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) 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 \(3\le5x\le4\Leftrightarrow\dfrac{3}{5}\le x\le\dfrac{4}{5}\)

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 

\(=\sqrt{\left(5x-2\right)^2}+\sqrt{\left(5x\right)^2}\)= \(\left|2-5x\right|+\left|5x\right|\ge2+5x-5x=2\)

min A=2 \(\Leftrightarrow\hept{\begin{cases}2-5x\ge0\\5x\ge0\end{cases}\Leftrightarrow0\le x\le\frac{2}{5}}\)

Chuẩn đấy

1) ĐKXĐ: \(x\ge-2\)

\(pt\Leftrightarrow\dfrac{\sqrt{x+2}}{2}+5\sqrt{x+2}-2\sqrt{x+2}=14\)

\(\Leftrightarrow\dfrac{\sqrt{x+2}+6\sqrt{x+2}}{2}=14\Leftrightarrow7\sqrt{x+2}=28\)

\(\Leftrightarrow\sqrt{x+2}=4\Leftrightarrow x+2=16\Leftrightarrow x=14\left(tm\right)\)

2) ĐKXĐ: \(x\ge0\)

\(pt\Leftrightarrow2x+3=x^2\Leftrightarrow\left(x-3\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=-1\left(ktm\right)\end{matrix}\right.\)

3) \(pt\Leftrightarrow\sqrt{\left(5x+2\right)^2}=1\Leftrightarrow\left|5x+2\right|=1\)

\(\Leftrightarrow\left[{}\begin{matrix}5x+2=1\\5x+2=-1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{5}\\x=-\dfrac{3}{5}\end{matrix}\right.\)

4) ĐKXĐ: \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x+1\ge0\\2x-1>0\end{matrix}\right.\\\left\{{}\begin{matrix}x+1\le0\\2x-1< 0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x>\dfrac{1}{2}\\x\le-1\end{matrix}\right.\)

\(pt\Leftrightarrow\dfrac{x+1}{2x-1}=4\Leftrightarrow x+1=8x-4\)

\(\Leftrightarrow7x=5\Leftrightarrow x=\dfrac{5}{7}\left(tm\right)\)

5) ĐKXĐ: \(x\ge2\)

\(pt\Leftrightarrow\dfrac{x-2}{3x+1}=36\)

\(\Leftrightarrow x-2=108x+36\Leftrightarrow107x=-38\Leftrightarrow x=-\dfrac{38}{107}\left(ktm\right)\)

Vậy \(S=\varnothing\)

Ta có : M =\(\sqrt{\left(5x\right)^2-2.2.5x+2^2}+\sqrt{\left(5x\right)^2}\)= \(\sqrt{\left(5x-2\right)^2}+\left|5x\right|\) = \(\left|5x-2\right|+\left|5x\right|=\left|2-5x\right|+\left|5x\right|\ge\left|2-5x+5x\right|=\left|2\right|=2\)

=> M ≥ 2

Min M = 2 : dấu"="xảy ra khi: ....

Mình bận rồi tự làm tí nhé!!

Lời giải:

Ta có:

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

\(=|5x-2|+|5x|=|2-5x|+|5x|\)

Áp dụng BĐT \(|a|+|b|\geq |a+b|\) ta có:

\(C=|2-5x|+|5x|\geq |2-5x+5x|=2\)

Vậy \(C_{\min}=2\). Dấu "=" xảy ra khi \(5x(2-5x)\geq 0\Leftrightarrow 0\leq x\leq \frac{2}{5}\)

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