}\\x+\dfrac{1}{2}=-\sqrt{\dfrac{1}{6}}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}+\sqrt{\dfrac{1}{6}}\\x=\dfrac{1}{2}-\sqrt{\dfrac{1}{6}}\end{matrix}\right.\)

\(A=1\dfrac{1}{15}.1\dfrac{1}{16}.1\dfrac{1}{17}......1\dfrac{1}{2016}.1\dfrac{1}{2017}\)

\(A=\dfrac{16}{15}.\dfrac{17}{16}.\dfrac{18}{17}......\dfrac{2017}{2016}.\dfrac{2018}{2017}\)

\(A=\dfrac{16.17.18......2017.2018}{15.16.17......2016.2017}\)

\(A=\dfrac{2018}{15}\)

\(\left(x+2\right)\left(x+5\right)< 0\)

Với mọi \(x\in R\) thì

\(x+2< x+5\)

\(\left\{{}\begin{matrix}x+2< 0\Rightarrow x< -2\\x+5>0\Rightarrow x>-5\end{matrix}\right.\)
Vậy \(-5< x< -2\)

\(\left(2x+5\right)\left(2x-1\right)< 0\)

Với mọi \(x\in R\) thì:

\(2x-1< 2x+5\)

Nên:

\(\left\{{}\begin{matrix}2x-1< 0\Rightarrow x< \dfrac{1}{2}\\2x+5>0\Rightarrow x>-\dfrac{5}{2}\end{matrix}\right.\)

Vậy \(-\dfrac{5}{2}< x< \dfrac{1}{2}\)

\(A=\left(\dfrac{1}{2^2}-1\right)\left(\dfrac{1}{3^2}-1\right)...\left(\dfrac{1}{10^2}-1\right)\)

\(A=\left(\dfrac{1}{4}-1\right)\left(\dfrac{1}{9}-1\right).....\left(\dfrac{1}{100}-1\right)\)

\(A=\left(\dfrac{1}{4}-\dfrac{4}{4}\right)\left(\dfrac{1}{9}-\dfrac{9}{9}\right)....\left(\dfrac{1}{100}-\dfrac{100}{100}\right)\)

\(A=\dfrac{-3}{4}.\dfrac{-8}{9}.....\dfrac{-99}{100}\)

\(A=\dfrac{\left(-1\right).3}{4}.\dfrac{\left(-1\right).8}{9}......\dfrac{\left(-1\right).99}{100}\)

\(A=\dfrac{\left(-1\right).1.3}{2.2}.\dfrac{-1.2.4}{3.3}....\dfrac{-1.9.11}{10.10}\)

\(A=\dfrac{-1.3}{2.2}.\dfrac{-2.4}{3.3}....\dfrac{-9.11}{10.10}\)

\(A=\dfrac{\left(-1\right)\left(-2\right)....\left(-9\right)}{2.3.....10}.\dfrac{3.4....11}{2.3.....10}\)

\(A=\dfrac{-1}{10}.\dfrac{11}{2}=-\dfrac{11}{20}\)

\(A=\left(\sqrt{64}+2\sqrt{\left(-3\right)^2}-7\sqrt{1,69}+3\sqrt{\dfrac{25}{16}}\right):\left(5\sqrt{\dfrac{2}{3}}\right)^2\)

\(A=\left(\sqrt{64}+2\sqrt{9}-7\sqrt{\dfrac{169}{100}}+3\sqrt{\dfrac{25}{16}}\right):\left(5\sqrt{\dfrac{2}{3}}\right)^2\)

\(A=\left(8+2.3-7.\dfrac{13}{10}+3.\dfrac{5}{4}\right):\left(25.\dfrac{2}{3}\right)\)

\(A=\left(8+6-\dfrac{91}{10}+\dfrac{15}{4}\right):\left(\dfrac{50}{3}\right)\)

\(A=\dfrac{519}{1000}\)

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