Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(C=\sqrt{25x^2-20x+4}+\sqrt{25x^2}\)
\(C=\sqrt{\left(5x-2\right)^2}+\sqrt{\left(5x\right)^2}\)
\(C=\left|5x-2\right|+\left|5x\right|=\left|2-5x\right|+\left|5x\right|\)
\(C\ge\left|2-5x+5x\right|=2\)
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
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
\(=\sqrt{\left(5x-2\right)^2}+\sqrt{\left(5x\right)^2}\)= \(\left|2-5x\right|+\left|5x\right|\ge2+5x-5x=2\)
min A=2 \(\Leftrightarrow\hept{\begin{cases}2-5x\ge0\\5x\ge0\end{cases}\Leftrightarrow0\le x\le\frac{2}{5}}\)
A có GTNN là 3
B có GTNN là 5
\(B = |5x-2| + | 5x -3|=|5x-2| +|3-5x| >=|5x-2+3-5x|=1 \)