TT
0
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
20 tháng 8 2016
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 \(M=\sqrt{2}+\sqrt{x-2}+\sqrt{x-2}-\sqrt{2}=2\sqrt{x-2}\)
27 tháng 4 2019
ĐKXĐ: x > 4
a, Có \(A=\sqrt{x+4\sqrt{x-4}}+\sqrt{x-4\sqrt{x-4}}\)
\(=\sqrt{x-4+4\sqrt{x-4}+4}+\sqrt{x-4-4\sqrt{x-4}+4}\)
\(=\sqrt{\left(\sqrt{x-4}+2\right)^2}+\sqrt{\left(\sqrt{x-4}-2\right)^2}\)
\(=\sqrt{x-4}+2+\left|\sqrt{x-4}-2\right|\)
\(\orbr{\begin{cases}=2\sqrt{x-4}\left(với\sqrt{x-4}\ge2\right)\\=4\left(với\sqrt{x-4}< 2\right)\end{cases}}\)
b, Xét \(A=2\sqrt{x-4}\)thì \(\sqrt{x-4}\ge2\)
\(\Leftrightarrow x-4\ge4\)
\(\Leftrightarrow x\ge8\)
Khi đó \(A=2\sqrt{x-4}\ge2\sqrt{8-4}=4\)
Nên \(A_{min}=4\Leftrightarrow x=8\)
c, Với \(x=\sqrt{15+\sqrt{6}}\)thì \(\sqrt{x-4}=\sqrt{\sqrt{15+\sqrt{6}}-4}< 2\)
Nên từ câu a => A = 4
21 tháng 7 2017
\(\sqrt{x-1-2\sqrt{x-1}+1}\)+\(\sqrt{x-1+4\sqrt{x-1}+4}\) (\(x\ge1\)
=\(\left|\sqrt{x-1}-1\right|+\left|\sqrt{x-1}-2\right|\)
dat \(\sqrt{x-1}=t\left(t\ge0\right)\)
ta co \(\left|t-1\right|+\left|t-2\right|\)
t |t-1| |t-2| 1 2 0 0 + - - +
nenta co voi0<= t<1 \(1-t+2-t=3-t=3-2\sqrt{x-1}\)
voi 1\(\le t\le2\) \(t-1+2-t=3\)
voi t>2 \(t-1+t-2=2t-3=2\sqrt{x-1}-3\)
b,\(\sqrt{x-4-4\sqrt{x-4}+4}\) =\(\left|\sqrt{x-4}-2\right|\)
3 tháng 7 2019
\(a,\sqrt{13+6\sqrt{4+\sqrt{9-4\sqrt{2}}}}\)
\(=\sqrt{13+6\sqrt{4+\sqrt{1-2.2\sqrt{2}+\left(2\sqrt{2}\right)^2}}}\)
\(=\sqrt{13+6\sqrt{4+\sqrt{\left(2\sqrt{2}-1\right)^2}}}\)
\(=\sqrt{13+6\sqrt{4+2\sqrt{2}-1}}\)
\(=\sqrt{13+6\sqrt{3+2\sqrt{2}}}\)
\(=\sqrt{13+6\sqrt{1+2\sqrt{2}+2}}\)
\(=\sqrt{13+6\sqrt{\left(1+\sqrt{2}\right)^2}}\)
\(=\sqrt{13+6\left(1+\sqrt{2}\right)}=\sqrt{13+6+\sqrt{12}}\)
\(=\sqrt{19+2\sqrt{3}}\)
3 tháng 7 2019
a) = \(\sqrt{13+6\sqrt{4+\sqrt{9-4\sqrt{2}}}}\)
= \(\sqrt{13+6\sqrt{4+\sqrt{8-2.2\sqrt{2}+1}}}\)
= \(\sqrt{13+6\sqrt{4+\sqrt{\left(2\sqrt{2}-1\right)^2}}}\)
= \(\sqrt{13+6\sqrt{4+2\sqrt{2}-1}}\)
= \(\sqrt{13+6\sqrt{2+2\sqrt{2}+1}}\)
= \(\sqrt{13+6\left(\sqrt{2}+1\right)}\)
= \(\sqrt{13+6\sqrt{2}+6}=\sqrt{19+6\sqrt{2}}\)
= \(\sqrt{18+2.3\sqrt{2}+1}\)
= \(\sqrt{\left(3\sqrt{2}+1\right)^2}\)
= \(3\sqrt{2}+1\)
NC
25 tháng 8 2020
Bài làm:
a) \(\sqrt{4-\sqrt{7}}=\frac{\sqrt{2\left(4-\sqrt{7}\right)}}{\sqrt{2}}=\sqrt{\frac{8-2\sqrt{7}}{2}}=\sqrt{\frac{7-2\sqrt{7}+1}{2}}\)
\(=\sqrt{\frac{\left(\sqrt{7}-1\right)^2}{2}}=\frac{\left(\sqrt{7}-1\right)\sqrt{2}}{2}=\frac{\sqrt{14}-\sqrt{2}}{2}\)
b) \(\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}\) (đề vậy chứ)
\(=\sqrt{3+2\sqrt{3}+1}-\sqrt{3-2\sqrt{3}+1}\)
\(=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(=\sqrt{3}+1-\sqrt{3}+1\)
\(=2\)
c) \(\sqrt{9-4\sqrt{5}}-\sqrt{9+4\sqrt{5}}\)
\(=\sqrt{5-4\sqrt{5}+4}-\sqrt{5+4\sqrt{5}+4}\)
\(=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{\left(\sqrt{5}+2\right)^2}\)
\(=\sqrt{5}-2-\sqrt{5}-2\)
\(=-4\)
d) \(\sqrt{x-2\sqrt{x-1}}\)
\(=\sqrt{\left(x-1\right)-2\sqrt{x-1}+1}\)
\(=\sqrt{\left(\sqrt{x-1}-1\right)^2}\)
\(=\left|\sqrt{x-1}-1\right|\)