\(C=\sqrt{25x^2-20x+4}+\sqrt{25x^2}\)

\(C=\sqrt{\left(5x-2\right)^2}+\sqrt{\left(5x\right)^2}\)

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

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

Dấu " = " xảy ra \(\Leftrightarrow\)( 2 - 5x ) . 5x \(\ge\)0

\(\Leftrightarrow\)\(\hept{\begin{cases}x\ge0\\2-5x\ge0\end{cases}}\)hoặc \(\hept{\begin{cases}x\le0\\2-5x\le0\end{cases}}\)

\(\Leftrightarrow\)\(0\le x\le\frac{2}{5}\)

Vậy GTNN của C là 2 \(\Leftrightarrow\)\(0\le x\le\frac{2}{5}\)

\(C=\sqrt{25x^2-20x+4}+\sqrt{25x^2}\)

\(C=\sqrt{\left(5x-2\right)^2}+\sqrt{\left(5x\right)^2}\)

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

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

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}2-5x\ge0\\5x\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}x\le\frac{2}{5}\\x\ge0\end{cases}\Leftrightarrow0\le}x\le\frac{2}{5}}\)

a) A = \(\sqrt{-x^2+x+\dfrac{3}{4}}=\sqrt{1-\left(x-\dfrac{1}{2}\right)^2}\le\sqrt{1}=1\) (dấu "=" xảy ra \(\Leftrightarrow x=\dfrac{1}{2}\))

Vậy max A = 1 (khi và chỉ khi x = \(\dfrac{1}{2}\))

b) B = \(\sqrt{\left(2x^2-x-1\right)^2+9}\ge\sqrt{9}=3\) (dấu "=" xảy ra \(\Leftrightarrow2x^2-x-1=0\)

\(\Leftrightarrow\left(2x+1\right)\left(x-1\right)=0\)

\(\Leftrightarrow x=1;x=-\dfrac{1}{2}\)).

Vậy min B = 3 (khi và chỉ khi x = 1 hoặc x = \(-\dfrac{1}{2}\))

c) C = \(\left|5x-2\right|+\left|5x\right|=\left|2-5x\right|+\left|5x\right|\);

C \(\ge\left|2-5x+5x\right|=\left|2\right|=2\) (dấu "=" xảy ra \(\Leftrightarrow\left(2-5x\right).5x\ge0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\2-5x\ge0\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}x\le0\\2-5x\le0\end{matrix}\right.\)

\(\Leftrightarrow0\le x\le\dfrac{2}{5}\)).

Vậy min C = 2 (khi và chỉ khi \(0\le x\le\dfrac{2}{5}\))

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

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

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

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

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

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

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

Dấu = xảy ra khi (5x-2)(5x-3)<=0

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

C = 25x² + 20x + 5/2

C = 25( x² + 4/5x + 4/25 ) - 3/2

C = 25( x + 2/5 )² - 3/2

25( x + 2/5 )² ≥ 0 ∀ x => 25( x + 2/5 )² - 3/2 ≥ -3/2

Đẳng thức xảy ra <=> x + 2/5 = 0 => x = -2/5

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

\(C=25x^2+20x+\frac{5}{2}=25x^2+20x+4-\frac{3}{2}\)

\(=25\left(x+\frac{2}{5}\right)^2-\frac{3}{2}\)

Vì \(\left(x+\frac{2}{5}\right)^2\ge0\forall x\)\(\Rightarrow25\left(x+\frac{2}{5}\right)^2-\frac{3}{2}\ge-\frac{3}{2}\)

Dấu "=" xảy ra \(\Leftrightarrow25\left(x+\frac{2}{5}\right)^2=0\Leftrightarrow x+\frac{2}{5}=0\Leftrightarrow x=-\frac{2}{5}\)

Vậy Cmin = -3/2 <=> x = -2/5

\(c=\left(5x\right)^2+2\cdot5x\cdot2+2^2-2^2+\frac{5}{2}\)    

=\(\left(5x+2\right)^2-\frac{3}{2}\)  

Ta có : \(\left(5x+2\right)^2\ge0\forall x\)      

\(\left(5x+2\right)^2-\frac{3}{2}\ge-\frac{3}{2}\)        

Dấu = xảy ra : 

\(\Leftrightarrow\left(5x+2\right)^2=0\)         

\(5x+2=0\)     

\(x=-\frac{2}{5}\)   

Vậy giá trị nhỏ nhất của c = -3/2 khi và chỉ khi x = -2/5 

chắc gõ dấu + nhưng quên ấn Shift thành dấu = r`

\(\sqrt{4x^2+4x+1}+\sqrt{25x^2+10x+1}\)

\(=\sqrt{\left(2x+1\right)^2}+\sqrt{\left(5x+1\right)^2}\)

\(=\left|2x+1\right|+\left|5x+1\right|\ge\frac{3}{5}\)

Dấu = khi \(x=-\frac{1}{5}\)

vào đây xem câu TL bạn nhé

https://www.youtube.com/watch?v=fvGaHwKrbUc

A= x²-4x+6 = (x-2)²+2 ≥ 2 

Dấu "=" xảy ra ⇔ x=2

B = 25x²+10x-3 = (5x+1)²-4 ≥ -4

Dấu "=" xảy ra \(\Leftrightarrow x=-\dfrac{1}{5}\)

C = 5-6x+4x² = \(\left(\dfrac{3}{2}-2x\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}\)

Dấu "=" xảy ra \(\Leftrightarrow x=\dfrac{3}{4}\)

A= 2x^2-4x+ 4+2

A=(x-2)² + 2

A có giá trị nhỏ nhất khi (x-2)² =0

x-2 =0

x=2

 B, C tự làm :>

B=(5x)²+2.5.x-1-2

B=(5x-1)²-2

B có giá trị nhỏ nhất khi (5x-1)²=0

5x-1=0

x=1/5

C=(2x)²-2.3.x+9-4

C=(2x-3)²-4

C có giá trị nhỏ nhất khi (2x-3)²=0

2x-3=0

2x=3/2