Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a/ Sai đề.
\(x+2\sqrt{2x-4}=\left(x-2\right)+2.\sqrt{2}.\sqrt{x-2}+2=\left(\sqrt{2}+\sqrt{x-2}\right)^2\)
b/ \(M=\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}=\sqrt{\left(\sqrt{2}+\sqrt{x-2}\right)^2}+\sqrt{\left(\sqrt{2}-\sqrt{x-2}\right)^2}\)
\(=\sqrt{2}+\sqrt{x-2}+\left|\sqrt{2}-\sqrt{x-2}\right|\)
1. Nếu \(2\le x\le4\) thì \(M=\sqrt{2}+\sqrt{x-2}+\sqrt{2}-\sqrt{x-2}=2\sqrt{2}\)
2. Nếu \(x>4\) thì \(M=\sqrt{2}+\sqrt{x-2}+\sqrt{x-2}-\sqrt{2}=2\sqrt{x-2}\)
a, \(\sqrt{4-2\sqrt{3}}-\sqrt{3}=\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{3}=\sqrt{3}-1-\sqrt{3}=-1\)
b,\(\sqrt{11+6\sqrt{2}}-3+\sqrt{2}=\sqrt{\left(\sqrt{2}+3\right)^2}-3+\sqrt{2}=\sqrt{2}+3-3+\sqrt{2}=2\sqrt{2}\)
c, \(\sqrt{9x^2}-2x=\sqrt{\left(3x\right)^2}-2x=3x-2x=x\)
d, câu này sai đề rồi , nếu sửa lại phải như này :
\(x-4+\sqrt{16-8x+x^2}=x-4+\sqrt{\left(4-x\right)^2}=x-4+4-x=0\)
a) \(\sqrt{4-2\sqrt{3}}-\sqrt{3}=\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{3}\)=\(\sqrt{3}-1-\sqrt{3}=-1\)
b) \(\sqrt{11+6\sqrt{2}}-3+\sqrt{2}\) = \(\sqrt{\left(3+\sqrt{2}\right)^2}-3+\sqrt{2}\)
= \(3+\sqrt{2}-3+\sqrt{2}\) = \(2\sqrt{2}\)
c) \(\sqrt{9x^2}-2x=\sqrt{\left(3x\right)^2}-2x\) = \(\left|3x\right|-2x=-3x-2x\) (x < 0)
= \(-5x\)
d) \(x-4+\sqrt{16-8x+x^2}\) \(\left(x>4\right)\) = \(x-4+\sqrt{\left(4-x\right)^2}\)
= \(x-4+\left|4-x\right|\) = \(x-4-4+x\) ( \(x>4\))
= \(2x-8\)
a, Ta có : \(4-2\sqrt{3}=3-2\sqrt{3}+1=\left(\sqrt{3}\right)^2-2\sqrt{3}\times1+1^2=\left(\sqrt{3}-1\right)^2\)
\(\Rightarrow\sqrt{4-2\sqrt{3}}-\sqrt{3}=\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{3}=\left|\sqrt{3}-1\right|-\sqrt{3}\)
Ta có : \(\sqrt{3}>\sqrt{1}\)(vì 3>1)
\(\Leftrightarrow\sqrt{3}>1\Leftrightarrow\sqrt{3}-1>0\Rightarrow\left|\sqrt{3}-1\right|=\sqrt{3}-1\)
Ta có: \(\sqrt{4-2\sqrt{3}}-\sqrt{3}=\left|\sqrt{3}-1\right|-\sqrt{3}=\sqrt{3}-1-\sqrt{3}=-1\)
a) \(\sqrt{4-2\sqrt{3}}-\sqrt{3}=\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{3}\)=\(\sqrt{3}-1-\sqrt{3}=-1\)
b) \(\sqrt{11+6\sqrt{2}}-3+\sqrt{2}\) = \(\sqrt{\left(3+\sqrt{2}\right)^2}-3+\sqrt{2}\)
= \(3+\sqrt{2}-3+\sqrt{2}\) = \(2\sqrt{2}\)
d) \(x-4+\sqrt{16-8x+x^2}\) \(\left(x>4\right)\) = \(x-4+\sqrt{\left(4-x\right)^2}\)
= \(x-4+\left|4-x\right|\) = \(x-4-4+x\) (vì \(x>4\))
= \(2x-8\)
