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\(M=\sqrt{x^2-4x+4}+2014\sqrt{x^2-6x+9}+\sqrt{x^2-10x+25}\)
\(M=\left|x-2\right|+2014\left|x-3\right|+\left|x-5\right|\)
\(M=\left|x-2\right|+\left|5-x\right|+2014\left|x-3\right|\)
\(M\ge\left|x-2+5-x\right|+2014\left|x-3\right|=3+2014\left|x-3\right|\ge3\)
\("="\Leftrightarrow x=3\)
a: ĐKXĐ: x>=5
\(\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\cdot\sqrt{9x-45}=4\)
=>\(2\sqrt{x-5}+\sqrt{x-5}-\dfrac{1}{3}\cdot3\sqrt{x-5}=4\)
=>\(2\sqrt{x-5}=4\)
=>\(\sqrt{x-5}=2\)
=>x-5=4
=>x=9(nhận)
b: ĐKXĐ: x>=1/2
\(\sqrt{2x-1}-\sqrt{8x-4}+5=0\)
=>\(\sqrt{2x-1}-2\sqrt{2x-1}+5=0\)
=>\(5-\sqrt{2x-1}=0\)
=>\(\sqrt{2x-1}=5\)
=>2x-1=25
=>2x=26
=>x=13(nhận)
c: \(\sqrt{x^2-10x+25}=2\)
=>\(\sqrt{\left(x-5\right)^2}=2\)
=>\(\left|x-5\right|=2\)
=>\(\left[{}\begin{matrix}x-5=2\\x-5=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=3\end{matrix}\right.\)
d: \(\sqrt{x^2-14x+49}-5=0\)
=>\(\sqrt{x^2-2\cdot x\cdot7+7^2}=5\)
=>\(\sqrt{\left(x-7\right)^2}=5\)
=>|x-7|=5
=>\(\left[{}\begin{matrix}x-7=5\\x-7=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=12\\x=2\end{matrix}\right.\)
\(a,\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9x-45}=4\left(đkxđ:x\ge5\right)\\ \Leftrightarrow\sqrt{4\left(x-5\right)}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9\left(x-5\right)}=4\\ \Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\\ \Leftrightarrow2\sqrt{x-5}=4\\ \Leftrightarrow\sqrt{x-5}=2\\ \Leftrightarrow x-5=4\\ \Leftrightarrow x=9\left(tm\right)\)
\(b,\sqrt{2x-1}-\sqrt{8x-4}+5=0\left(đkxđ:x\ge\dfrac{1}{2}\right)\\ \Leftrightarrow\sqrt{2x-1}-\sqrt{4\left(2x-1\right)}=-5\\ \Leftrightarrow\sqrt{2x-1}-2\sqrt{2x-1}=-5\\ \Leftrightarrow-\sqrt{2x-1}=-5\\ \Leftrightarrow\sqrt{2x-1}=5\\ \Leftrightarrow2x-1=25\\ \Leftrightarrow2x=26\\ \Leftrightarrow x=13\left(tm\right)\)
\(c,\sqrt{x^2-10x+25}=2\\ \Leftrightarrow\sqrt{\left(x-5\right)^2}=2\\ \Leftrightarrow\left|x-5\right|=2\\ \Leftrightarrow\left[{}\begin{matrix}x-5=2\\x-5=-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=7\\x=3\end{matrix}\right.\)
\(d,\sqrt{x^2-14x+49}-5=0\\ \Leftrightarrow\sqrt{\left(x-7\right)^2}=5\\ \Leftrightarrow\left|x-7\right|=5\\ \Leftrightarrow\left[{}\begin{matrix}x-7=5\\x-7=-5\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=12\\x=2\end{matrix}\right.\)
1, \(\sqrt{4-4x+x^2}=3\)
\(\Leftrightarrow\sqrt{\left(2+x\right)^2}=3\)
\(\Leftrightarrow\left|2+x\right|=3\)
TH1: \(\left|2-x\right|=2-x\) với \(2-x\ge0\Leftrightarrow x\le2\)
Pt trở thành:
\(2-x=3\) (ĐK: \(x\le2\) )
\(\Leftrightarrow x=2-3\)
\(\Leftrightarrow x=-1\left(tm\right)\)
TH2: \(\left|2-x\right|=-\left(2-x\right)\) với \(2-x< 0\Leftrightarrow x>2\)
Pt trở thành:
\(-\left(2-x\right)=3\) (ĐK: \(x>2\))
\(\Leftrightarrow-2+x=3\)
\(\Leftrightarrow x=3+2\)
\(\Leftrightarrow x=5\left(tm\right)\)
Vậy \(S=\left\{-1;5\right\}\)
c/ \(C=\sqrt{x^2-6x+9}+\sqrt{x^2+10x+25}\)
\(=\sqrt{\left(x-3\right)^2}+\sqrt{\left(x+5\right)^2}\)
\(=|3-x|+|x+5|\ge|3-x+x+5|=8\)
d/ \(D=\sqrt{x^2-6x+9}+\sqrt{4x^2+24x+36}\)
\(=\sqrt{\left(x-3\right)^2}+\sqrt{4\left(x+3\right)^2}\)
\(=|3-x|+|x+3|+|x+3|\ge|3-x+x+3|+0=6\)
e/ \(2E=\sqrt{x^2}+2\sqrt{x^2-2x+1}\)
\(=\sqrt{x^2}+2\sqrt{\left(x-1\right)^2}\)
\(=|x|+|1-x|+|x-1|\ge|x+1-x|+0=1\)
\(\Rightarrow E\ge\frac{1}{2}\)
Ngại làm lần 2 quá bạn ơi
Câu hỏi của Chuột yêu Gạo - Toán lớp 9 | Học trực tuyến
a) \(\Leftrightarrow\sqrt{\left(x+3\right)^2}=4\)
\(\Leftrightarrow\left|x+3\right|=4\) \(\Leftrightarrow\left[{}\begin{matrix}x+3=4\\x+3=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-7\end{matrix}\right.\) ( TM )
b) \(\Leftrightarrow\sqrt{\left(2x-1\right)^2}=5x+3\)
\(\Leftrightarrow\left|2x-1\right|=5x+3\)
\(\Leftrightarrow\left\{{}\begin{matrix}5x+3\ge0\\\left[{}\begin{matrix}2x-1=5x+3\\2x-1=-5x-3\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-\frac{3}{5}\\\left[{}\begin{matrix}x=-\frac{4}{3}\left(KTM\right)\\x=-\frac{2}{7}\left(TM\right)\end{matrix}\right.\end{matrix}\right.\)
a \(\sqrt{x^2+6x+9}=4\Leftrightarrow\sqrt{\left(x+3\right)^2=4}\)
\(\Leftrightarrow x+3=4\)
\(\Rightarrow x=1\)
\(\sqrt{x^2+2x+1}+\sqrt{x^2-2x+1}=\sqrt{\left(x+1\right)^2}-\sqrt{\left(1-x\right)^2}\)
= | x+1 | - | 1-x | \(\ge\left|x+1+1-x\right|=\left|2\right|=2\)
dấu "=" xảy ra <=> \(\left(x+1\right)\left(1-x\right)\ge0\)
\(\Leftrightarrow\) \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x+1\ge0\\1-x\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x+1\le0\\1-x\le0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge-1\\x\le1\end{matrix}\right.\\\left\{{}\begin{matrix}x\le-1\\x\ge1\end{matrix}\right.\end{matrix}\right.\)
<=> \(-1\le x\le1\)
Vậy min C = 1 khi và chỉ khi \(-1\le x\le1\)