Tìm ĐKXĐ và rút gọn A:
A=\(\sqrt{x+4\sqrt{x-4}}+\sqrt{x-\sqrt{x-4}}\)
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V
1
31 tháng 5 2021
Đk: \(x\ge4\)
\(A=\sqrt{x+4\sqrt{x-4}}+\sqrt{x-4\sqrt{x-4}}\)
\(=\sqrt{\left(x-4\right)+4\sqrt{x-4}+4}+\sqrt{\left(x-4\right)-4\sqrt{x-4}+4}\)
\(=\sqrt{\left(\sqrt{x-4}+2\right)^2}+\sqrt{\left(\sqrt{x-4}-2\right)^2}\)
\(=\sqrt{x-4}+2+\left|\sqrt{x-4}-2\right|\)
TH1:\(\sqrt{x-4}>2\Leftrightarrow x>8\)
\(A=\sqrt{x-4}+2+\sqrt{x-4}-2=2\sqrt{x-4}\)
TH2:\(\sqrt{x-4}\le2\Leftrightarrow4\le x\le8\)
\(A=\sqrt{x-4}+2-\left(\sqrt{x-4}-2\right)=4\)
Vậy...
5 tháng 8 2023
Sửa đề: \(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{\sqrt{x}+2}\right):\dfrac{2\sqrt{x}}{x-4}\)
ĐKXĐ: x>0; x<>4
\(A=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)+\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{2\sqrt{x}}\)
\(=\dfrac{x+2\sqrt{x}+x-2\sqrt{x}}{2\sqrt{x}}=\dfrac{2x}{2\sqrt{x}}=\sqrt{x}\)
HQ
Hà Quang Minh
Giáo viên
5 tháng 8 2023
Điều kiện: x>2, \(x\ne4\)
\(A=\left(\dfrac{\sqrt{x}}{\sqrt{x-2}}+\dfrac{\sqrt{x}}{\sqrt{x+2}}\right):\dfrac{2\sqrt{x}}{x-4}\\ \Rightarrow A=\sqrt{x}\cdot\dfrac{\sqrt{x+2}+\sqrt{x-2}}{\sqrt{x^2-4}}\cdot\dfrac{x-4}{2\sqrt{x}}\\ \Rightarrow A=\dfrac{\left(x-4\right)\left(\sqrt{x+2}+\sqrt{x-2}\right)}{2\sqrt{x^2-4}}\)
29 tháng 6 2018
dk , x lơn hơn hoặc = 0 , x khác 4
\(\frac{\sqrt{x}}{\sqrt{x-2}}\times\frac{x-4}{2\sqrt{x}}+\frac{\sqrt{x}}{\sqrt{x+2}}\times\frac{x-4}{2\sqrt{x}}.\)
có \(x-4=\left(\sqrt{x}-2\right)\left(\sqrt{x+2}\right)\)
\(\frac{\sqrt{x}}{\sqrt{x}-2}\times\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{2\sqrt{x}}+\frac{\sqrt{x}}{\sqrt{x}+2}+\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{2\sqrt{x}}\)
rút gọn
\(\frac{\left(\sqrt{x}+2\right)}{2}+\frac{\left(\sqrt{x}-2\right)}{2}\)
\(\frac{2\sqrt{x}}{2}\)
3 tháng 2 2021
Điều kiện: x>2
P= \(\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{\sqrt{2}+2}{\sqrt{x}-1}\right)\)
P= \(\left(\dfrac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\dfrac{x-1-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\right)\)
P= \(\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{3}\)
P= \(\dfrac{\sqrt{x}-2}{3\sqrt{x}}\)
b) P= \(\dfrac{1}{4}\)
⇔\(\dfrac{\sqrt{x}-2}{3\sqrt{x}}\) =\(\dfrac{1}{4}\)
⇔\(4\sqrt{x}-8=3\sqrt{x}\)
⇔\(\sqrt{x}=8\)
⇔x=64 (TM)
Vậy X=64(TMĐK) thì P=\(\dfrac{1}{4}\)
11 tháng 7 2021
a, \(P=\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{2+5\sqrt{x}}{4-x}\)ĐK : \(x\ge0;x\ne4\)
\(=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)+2\sqrt{x}\left(\sqrt{x}-2\right)-2-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x+3\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{3x-6\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{3\sqrt{x}}{\sqrt{x}+2}\)