\(a.A=\dfrac{\sqrt{x-2\sqrt{2x-4}}}{\sqrt{2}}=\dfrac{\sqrt{x-2-2.\sqrt{2}.\sqrt{x-2}+2}}{\sqrt{2}}=\dfrac{\sqrt{x-2}-\sqrt{2}}{\sqrt{2}}\) \(b.A=\dfrac{\sqrt{x-2\sqrt{2x-4}}}{\sqrt{2}}=\dfrac{\sqrt{x-2-2.\sqrt{2}.\sqrt{x-2}+2}}{\sqrt{2}}=\dfrac{\sqrt{2}-\sqrt{x-2}}{\sqrt{2}}\)
1/ \(\sqrt{2x-1+2\sqrt{2x-1}+1}+\sqrt{2x-1-2\sqrt{2x-1}+1}\)
\(=\sqrt{\left(\sqrt{2x-1}+1\right)^2}+\sqrt{\left(\sqrt{2x-1}-1\right)^2}\)
\(=\left|\sqrt{2x-1}+1\right|+\left|\sqrt{2x-1}-1\right|\)
\(=\sqrt{2x-1}+1+1-\sqrt{2x-1}\)
\(=2\)
2/ ĐKXĐ: \(a^2-1\ge0\Rightarrow a^2\ge1\Rightarrow\left[{}\begin{matrix}a\ge1\\a\le-1\end{matrix}\right.\)
3/ \(4\left|x\right|-\sqrt{\left(5x-1\right)^2}=4\left|x\right|-\left|5x-1\right|\)
\(=4x-\left(5x-1\right)=1-x\)
4/ \(\left\{{}\begin{matrix}x\ge0\\\sqrt{x}< \sqrt{7}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\ge0\\x< 7\end{matrix}\right.\) \(\Rightarrow0\le x< 7\)
5/ \(M=\sqrt{3-2\sqrt{2.3}+2}=\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\)
\(=\left|\sqrt{3}-\sqrt{2}\right|=\sqrt{3}-\sqrt{2}\)
6/ \(\left|x\right|-\sqrt{\left(x-1\right)^2}=\left|x\right|-\left|x-1\right|=x-\left(x-1\right)=1\)
1.
\(\sqrt{2x+2\sqrt{2x-1}}+\sqrt{2x-2\sqrt{2x-1}}\)
\(=\sqrt{2x-1+2\sqrt{2x-1}+1}+\sqrt{2x-1-2\sqrt{2x-1}+1}\)
\(=\sqrt{\left(\sqrt{2x-1}+1\right)^2}+\sqrt{\left(\sqrt{2x-1}-1\right)^2}\)
\(=\left|\sqrt{2x-1}+1\right|+\left|\sqrt{2x-1}-1\right|\)
\(=\sqrt{2x-1}+1+1-\sqrt{2x-1}=2\)
2.
\(\sqrt{a^2-1}\text{ xác định }\Leftrightarrow a^2-1\ge0\)
\(\Leftrightarrow\left(a-1\right)\left(a+1\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a-1\ge0\\a+1\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}a-1\le0\\a+1\le0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a\ge1\\a\le-1\end{matrix}\right.\)
3.
\(4\left|x\right|-\sqrt{1+25x^2-10x}\)
\(=4\left|x\right|-\sqrt{\left(5x-1\right)^2}\)
\(=4\left|x\right|-\left|5x-1\right|\)
\(=4x-5x+1=1-x\)
4.
ĐKXĐ: \(x\ge0\)
\(-\sqrt{x}>-\sqrt{7}\)
\(\Leftrightarrow\sqrt{x}< \sqrt{7}\)
\(\Leftrightarrow\text{ }x< 7\)
Vậy bât phương trình có nghiệm \(0\le x< 7\)
5.
\(\sqrt{5-2\sqrt{6}}=\sqrt{2-2\sqrt{2}.\sqrt{3}+3}\)
\(=\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}\)
\(=\sqrt{3}-\sqrt{2}\)
6.
\(\left|x\right|-\sqrt{1-2x+x^2}\)
\(=\left|x\right|-\sqrt{\left(1-x\right)^2}\)
\(=\left|x\right|-\left|x-1\right|\)
\(=x-x+1=1\)