rút gọn biểu thức: \(\sqrt{x+\sqrt{x-4}}+\sqrt{x-\sqrt{x-4}}\)
K
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Những câu hỏi liên quan
25 tháng 7 2021
\(A=\left(\sqrt{x}+\dfrac{4\sqrt{x}}{\sqrt{x}-2}\right):\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{4}{2\sqrt{x}-x}\right)\)ĐK : x > 0 ; x \(\ne\)4
\(=\left(\dfrac{x+2\sqrt{x}}{\sqrt{x}-2}\right):\left(\dfrac{x-4}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)=\dfrac{x\left(x-4\right)}{\left(\sqrt{x}-2\right)\left(x-4\right)}\)
\(=\dfrac{x}{\sqrt{x}-2}\)
LD
26 tháng 7 2021
đk \(\left\{{}\begin{matrix}x\ne1\\x>0\end{matrix}\right.\)
A= \(\dfrac{-x\left(1+\sqrt{x}\right)}{\sqrt{x}\left(1-x\right)}\)+\(\dfrac{3\sqrt{x}\left(1-\sqrt{x}\right)}{\left(1-x\right)\sqrt{x}}\)+\(\dfrac{\left(6\sqrt{x}-4\right)\sqrt{x}}{\left(1-x\right)\sqrt{x}}\)
=\(\dfrac{-x-x\sqrt{x}+3\sqrt{x}-3x+6x-4\sqrt{x}}{\left(1-x\right)\sqrt{x}}\)
=\(\dfrac{-\left(x-2\sqrt{x}=1\right)}{1-x}\)=-\(\dfrac{\left(\sqrt{x}-1\right)^2}{1-x}\)=\(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
26 tháng 7 2021
Ta có: \(A=\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3\sqrt{x}}{x+\sqrt{x}}+\dfrac{6\sqrt{x}-4}{1-x}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}+1}-\dfrac{6\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
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 \(0\le x< 4\Rightarrow\sqrt{x-2}< \sqrt{2}\)
\(\Rightarrow A=\sqrt{x-2}+\sqrt{2}-\sqrt{x-2}+\sqrt{2}=2\sqrt{2}\)
20 tháng 12 2022
\(=\dfrac{x-4\sqrt{x}+4\sqrt{x}+16}{x-16}\cdot\dfrac{\sqrt{x}+2}{\sqrt{x}+16}=\dfrac{\left(x+16\right)\left(\sqrt{x}+2\right)}{\left(x-16\right)\left(\sqrt{x}+16\right)}\)
18 tháng 2 2022
đk : x >= 0, x khác 4
\(=\dfrac{x+2\sqrt{x}-\left(x-\sqrt{x}-2\right)-\sqrt{x}-4}{x-4}\)
\(=\dfrac{2\sqrt{x}-2}{x-4}=\dfrac{2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
BC
2
9 tháng 9 2021
\(\sqrt{x}-\sqrt{x-\sqrt{x}+\dfrac{1}{4}}\left(đk:x\ge0\right)\left(1\right)\)
\(=\sqrt{x}-\sqrt{\left(\sqrt{x}-\dfrac{1}{2}\right)^2}\)
\(=\sqrt{x}-\left|\sqrt{x}-\dfrac{1}{2}\right|\)
TH1: \(x\ge\dfrac{1}{4}\)
\(\left(1\right)=\sqrt{x}-\sqrt{x}+\dfrac{1}{2}=\dfrac{1}{2}\)
TH2: \(0\le x< \dfrac{1}{4}\)
\(\left(1\right)=\sqrt{x}+\sqrt{x}-\dfrac{1}{2}=2\sqrt{x}-\dfrac{1}{2}\)
NV
Nguyễn Việt Lâm
Giáo viên
9 tháng 9 2021
\(=\sqrt{x}-\sqrt{\left(\sqrt{x}-\dfrac{1}{2}\right)^2}=\sqrt{x}-\left|\sqrt{x}-\dfrac{1}{2}\right|\)
\(=\left[{}\begin{matrix}\dfrac{1}{2}\text{ nếu }x\ge\dfrac{1}{4}\\2\sqrt{x}-\dfrac{1}{2}\text{ nếu }0\le x< \dfrac{1}{4}\end{matrix}\right.\)
6 tháng 2 2022
