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N
1
15 tháng 7 2018
\(A=\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}=\sqrt{x-2+2\sqrt{2}.\sqrt{x-2}+2}+\sqrt{x-2-2\sqrt{2}.\sqrt{x-2}+2}=\text{| }\sqrt{x-2}+\sqrt{2}\text{| }+\text{| }\sqrt{x-2}-\sqrt{2}\text{| }\) +) \(A=\sqrt{x-2}+\sqrt{2}+\sqrt{x-2}-\sqrt{2}=2\sqrt{x-2}\left(\text{ }\sqrt{x-2}\text{≥}\sqrt{2}\right)\)
+) \(A=\sqrt{x-2}+\sqrt{2}+\sqrt{2}-\sqrt{x-2}=2\sqrt{2}\left(\sqrt{x-2}< \sqrt{2}\right)\)
NP
3
VT
15 tháng 7 2018
Ta có \(A\sqrt{2}=\sqrt{2x+4\sqrt{2x-4}}+\sqrt{2x-4\sqrt{2x-4}}=\sqrt{2x-4+4\sqrt{2x-4}+4}\)
+\(\sqrt{2x-4-4\sqrt{2x-4}+4}=\sqrt{2x-4}+2+\sqrt{2x-4}-2=2\sqrt{2x-4}\)
=> A=\(2\sqrt{x-2}\)
VV
15 tháng 7 2018
A = \(\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}\)
\(\sqrt{2}A=\sqrt{2}\left(\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}\right)\)
\(\sqrt{2}A=\sqrt{2\left(x+2\sqrt{2x-4}\right)}+\sqrt{2\left(x-2\sqrt{2x-4}\right)}\)
\(\sqrt{2}A=\sqrt{2x+4\sqrt{2x-4}}+\sqrt{2x-4\sqrt{2x-4}}\)
\(\sqrt{2}A=\sqrt{\left(2x-4\right)+2.2\sqrt{2x-4}+4}+\sqrt{\left(2x-4\right)-2.2\sqrt{2x-4}+4}\)
\(\sqrt{2}A=\sqrt{\left(\sqrt{2x-4}+2\right)^2}+\sqrt{\left(\sqrt{2x-4}-2\right)^2}\)
\(\sqrt{2}A=|\sqrt{2x-4}+2|+|\sqrt{2x-4}+2|\)
\(\sqrt{2}A=\sqrt{2x-4}+2+|\sqrt{2x-4}-2|\)
Xét 2 trường hợp:
+)\(\sqrt{2x-4}\ge2\)
\(\sqrt{2}A=\sqrt{2x-4}+2+\sqrt{2x-4}-2\)
\(\sqrt{2}A=2\sqrt{2x-4}\)
\(A=\sqrt{2}\sqrt{2x-4}=\sqrt{4x-8}\)
+)\(\sqrt{2x-4}< 2\)
\(\sqrt{2}A=\sqrt{2x-4}+2+2-\sqrt{2x-4}=4\)
Vậy...
( Bạn có thể bình phương lên cũng đc)
8 tháng 9 2020
Ta có: \(\frac{2x+2}{\sqrt{x}}+\frac{x\sqrt{x}-1}{x-\sqrt{x}}-\frac{x^2+\sqrt{x}}{x\sqrt{x}+x}\)
\(=\frac{2\left(x+1\right)}{\sqrt{x}}+\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\cdot\left(\sqrt{x}-1\right)}-\frac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x\left(\sqrt{x}+1\right)}\)
\(=\frac{2\left(x+1\right)}{\sqrt{x}}+\frac{x+\sqrt{x}+1}{\sqrt{x}}-\frac{x-\sqrt{x}+1}{\sqrt{x}}\)
\(=\frac{2x+2+x+\sqrt{x}+1-x+\sqrt{x}-1}{\sqrt{x}}\)
\(=\frac{2x+2\sqrt{x}+2}{\sqrt{x}}\)
DK
29 tháng 7 2016
a) \(x+3+\sqrt{x^2-6x+9}=x+3+\sqrt{\left(x-3\right)^2}=x+3+x-3=2x\)
b) \(\sqrt{x^2+4x+4}-\sqrt{x^2}=\sqrt{\left(x+2\right)^2}-\sqrt{x^2}=x+2-x=2\)
c) \(\sqrt{\frac{x^2-2x+1}{x-1}}=\sqrt{\frac{\left(x-1\right)^2}{x-1}}=\sqrt{x-1}\)
(Nhớ k cho mình với nhá!)
19 tháng 6 2015
\(P=\frac{\sqrt{x}\left(\sqrt{x^3}+1\right)}{\left(x-\sqrt{x}+1\right)}+1-\frac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}\)
\(P=\frac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}+1-2\sqrt{x}-1\)
\(P=\sqrt{x}\left(\sqrt{x}+1\right)-2\sqrt{x}=x+\sqrt{x}-2\sqrt{x}=x-\sqrt{x}\)
3 tháng 8 2019
\(\sqrt{3}+\sqrt{8-2\sqrt{15}}\\ =\sqrt{3}+\sqrt{5-2\sqrt{5\cdot3}+3}\\ =\sqrt{3}+\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\\ =\sqrt{3}+\sqrt{5}-\sqrt{3}=\sqrt{5}\)
\(\sqrt{x-1-2\sqrt{x-2}}\left(x\ge2\right)\\ =\sqrt{x-2-2\sqrt{x-2}+1}\\ =\sqrt{\left(\sqrt{x-2}-1\right)^2}\\ =\left|\sqrt{x-2}-1\right|\\ =\left[{}\begin{matrix}\sqrt{x-2}-1\left(\sqrt{x-2}\ge1\Leftrightarrow x\ge3\right)\\1-\sqrt{x-2}\left(\sqrt{x-2}< 1\Leftrightarrow2\le x< 3\right)\end{matrix}\right.\)
Chúc bạn học tốt nha
.
S
21 tháng 8 2018
\(A=\dfrac{\sqrt{2+\sqrt{4-x^2}}\left(\sqrt{\left(2+x\right)^3}-\sqrt{\left(2-x\right)^3}\right)}{4+\sqrt{4-x^2}}\)
\(\Rightarrow A=\sqrt{\left(2+x\right)^{^{ }3}}-\sqrt{\left(2-x\right)^3}=\left(\sqrt{2+x}-\sqrt{2-x}\right)\left(4+\sqrt{4-x^2}\right)\)
\(\Rightarrow A=\dfrac{\sqrt{4+2\sqrt{4-x^2}}\left(\sqrt{2+x}-\sqrt{2-x}\right)\left(4+\sqrt{4-x^2}\right)}{\sqrt{2}\left(4+\sqrt{4-x^2}\right)}\)
\(\Rightarrow A=\dfrac{\left(\sqrt{2+x}+\sqrt{2-x}\right)\left(\sqrt{2+x}-\sqrt{2-x}\right)}{\sqrt{2}}=2\sqrt{2}\)
lm trên symbolab.com (thêm simplify là ra)
\(\sqrt{2x-2\sqrt{x^2-4}}+\sqrt{x-2}=\sqrt{\left(\sqrt{x-2}-\sqrt{x+2}\right)^2}+\sqrt{x-2}\)
\(=\left|\sqrt{x-2}-\sqrt{x+2}\right|+\sqrt{x-2}\)
\(=-\sqrt{x-2}+\sqrt{x+2}+\sqrt{x-2}\)
\(=\sqrt{x+2}\)