\(A=\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+...+\dfrac{1}{2019^2}>\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+...+\dfrac{1}{2019\cdot2020}=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{2019}-\dfrac{1}{2020}=\dfrac{1}{5}-\dfrac{1}{2020}=\dfrac{404-1}{2020}=\dfrac{403}{2020}>\dfrac{40}{2020}=\dfrac{20}{101}\left(1\right)\) \(A=\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+...+\dfrac{1}{2019^2}< \dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+...+\dfrac{1}{2018\cdot2019}=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{2018}-\dfrac{1}{2019}=\dfrac{1}{4}-\dfrac{1}{2019}=\dfrac{2019-4}{4\cdot2019}=\dfrac{2015}{4\cdot2019}< \dfrac{2019}{4\cdot2019}=\dfrac{1}{4}\left(2\right)\) Từ (1) và (2) \(\Rightarrow\dfrac{20}{101}< A< \dfrac{1}{4}\)

281:

Ta có:\(ab+bc+ca=3abc\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=3\)

\(\dfrac{1}{\sqrt{a^3+b}}\le\dfrac{1}{\sqrt{2\sqrt{a^3b}}}=\dfrac{1}{\sqrt{2a}\cdot\sqrt[4]{ab}}\le\dfrac{1}{2\sqrt{2a}}\cdot\left(\dfrac{1}{\sqrt{a}}+\dfrac{1}{\sqrt{b}}\right)=\dfrac{1}{2\sqrt{2}}\left(\dfrac{1}{a}+\dfrac{1}{\sqrt{ab}}\right)\le\dfrac{1}{2\sqrt{2}}\cdot\left[\dfrac{1}{a}+\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\right]=\dfrac{1}{2\sqrt{2}}\cdot\left(\dfrac{1}{a}+\dfrac{1}{2a}+\dfrac{1}{2b}\right)\) Chứng minh tương tự:

\(\dfrac{1}{\sqrt{b^3+c}}\le\dfrac{1}{2\sqrt{2}}\cdot\left(\dfrac{1}{b}+\dfrac{1}{2b}+\dfrac{1}{2c}\right);\dfrac{1}{\sqrt{c^3+a}}\le\dfrac{1}{2\sqrt{2}}\cdot\left(\dfrac{1}{c}+\dfrac{1}{2c}+\dfrac{1}{2a}\right)\)\(\Rightarrow\dfrac{1}{\sqrt{a^3+b}}+\dfrac{1}{\sqrt{b^3+c}}+\dfrac{1}{\sqrt{c^3+a}}\le\dfrac{1}{2\sqrt{2}}\left(\dfrac{1}{a}+\dfrac{1}{2a}+\dfrac{1}{2b}+\dfrac{1}{b}+\dfrac{1}{2b}+\dfrac{1}{2c}+\dfrac{1}{c}+\dfrac{1}{2c}+\dfrac{1}{2a}\right)=\dfrac{1}{2\sqrt{2}}\left(\dfrac{2}{a}+\dfrac{2}{b}+\dfrac{2}{c}\right)=\dfrac{3}{\sqrt{2}}\) Dấu = xảy ra \(\Leftrightarrow a=b=c=1\)

290

Ta có \(\dfrac{a^4b}{a^2+1}=a^2b-\dfrac{a^2b}{a^2+1}\ge a^2b-\dfrac{a^2b}{2a}=a^2b-\dfrac{ab}{2}\)

Chứng minh tương tự ta được:  

\(\dfrac{b^4c}{b^2+1}\ge b^2c-\dfrac{bc}{2};\dfrac{c^4a}{c^2+1}\ge c^2a-\dfrac{ca}{2}\)

\(\Rightarrow\dfrac{a^4b}{a^2+1}+\dfrac{b^4c}{b^2+1}+\dfrac{c^4a}{c^2+1}\ge a^2b+b^2c+c^2a-\dfrac{ab}{2}-\dfrac{bc}{2}-\dfrac{ca}{2}\)

Áp dụng bđt Cô-si:

