A = \(x^2\) - 20\(x\) + 2024

A = (\(x^2\) - 20\(x\) + 100) + 1924

A = (\(x\) - 10)² + 1924

Vì ( \(x-10\))² ≥ 0 ⇒ (\(x-10\))² + 1924 ≥ 1924

     Vậy A_min  =  1924 ⇔ \(x\) = 10 

Kết luận giá trị nhỏ nhất của A là 1924 xảy ra khi \(x\) = 10

\(M=\sqrt{x^2+y^2-2xy+2x-2y+10}+2y^2-8y+2024\\ =\sqrt{\left(x^2+y^2+1-2xy+2x-2y\right)+9}+\left(2y^2-8y+8\right)+2016\\ =\sqrt{\left(x-y+1\right)^2+9}+2\left(y^2-4y+4\right)+2016\\ =\sqrt{\left(x-y+1\right)^2+9}+2\left(y-2\right)^2+2016\) \(\text{Do }\left(x-y+1\right)^2\ge0\forall x;y\\ \Rightarrow\left(x-y+1\right)^2+9\ge9\forall x;y\\ \Rightarrow\sqrt{\left(x-y+1\right)^2+9}\ge3\forall x;y\\ Mà\text{ }2\left(y-2\right)^2\ge0\forall y\\ \Rightarrow\sqrt{\left(x-y+1\right)^2+9}+2\left(y-2\right)^2\ge3\forall x;y\\ M=\sqrt{\left(x-y+1\right)^2+9}+2\left(y-2\right)^2+2016\ge2019\forall x;y\)

Dấu "=" xảy ra khi:

\(\left\{{}\begin{matrix}2\left(y-2\right)^2=0\\\left(x-y+1\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y-2=0\\x-y+1=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=2\\x=y-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\x=1\end{matrix}\right.\)

Vậy \(M_{Min}=2019\) khi \(\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)

\(Q=\sqrt{25x^2-20x+4}+\sqrt{25x^2-30x+9}\\ =\sqrt{\left(5x-2\right)^2}+\sqrt{\left(5x-3\right)^2}\\ =\left|5x-2\right|+\left|5x-3\right|\\ =\left|5x-2\right|+\left|3-5x\right|\)

Áp dụng BDT: \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\)

\(\Rightarrow\left|5x-2\right|+\left|3-5x\right|\ge\left|5x-2+3-5x\right|=\left|1\right|=1\)

Dấu "=" xảy ra khi:

\(\left(5x-2\right)\left(3-5x\right)\ge0\\\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}5x-2\ge0\\3-5x\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}5x-2\le0\\3-5x\le0\end{matrix}\right.\end{matrix}\right. \) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}5x\ge2\\5x\le3\end{matrix}\right.\\\left\{{}\begin{matrix}5x\le2\\5x\ge3\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge\dfrac{2}{5}\\x\le\dfrac{3}{5}\end{matrix}\right.\left(T/m\right)\\\left\{{}\begin{matrix}x\le\dfrac{2}{5}\\x\ge\dfrac{3}{5}\end{matrix}\right.\left(K^0\text{ }T/m\right)\end{matrix}\right.\)

\(\Leftrightarrow\dfrac{2}{5}\le x\le\dfrac{3}{5}\)

Vậy \(Q_{Min}=1\) khi \(\dfrac{2}{5}\le x\le\dfrac{3}{5}\)

Có: 2(x - 1)² ≥ 0 ∀ x 

=> A = 2(x - 1)² - 2024 ≥ -2024

Dấu "=" xảy ra \(\Leftrightarrow\)2(x - 1)² = 0 \(\Leftrightarrow\)x = 1.

Vậy A_min = -2024 khi x = 1.

A = 2(x - 1)² - 2024

Ta có:

(x - 1)² ≥ 0 với mọi x ∈ R

⇒ 2(x - 1)² ≥ 0 với mọi x ∈ R

⇒ 2(x - 1)² - 2024 ≥ -2024 với mọi x ∈ R

Vậy GTNN của A là -2024 khi x = 1

\(A=\dfrac{4}{2-x}+\dfrac{100}{x}+2021=36\left(2-x\right)+\dfrac{4}{2-x}+36x+\dfrac{100}{x}+1949\)

\(0< x< 2\Rightarrow\left\{{}\begin{matrix}x>0\\x< 2\Rightarrow-x>-2\Leftrightarrow2-x>0\end{matrix}\right.\)

\(\Rightarrow A\ge2\sqrt{36\left(2-x\right).\dfrac{4}{\left(2-x\right)}}+2\sqrt{36x.\dfrac{100}{x}}+1985=2\sqrt{4.36}+2\sqrt{36.100}+1949=2093\Rightarrow A_{min}=2093\Leftrightarrow\left\{{}\begin{matrix}36\left(2-x\right)=\dfrac{4}{2-x}\\36x=\dfrac{100}{x}\end{matrix}\right.\)

\(\Leftrightarrow x=\dfrac{5}{3}\left(tm\right)\)

Tham Khảo

\(y=5-2cos^2x=4-\left(2cos^2x-1\right)=4-cos2x\)

Vì \(cos2x\in\left[-1;1\right]\Rightarrow y=4-cos2x\in\left[3;5\right]\)

\(\Rightarrow y_{min}=3\Leftrightarrow cos2x=1\Leftrightarrow x=k\pi\)

\(y_{max}=5\Leftrightarrow cos2x=-1\Leftrightarrow x=\dfrac{\pi}{2}+k\pi\)

\(\Leftrightarrow x=2-\sqrt{3}\)

Dễ thấy x là nghiệm của PT \(x^2-4x+1\)

\(H=\left(x^5-4x^4+x^3\right)+\left(x^4-4x^3+x^2\right)+\left(5x^2-20x+5\right)+2019\\ H=\left(x^2-4x+1\right)\left(x^3+x^2+5\right)+2019\\ H=2019\)

Để Amin.

=>20x^2-20x min.

=>>x^2-x min.

Mà nếu x âm =>x^2 dương và -x sẽ thành cộng với 1 số dương.

=>x ko âm.

=>x =1 hoặc x=0

Thử lại:

Với 2 số trên A đểu =1.

Vậy........

\(A=20x^2-20x+1=20\left(x^2-x+\frac{1}{4}\right)-20.\frac{1}{4}+1\)

\(=20\left(x-\frac{1}{2}\right)^2-4\ge-4\)( Vì \(20\left(x-\frac{1}{2}\right)^2\ge0\))

Vậy \(MinA=-4\Leftrightarrow\left(x-\frac{1}{2}\right)^2=0\Leftrightarrow x-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{2}\)