Câu 1: \(\sqrt{8}\) − \(\sqrt{18}\) + \(2\sqrt{32}\) = \(\sqrt{4\text{×}2}\) − \(\sqrt{\text{9×2}}\) + 2\(\sqrt{\text{16×2}}\)
=2\(\sqrt{2}\) − 3\(\sqrt{2}\) + 2×4\(\sqrt{2}\)
=(2− 3+ 8)\(\sqrt{2}\)
=7\(\sqrt{2}\)
Câu 2: Mik ko chắc làm đúng hay ko 
nên ko làm
7 tháng 10 2021
a) \(P=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
b) \(P=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=-1\)
\(\Leftrightarrow-\sqrt{x}-1=\sqrt{x}-1\Leftrightarrow2\sqrt{x}=0\Leftrightarrow x=0\left(tm\right)\)
c) \(P=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}\in Z\)
\(\Rightarrow\sqrt{x}+1\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\)
Kết hợp đk:
\(\Rightarrow x\in\left\{0\right\}\)
d) \(P=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=\dfrac{\left(\sqrt{x}+1\right)-2}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}< 1\)
7 tháng 10 2021
\(a,P=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ P=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\\ b,P=-1\Leftrightarrow\sqrt{x}-1=-\sqrt{x}-1\\ \Leftrightarrow2\sqrt{x}=0\Leftrightarrow x=0\left(tm\right)\\ c,P\in Z\Leftrightarrow\dfrac{\sqrt{x}+1-2}{\sqrt{x}+1}\in Z\Leftrightarrow1-\dfrac{2}{\sqrt{x}+1}\in Z\\ \Leftrightarrow2⋮\sqrt{x}+1\\ \Leftrightarrow\sqrt{x}+1\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\\ \Leftrightarrow\sqrt{x}+1\in\left\{1;2\right\}\left(\sqrt{x}+1\ge1\right)\\ \Leftrightarrow\sqrt{x}\in\left\{0;1\right\}\\ \Leftrightarrow x\in\left\{0;1\right\}\)
\(d,P=\dfrac{\sqrt{x}+1-2}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}\)
Có \(\dfrac{2}{\sqrt{x}+1}>0\left(2>0;\sqrt{x}+1>0\right)\Leftrightarrow1-\dfrac{2}{\sqrt{x}+1}< 1\Leftrightarrow P< 1\)
\(e,P=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}\)
Có \(\sqrt{x}+1\ge1\Leftrightarrow\dfrac{2}{\sqrt{x}+1}\le2\Leftrightarrow1-\dfrac{2}{\sqrt{x}+1}\ge1-2=-1\)
\(P_{min}=-1\Leftrightarrow x=0\)
N
23 tháng 11 2021
\(\dfrac{2\left(2-\sqrt{x}\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}+\dfrac{2\sqrt{x}\left(2+\sqrt{x}\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}-\dfrac{2x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}=\dfrac{4-2\sqrt{x}+4\sqrt{x}+2x-2x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}=\dfrac{4-2\sqrt{x}}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}=\dfrac{2\left(2-\sqrt{x}\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}=\dfrac{2}{2+\sqrt{x}}\)
ĐKXĐ: x >= 4
Bình phương hai vế ta có : x + \(\sqrt{x-4}\)+ x - \(\sqrt{x-4}\)
= 2x
Theo Bình phương 2 vế ta có :
\(\sqrt{x-4}\)+ X - \(\sqrt{x-4}\)
~ Hok tốt ~
#Gumball