b, Ta có :
\(P=2\Rightarrow\dfrac{3\sqrt{x}}{\sqrt{x}+2}=2\Rightarrow3\sqrt{x}=2\sqrt{x}+4\Leftrightarrow\sqrt{x}=4\Leftrightarrow x=16\)( tmđk )
Vậy P = 2 thì x = 16
TH
2
TG
12 tháng 7 2021
a) a ≠ 1; a ≥ 0
\(\dfrac{a-5\sqrt{a}+4}{a-1}=\dfrac{a-\sqrt{a}-4\sqrt{a}+4}{a-1}=\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)-4\left(\sqrt{a}-1\right)}{a-1}=\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-4\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}=\dfrac{\sqrt{a}-4}{\sqrt{a}+1}\)
b) a ≥ 0; \(x\ne\pm\sqrt{3}\)
\(\dfrac{\sqrt{x^2+2\sqrt{3x}+3}}{x^2-3}=\dfrac{x+\sqrt{3}}{\left(x+\sqrt{3}\right)\left(x-\sqrt{3}\right)}=\dfrac{1}{x-\sqrt{3}}\)
12 tháng 7 2021
1) ĐKXĐ: \(\left\{{}\begin{matrix}a\ge0\\a\ne1\end{matrix}\right.\)
Ta có: \(\dfrac{a-5\sqrt{a}+4}{a-1}\)
\(=\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-4\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\)
\(=\dfrac{\sqrt{a}-4}{\sqrt{a}+1}\)
2) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne\sqrt{3}\end{matrix}\right.\)
Ta có: \(\dfrac{\sqrt{x^2+2\sqrt{3x}+3}}{x^2-3}\)
\(=\dfrac{x+\sqrt{3}}{\left(x+\sqrt{3}\right)\left(x-\sqrt{3}\right)}\)
\(=\dfrac{1}{x-\sqrt{3}}\)
14 tháng 7 2016
a) ĐKXĐ : \(0\le x\ne4\)
b) \(A=\left(\frac{\sqrt{x}}{\sqrt{x}+2}+\frac{\sqrt{x}}{2-\sqrt{x}}+\frac{4\sqrt{x}-1}{x-4}\right):\frac{1}{x-4}\)
\(=\left[\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{4\sqrt{x}-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right].\left(x-4\right)\)
\(=\frac{x-2\sqrt{x}-x-2\sqrt{x}+4\sqrt{x}-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)\)
\(=\frac{-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)=-1\)
13 tháng 7 2016
\(A=\left[\frac{\left(\sqrt{x}-2\right)\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{4\sqrt{x}-1}{x-4}\right]:\frac{1}{x-4}\)
\(=\frac{x-2\sqrt{x}-x-2\sqrt{x}+4\sqrt{x}-1}{x-4}.\left(x-4\right)\)=\(=\frac{-1}{x-4}.\left(x-4\right)=-1\)
Vậy giá trị của A thỏa mãn mọi x và rút gọn lại còn -1
15 tháng 8 2023
1) ĐKXĐ của phân thức là : \(\left\{{}\begin{matrix}\sqrt{x}\ge0\\\sqrt{x}-3\ne0\\x-9\ne0\\\sqrt{x}+3\ne0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\sqrt{x}\ne3\\\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)\ne0\\\sqrt{x}\ne-3\left(LĐ\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne9\end{matrix}\right.\)
Ta có : \(P=\dfrac{\sqrt{x}}{\sqrt{x}-3}:\left(\dfrac{x+3}{x-9}+\dfrac{1}{\sqrt{x}+3}\right)\)
