Đặt \(A=x^2+x+1\)

Ta có: \(x^2+x+1\)

\(=x^2+2\cdot x\cdot\dfrac{1}{2}+1\)

\(=\left(x+\dfrac{1}{2}\right)+\dfrac{3}{2}\ge\dfrac{3}{2}\)

Dấu "=" xảy ra khi \(x=-\dfrac{1}{2}\)

Vậy \(MIN_A=\dfrac{3}{2}\) khi \(x=-\dfrac{1}{2}\)

cái này k thể phân tích ra đc, làm ra approximate value thôi

\(\left|x-2,5\right|+\left|3,5-x\right|=0\)

\(\Leftrightarrow\left|x-2,5\right|=-\left|3,5-x\right|\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left|x-2,5\right|=0\\-\left|3,5-x\right|=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=2,5\\x=3,5\end{matrix}\right.\)

Vậy \(x\in\varnothing\)

+) \(2^6\) có 7 ước.

+) \(2^5\cdot3=96\) có 12 ước.

+) \(2^4\cdot3\cdot5\) quá 100 ước mất rồi.

Vậy 96 là số có nhiều ước nhất.

không sửa đề giải luôn:

\(\dfrac{\dfrac{1}{13^2}(13^{2019}+69)+69-\dfrac{69}{13^2}}{13^{2019}+69} \)

\(\\\Rightarrow A=\dfrac{1}{13^2}+\dfrac{69(1-\dfrac{1}{13^2})}{13^{2019}+69}\)

Tương tự: \(B=\dfrac{1}{13^2}+\dfrac{1-\dfrac{1}{13^2}}{13^{2017}+1}\)

Giả sử: \(\dfrac{69}{13^{2019}+69}<\dfrac{1}{13^{2017}+1}\)

\(\\\Rightarrow 13^{2019}>69.13^{2017}\)

Điều này hiển nhiên đúng do \(13^2>69\)

Vậy \(B>A\)

trùng?

\(\left(2x+3y\right)^2+2\left(2x+3y\right)\)

\(=\left(2x+3y\right)\left(2x+3y+2\right)\)

\(2,25:\dfrac{-6}{25}=\dfrac{9}{4}:\dfrac{-6}{25}\\ =-\dfrac{9}{4}\cdot\dfrac{25}{6}=-\dfrac{3}{4}\cdot\dfrac{25}{2}\\ =-\dfrac{75}{8}\)

\(3x^3-15x^2+2x-10< 0\)

\(\Leftrightarrow3x^2\left(x-5\right)+2\left(x-5\right)< 0\)

\(\Leftrightarrow\left(3x^2+2\right)\left(x-5\right)< 0\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x^2+2< 0\\x-5>0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}3x^2+2>0\\x-5< 0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\in\varnothing\\x>5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x\in R\\x< 5\end{matrix}\right.\)