Rút gọn B=\(\sqrt{2x-2\sqrt{x^2-4}}+\sqrt{x-2}\)
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ND
1
AT
1 tháng 7 2021
ĐKXĐ: \(x\ge2\)
\(A=\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}\)
\(=\sqrt{x-2+2.\sqrt{x-2}.\sqrt{2}+2}+\sqrt{x-2-2.\sqrt{x-2}.\sqrt{2}+2}\)
\(=\sqrt{\left(\sqrt{x-2}+\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{x-2}-\sqrt{2}\right)^2}\)
\(=\left|\sqrt{x-2}+\sqrt{2}\right|+\left|\sqrt{x-2}-\sqrt{2}\right|=\sqrt{x-2}+\sqrt{2}+\left|\sqrt{x-2}-\sqrt{2}\right|\)
Xét \(x\ge4\Rightarrow\sqrt{x-2}\ge\sqrt{2}\)
\(\Rightarrow A=\sqrt{x-2}+\sqrt{2}+\sqrt{x-2}-\sqrt{2}=2\sqrt{x-2}\)
Xét \(0\le x< 4\Rightarrow\sqrt{x-2}< \sqrt{2}\)
\(\Rightarrow A=\sqrt{x-2}+\sqrt{2}-\sqrt{x-2}+\sqrt{2}=2\sqrt{2}\)
H9
HT.Phong (9A5)
CTVHS
7 tháng 9 2023
a) \(3\sqrt{2x}-4\sqrt{2x}+8-2\sqrt{x}\)
\(=-\left(4\sqrt{2x}-3\sqrt{2x}\right)+8-2\sqrt{x}\)
\(=-\sqrt{2x}-2\sqrt{x}+8\)
b) \(3\sqrt{2x}-\sqrt{72x}+3\sqrt{18x}+18\)
\(=3\sqrt{2x}-6\sqrt{2x}+3\cdot3\sqrt{2x}+18\)
\(=3\sqrt{2x}-6\sqrt{2x}+9\sqrt{2x}+18\)
\(=\left(3+9-6\right)\sqrt{2x}+18\)
\(=6\sqrt{2x}+18\)
em cảm mơn nhìu nhìu 
KH
4 tháng 7 2021
\(1.\\ A=\sqrt{\left(2+\sqrt{3}\right)^2}+\sqrt{\left(2-\sqrt{3}\right)^2}\\ =\left|2+\sqrt{3}\right|+\left|2-\sqrt{3}\right|\\ =2+\sqrt{3}+2-\sqrt{3}=4\)
\(2.\\a.\\ P=3x-\sqrt{\left(x-5\right)^2}=3x-\left|x-5\right|\\ b.\\ x=2\Rightarrow P=3\)
\(3.\\ M=\dfrac{\sqrt{\left(x-1\right)^2}}{x-1}=\dfrac{\left|x-1\right|}{x-1}\)
\(\cdot x>1\Rightarrow M=1\\ \cdot x=1\Rightarrow M=0\\\cdot x< 1\Rightarrow M=-1\)
4 tháng 7 2021
B1.
Ta có:A\(=\sqrt{3+4\sqrt{3}+4}+\sqrt{3-4\sqrt{3}+4}\)
\(=\sqrt{\left(\sqrt{3}+2\right)^2}+\sqrt{\left(\sqrt{3}-2\right)^2}\)
\(=\sqrt{3}+2+\sqrt{3}-2=2\sqrt{3}\)
5 tháng 5 2021
Câu 1:
Sửa đề: \(B=\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)
Ta có: \(B=\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)
\(=\left(\dfrac{x}{\sqrt{x}\left(\sqrt{x}+3\right)}+\dfrac{1}{\sqrt{x}+3}\right):\left(\dfrac{x+3\sqrt{x}-2\left(\sqrt{x}+3\right)+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\right)\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}:\dfrac{x+3\sqrt{x}-2\sqrt{x}-6+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{x+\sqrt{x}}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}=1\)
5 tháng 5 2021
Câu 3:
Ta có: \(Q=\left(\dfrac{a}{a-2\sqrt{a}}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{a-4\sqrt{a}+4}\)
\(=\left(\dfrac{a}{\sqrt{a}\left(\sqrt{a}-2\right)}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{\left(\sqrt{a}-2\right)^2}\)
\(=\dfrac{a+\sqrt{a}}{\sqrt{a}-2}\cdot\dfrac{\sqrt{a}-2}{\sqrt{a}+1}\cdot\dfrac{\sqrt{a}-2}{1}\)
\(=\sqrt{a}\left(\sqrt{a}-2\right)\)
\(=a-2\sqrt{a}\)
8 tháng 10 2017
1.