\(a^2b+a^2b+b^2c\ge3\sqrt[3]{a^2b\cdot a^2b\cdot b^2c}=3\sqrt[3]{a^3b^3\cdot abc}=3ab\)

Tương tự: \(b^2c+b^2c+c^2a\ge3bc;c^2a+c^2a+a^2b\ge3ca\)

\(\Rightarrow a^2b+a^2b+b^2c+b^2c+b^2c+c^2a+c^2a+c^2a+a^2b\ge3ab+3bc+3ca\Rightarrow3\left(a^2b+b^2c+c^2a\right)\ge3\left(ab+bc+ca\right)\Rightarrow a^2b+b^2c+c^2a\ge ab+bc+ca\)

\(\Rightarrow\dfrac{a^4b}{a^2+1}+\dfrac{b^4c}{b^2+1}+\dfrac{c^4a}{c^2+1}\ge a^2b+b^2c+c^2a-\dfrac{1}{2}\left(ab+bc+ca\right)\ge ab+bc+ca-\dfrac{1}{2}\left(ab+bc+ca\right)=\dfrac{1}{2}\left(ab+bc+ca\right)\ge\dfrac{3}{2}\sqrt[3]{\left(abc\right)^2}=\dfrac{3}{2}\) Dấu = xảy ra \(\Leftrightarrow a=b=c=1\)

a \(\Rightarrow\left|x\right|=\dfrac{19}{20}\Rightarrow\left[{}\begin{matrix}x=\dfrac{19}{20}\\x=-\dfrac{19}{20}\end{matrix}\right.\)

b \(\Rightarrow\left|x-5\right|=-\dfrac{17}{12}\) vô lí vì\(VT=\left|x-5\right|\ge0\) mà \(VP=-\dfrac{17}{20}< 0\) 

\(\Rightarrow\) ko có x

GPT \(\sqrt{9-\dfrac{9}{x}}=x-\sqrt{x-\dfrac{9}{x}}ĐKXĐ:x\ne0;x\ge1\)

\(\Rightarrow9-\dfrac{9}{x}=x^2+x-\dfrac{9}{x}-2x\sqrt{x-\dfrac{9}{x}}\Leftrightarrow x^2+x-9-2x\sqrt{x-\dfrac{9}{x}}=0\)

\(\Rightarrow x-\dfrac{9}{x}+1-2\sqrt{x-\dfrac{9}{x}}=0\Leftrightarrow\left(\sqrt{x-\dfrac{9}{x}}-1\right)^2=0\Leftrightarrow\sqrt{x-\dfrac{9}{x}}=1\)

\(\Leftrightarrow x-\dfrac{9}{x}=1\Rightarrow x^2-9=x\Leftrightarrow x^2-x+\dfrac{1}{4}-\dfrac{37}{4}=0\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2=\dfrac{37}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1+\sqrt{37}}{2}\left(TM\right)\\x=\dfrac{1-\sqrt{37}}{2}\left(L\right)\end{matrix}\right.\) Vậy...

Câu 1 

Để pt vô nghiệm \(\Rightarrow\Delta'=\left(m+2\right)^2-\left(3m+2\right)m=m^2+4m+4-3m^2-2m=-2m^2+2m+4=-2\left(m^2-m-2\right)=-2\left(m+1\right)\left(m-2\right)< 0\) \(\Leftrightarrow\left(m+1\right)\left(m-2\right)>0\Leftrightarrow\left[{}\begin{matrix}m< -1\\m>2\end{matrix}\right.\)

Áp dụng bđt Cô-si vào 3 số a,b,c:

\(\Rightarrow a+b+c\ge3\sqrt[3]{abc}\)

\(\Leftrightarrow4x\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

\(\dfrac{n}{n-2}=\dfrac{n-2+2}{n-2}=1+\dfrac{2}{n-2}\)

Để \(\dfrac{n}{n-2}\in Z\Leftrightarrow\dfrac{2}{n-2}\in Z\Leftrightarrow2⋮n-2\Rightarrow n-2\in\left\{-2;-1;1;2\right\}\Rightarrow n\in\left\{0;1;3;4\right\}\)