\(P=\dfrac{\sqrt{x}}{\sqrt{x}-3}:\left(\dfrac{x+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\dfrac{\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right)\)
\(P=\dfrac{\sqrt{x}}{\sqrt{x}-3}:\dfrac{x+3+\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(P=\dfrac{\sqrt{x}}{\sqrt{x}-3}:\dfrac{x+\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(P=\dfrac{\sqrt{x}}{\sqrt{x}-3}.\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\sqrt{x}.\left(\sqrt{x}+1\right)}\)
\(P=\dfrac{\sqrt{x}+3}{\sqrt{x}+1}\)
2) Với \(x=4-2\sqrt{3}=3-2\sqrt{3}+1=\left(\sqrt{3}-1\right)^2\)
\(\Rightarrow\sqrt{x}=\sqrt{3}-1\)
Do đó : \(P=\dfrac{\sqrt{3}-1+3}{\sqrt{3}-1+1}\)
\(P=\dfrac{\sqrt{3}+2}{\sqrt{3}}=\dfrac{3+2\sqrt{3}}{3}\)
3) Xét hiệu của : P với 3
\(\dfrac{\sqrt{x}+3}{\sqrt{x}+1}-3\)
\(=\dfrac{-2\sqrt{x}}{\sqrt{x}+1}\)
Ta thấy : \(\sqrt{x}+1\ge1;-2\sqrt{x}\le0\)
\(\Rightarrow\dfrac{-2\sqrt{x}}{\sqrt{x}+1}\le0\)
\(\Rightarrow P\le3\)
Dấu bằng xảy ra : \(\Leftrightarrow x=0\). Thế lại ta thấy ktm nên P<3
CT
18 tháng 8 2021
a. \(N=\left(\dfrac{x+2}{x\sqrt{x}+1}-\dfrac{1}{\sqrt{x}+1}\right).\dfrac{4\sqrt{x}}{3}\) \(\left(ĐKXĐ:x\ge0\right)\)
\(N=\left(\dfrac{x+2}{x\sqrt{x}+1}-\dfrac{x-\sqrt{x}+1}{x\sqrt{x}+1}\right).\dfrac{4\sqrt{x}}{3}\)
\(\text{}\text{}N=\dfrac{\sqrt{x}+1}{x\sqrt{x}+1}.\dfrac{4\sqrt{x}}{3}\)
\(N=\dfrac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}\)
b.\(N=\dfrac{8}{9}\Leftrightarrow\dfrac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}=\dfrac{8}{9}\)
\(\Leftrightarrow3\sqrt{x}=2x-2\sqrt{x}+2\)
\(\Leftrightarrow\left(2\sqrt{x}-1\right)\left(\sqrt{x}-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=4\end{matrix}\right.\)
c.\(\dfrac{1}{N}>\dfrac{3\sqrt{x}}{4}\Leftrightarrow\dfrac{3\left(x-\sqrt{x}+1\right)}{4\sqrt{x}}>\dfrac{3\sqrt{x}}{4}\)
\(\Leftrightarrow x-\sqrt{x}+1>x\)
\(\Leftrightarrow x< 1\)
18 tháng 8 2021
a: ĐKXĐ: \(x\ge0\)
Ta có: \(N=\left(\dfrac{x+2}{x\sqrt{x}+1}-\dfrac{1}{\sqrt{x}+1}\right)\cdot\dfrac{4\sqrt{x}}{3}\)
\(=\dfrac{x+2-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\dfrac{4\sqrt{x}}{3}\)
\(=\dfrac{4\sqrt{x}}{3x-3\sqrt{x}+3}\)
Sửa đề : \(A=\sqrt{x+4\sqrt{x-4}+}\sqrt{x-4\sqrt{x-4}}\)
\(ĐKXĐ:x\ge4\)
\(A=\sqrt{x+4\sqrt{x-4}}+\sqrt{x-4\sqrt{x-4}}=\sqrt{x+2.2\sqrt{x-4}}+\sqrt{x-2.2\sqrt{x-4}}\)
\(A=\sqrt{x-4+2.2\sqrt{x-4+4}}+\sqrt{x-4-2.2\sqrt{x-4+4}}\)
\(A=\sqrt{\left(\sqrt{x-4}\right)^2+2.2\sqrt{x-4}+2^2}+\sqrt{\left(\sqrt{x-4}^2\right)-2.2\sqrt{x-4}+2^2}\)
\(A=\sqrt{\left(\sqrt{x-4}+2\right)^2}+\sqrt{\left(\sqrt{x-4}-2\right)^2}=\sqrt{x-4}+2+\left|\sqrt{x-4}-2\right|\)
\(A=\hept{\begin{cases}2\sqrt{x-4}\\4\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge8\\4\le x< 8\end{cases}}\)