a. ĐKXĐ : x lớn hơn hoặc bằng 1/2
b. A\(\sqrt{2}\)= \(\sqrt{2x+2\sqrt{2x-1}}-\sqrt{2x-2\sqrt{2x-1}}\)
= \(\sqrt{2x-1+1+2\sqrt{2x-1}}-\sqrt{2x-1+1-2\sqrt{2x-1}}\)
=\(\sqrt{\left(\sqrt{2x-1}+1\right)^2}-\sqrt{\left(\sqrt{2x-1}-1\right)^2}\)
= \(\sqrt{2x-1}+1-\left|\sqrt{2x-1}-1\right|\)
Nếu \(x\ge1thìA\sqrt{2}=\sqrt{2x-1}+1-\left(\sqrt{2x-1}-1\right)=2\)
\(\Rightarrow A=2\)
Nếu 1/2 \(\le x< 1thìA\sqrt{2}=\sqrt{2x-1}+1-\left(1-\sqrt{2x-1}\right)=2\sqrt{2x-1}\)
Do đó : A= \(\sqrt{4x-2}\)
Vậy ............
8 tháng 10 2017
2.
a. \(x\ge2\)hoặc x<0
b. A= \(2\sqrt{x^2-2x}\)
c. A<2 \(\Leftrightarrow\)\(2\sqrt{x^2-2x}< 2\Leftrightarrow\sqrt{x^2-2x}< 1\Leftrightarrow x^2-2x< 1\Leftrightarrow\left(x-1\right)^2< 2\)
\(-\sqrt{2}< x-1< \sqrt{2}\Leftrightarrow1-\sqrt{2}< x< 1+\sqrt{2}\)
Kết hợp vs đk câu a , ta đc : \(1-\sqrt{2}< x< 0và2\le x< 1+\sqrt{2}\)
Vậy...........
AH
Akai Haruma
Giáo viên
26 tháng 6 2021
Câu a, bạn coi lại đề xem $a^2=6-3\sqrt{3}$ hay $a=6-3\sqrt{3}$???
AH
Akai Haruma
Giáo viên
26 tháng 6 2021
b.
\(B=\frac{\sqrt{(x-2)+(x+2)+2\sqrt{(x-2)(x+2)}}}{\sqrt{x^2-4}+x+2}\)
\(=\frac{\sqrt{(\sqrt{x-2}+\sqrt{x+2})^2}}{\sqrt{x^2-4}+x+2}=\frac{\sqrt{x-2}+\sqrt{x+2}}{\sqrt{x^2-4}+x+2}=\frac{\sqrt{x-2}+\sqrt{x+2}}{\sqrt{x+2}(\sqrt{x-2}+\sqrt{x+2})}=\frac{1}{\sqrt{x+2}}\)
\(=\frac{1}{\sqrt{3+\sqrt{5}}}=\frac{\sqrt{2}}{\sqrt{6+2\sqrt{5}}}=\frac{\sqrt{2}}{\sqrt{(\sqrt{5}+1)^2}}=\frac{\sqrt{2}}{\sqrt{5}+1}\)
30 tháng 6 2019
\(A=\sqrt{3+\sqrt{5}}+\sqrt{7-3.\sqrt{5}}-\sqrt{2}\)
\(\sqrt{2}.A=\sqrt{5+2\sqrt{5}.1+1}+\sqrt{9-2.3.\sqrt{5}+5}-2\)
\(\sqrt{2}.A=\sqrt{5}+1+3-\sqrt{5}-2=2\)
\(\Rightarrow A=\sqrt{2}\)
ĐKXĐ: \(\hept{\begin{cases}2x-4\ge0\\x+2.\sqrt{2x-4}\ge0\\x-2\sqrt{2x-4}\end{cases}}\Leftrightarrow x\ge2\)
\(\sqrt{x+2.\sqrt{2x-4}}+\sqrt{x-2.\sqrt{2x-4}}\)
\(=\sqrt{x-2+2.\sqrt{x-2}.\sqrt{2}+2}+\sqrt{x-2-2.\sqrt{x-2}.\sqrt{2}+2}\)
\(=\sqrt{x-2}+\sqrt{2}+\left|\sqrt{x-2}-\sqrt{2}\right|\)
Tự phá trị tuyệt đối
PJ
2
2
18 tháng 5 2022
`[2x+\sqrt{2}]/[4x^2+4\sqrt{2}x+\sqrt{2}]`
`=[\sqrt{2}(\sqrt{2}x+1)]/[\sqrt{2}(2\sqrt{2}x^2+4x+1)]`
`=[\sqrt{2}x+1]/[2\sqrt{2}x^2+4x+1]`
ĐKXĐ: \(x\ge2\)
\(\sqrt{2x-2\sqrt{x^2-4}}+\sqrt{x-2}\)
\(=\sqrt{x+2-2\sqrt{\left(x+2\right)\left(x-2\right)}+x-2}+\sqrt{x-2}\)
\(=\sqrt{\left(\sqrt{x+2}-\sqrt{x-2}\right)^2}+\sqrt{x-2}\)
\(=\sqrt{x+2}-\sqrt{x-2}+\sqrt{x-2}\)
\(=\sqrt{x+2}\)