\(B=\sqrt{\left(5x-3\right)^2}+\sqrt{\left(5x-4\right)^2}\ge\left|5x-3\right|+\left|4-5x\right|\ge5x-3+4-5x=1\).

Dấu "=" xảy ra khi và chỉ khi \(3\le5x\le4\Leftrightarrow\dfrac{3}{5}\le x\le\dfrac{4}{5}\)

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

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

=>B>=|5x-2+3-5x|=1

Dấu = xảy ra khi (5x-2)(5x-3)<=0

=>2/5<=x<=3/5

Ta có: \(\sqrt{25x^2-30x+9}=x+7\)

\(\Leftrightarrow\sqrt{25x^2-15x-15x+9}=x+7\)

\(\Leftrightarrow\sqrt{5x\left(5x-3\right)-3\left(5x-3\right)}=x+7\)

\(\Leftrightarrow\sqrt{\left(5x-3\right)^2}=x+7\)

\(\Leftrightarrow5x-3=x+7\)

\(\Leftrightarrow5x-x=3+7\)

\(\Leftrightarrow4x=10\)

\(\Leftrightarrow x=\dfrac{5}{2}\)

Vậy \(x=\dfrac{5}{2}.\)

\(\sqrt{25x^2-30x+9}=x+7\) (ĐK: \(x\ge-7\))

\(\Leftrightarrow\sqrt{\left(5x-3\right)^2}=x+7\)

\(\Leftrightarrow\left|5x-3\right|=x+7\)

\(\Leftrightarrow\left[{}\begin{matrix}5x-3=x+7\\5x-3=-x-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\left(n\right)\\x=-\dfrac{2}{3}\left(n\right)\end{matrix}\right.\)

Vậy \(S=\left\{\dfrac{5}{2};-\dfrac{2}{3}\right\}\)

\(\sqrt{25x^2-30x+9}=x+7\left(x\ge-7\right)\)

\(\Rightarrow\sqrt{\left(5x+3\right)^2}=x+7\)

\(\Rightarrow\left|5x+3\right|=x+7\)

Xét trường hợp \(x\ge-\dfrac{5}{3}\) và \(x< \dfrac{5}{3}\) nha

\(A=25x^2-30x+9=\left(5x-3\right)^2\)

\(B=\left(2x-7\right)^2\)

A có GTNN là 3

B có GTNN là 5

\(B = |5x-2| + | 5x -3|=|5x-2| +|3-5x| >=|5x-2+3-5x|=1 \)

\(A=x^2-4x+7=x^2-4x+4+3=\left(x-2\right)^2+3\ge3\)

Vậy \(A_{min}=3\Leftrightarrow x-2=0\Leftrightarrow x=2\)

`A = -25x^2 +30x -2 = -(25x^2 -30x +2)`

`= -[(5x)^2 - 2*5x*3 +3^2 +2-3^2]`

`=-[(5x-3)^2 -7] = 7-(5x-3)^2`

Do `-(5x-3)^2 <= 0 AA x`

`=> 7- (5x-3)^2 <0 AA x `

hay `A<0 AA x (đpcm)`

A(\(x\)) = -25\(x^2\) + 30\(x\) - 2 < 0  ∀ \(x\)

Giả sử A(\(x\)) < 0 ∀  \(x\) ta có :

A(1) < 0 ⇔ -25 \(\times\) 1² + 30 \(\times\) 1 -  2 < 0 ⇔ 3 < 0 ( vô lý)

Vậy điều giả sử là sai vậy  A(\(x\)) < 0 ∀ \(x\) là điều không thể xảy ra.

B = \(-x^2+4x+5=-\left(x^2-4x-5\right)=-\left[\left(x^2-4x+4\right)-9\right]=-\left(x-2\right)^2+9\)

Có: \(-\left(x-2\right)^2\le0\forall x\Rightarrow-\left(x-2\right)^2+9\le9\)

Vậy Max_B = 9 <=> x = 2

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C = \(x^2-4x+9=\left(x^2-4x+4\right)+5=\left(x-2\right)^2+5\)

Có: \(\left(x-2\right)^2\ge0\Rightarrow\left(x-2\right)^2+5\ge5\)

Dấu ''='' xảy ra khi x = 2

Vậy Min_C = 5 <=> x = 2

--------

D = \(9+30x^2+25x^2=9+55x^2\ge9\)

dấu ''='' xảy ra khi x = 0

vậy min_C = 9 <=> x = 0

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

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