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 \(5x(2-5x)\geq 0\Leftrightarrow 0\leq x\leq \frac{2}{5}\)

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

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

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

Dấu = xảy ra khi (5x-2)(5x-3)<=0

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

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

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

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

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

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

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

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

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 \(S=\varnothing\)

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

câu a) rút x theo y thế vào A rồi áp dụng HĐT

b)rút xy thế vào B 

c)HĐT

d)rút x theo y thé vào C

rồi dùng BĐT cô-si

e)BĐT chưa dấu giá trị tuyệt đối