\(\sqrt{25x^2-30x+9}\) = x + 7
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PD
4
ND
1
2 tháng 7 2021
\(B=\sqrt{\left(5x-3\right)^2}+\sqrt{\left(5x-4\right)^2}\ge\left|5x-3\right|+\left|4-5x\right|\ge5x-3+4-5x=1\).
Dấu "=" xảy ra khi và chỉ khi \(3\le5x\le4\Leftrightarrow\dfrac{3}{5}\le x\le\dfrac{4}{5}\)
28 tháng 8 2022
\(B=\left|5x-2\right|+\left|5x-3\right|\)
\(=\left|5x-2\right|+\left|3-5x\right|\)
=>B>=|5x-2+3-5x|=1
Dấu = xảy ra khi (5x-2)(5x-3)<=0
=>2/5<=x<=3/5
11 tháng 6 2018
\(M=\sqrt{x^2+y^2-2xy+2x-2y+10}+2y^2-8y+2024\\ =\sqrt{\left(x^2+y^2+1-2xy+2x-2y\right)+9}+\left(2y^2-8y+8\right)+2016\\ =\sqrt{\left(x-y+1\right)^2+9}+2\left(y^2-4y+4\right)+2016\\ =\sqrt{\left(x-y+1\right)^2+9}+2\left(y-2\right)^2+2016\) \(\text{Do }\left(x-y+1\right)^2\ge0\forall x;y\\ \Rightarrow\left(x-y+1\right)^2+9\ge9\forall x;y\\ \Rightarrow\sqrt{\left(x-y+1\right)^2+9}\ge3\forall x;y\\ Mà\text{ }2\left(y-2\right)^2\ge0\forall y\\ \Rightarrow\sqrt{\left(x-y+1\right)^2+9}+2\left(y-2\right)^2\ge3\forall x;y\\ M=\sqrt{\left(x-y+1\right)^2+9}+2\left(y-2\right)^2+2016\ge2019\forall x;y\)
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 \(\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)
11 tháng 6 2018
\(Q=\sqrt{25x^2-20x+4}+\sqrt{25x^2-30x+9}\\ =\sqrt{\left(5x-2\right)^2}+\sqrt{\left(5x-3\right)^2}\\ =\left|5x-2\right|+\left|5x-3\right|\\ =\left|5x-2\right|+\left|3-5x\right|\)
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 \(\dfrac{2}{5}\le x\le\dfrac{3}{5}\)
NV
Nguyễn Việt Lâm
Giáo viên
19 tháng 10 2019
1/
\(\Leftrightarrow\left\{{}\begin{matrix}2x-y=0\\y-2=0\\x+y+z=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=1\\y=2\\z=-3\end{matrix}\right.\)
2/ \(P=\sqrt{\left(5x-2\right)^2}+\sqrt{\left(3-5x\right)^2}\)
\(P=\left|5x-2\right|+\left|3-5x\right|\ge\left|5x-2+3-5x\right|=1\)
\(\Rightarrow P_{min}=1\) khi \(\frac{2}{5}\le x\le\frac{3}{5}\)
3/ ĐKXĐ: \(\left|x\right|\ge1\)
\(x^2-1-\sqrt{x^2-1}=0\)
\(\Leftrightarrow\sqrt{x^2-1}\left(\sqrt{x^2-1}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-1}=0\\\sqrt{x^2-1}=1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2-1=0\\x^2-1=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\pm1\\x=\pm\sqrt{2}\end{matrix}\right.\)
HT
15 tháng 6 2019
\(f\left(a\right)=\sqrt{25a^2-30a+9}=\sqrt{\left(5a-3\right)^2}=\left|5a-3\right|\)
\(g\left(a\right)=a\)
\(\Rightarrow\left|5a-3\right|=a+7\)
Nếu \(a\ge\frac{3}{5}\Rightarrow\left|5a-3\right|=5a-3\)
\(\Rightarrow5a-3=a+7\Leftrightarrow a=\frac{5}{2}\left(tm\right)\)
Nếu \(a< \frac{3}{5}\Rightarrow\left|5a-3\right|=3-5a\)
\(\Rightarrow3-5a=a+7\Leftrightarrow a=\frac{-2}{3}\left(tm\right)\)
AT
1
AT
10 tháng 6 2018
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 MaxB = 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 MinC = 5 <=> x = 2
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D = \(9+30x^2+25x^2=9+55x^2\ge9\)
dấu ''='' xảy ra khi x = 0
vậy minC = 9 <=> x = 0
\(\sqrt{25x^2-30x+9}=x+7\) (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