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LT
2
TT
6 tháng 8 2020
Sửa đề :
\(P=\sqrt{x+5+2\sqrt{x+4}}-\sqrt{x+5-2\sqrt{x+4}}\)\(\left(x\ge-4\right)\)
\(=\sqrt{\left(x+4\right)+2\sqrt{x+4}+1}-\sqrt{\left(x+4\right)-2\sqrt{x+4}+1}\)
\(=\sqrt{\left(\sqrt{x+4}+1\right)^2}-\sqrt{\left(\sqrt{x+4}-1\right)^2}\)
\(=\left|\sqrt{x+4}+1\right|-\left|\sqrt{x+4}-1\right|\)
\(=\sqrt{x+4}+1-\sqrt{x+4}+1=2\)
Vậy \(P=2\)
6 tháng 11 2021
a, \(\Rightarrow M=\dfrac{x}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{4\sqrt{x}-4}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(\Rightarrow M=\dfrac{x-4\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(\Rightarrow M=\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(\Rightarrow M=\dfrac{\sqrt{x}-2}{\sqrt{x}}\)
b, \(x=3+2\sqrt{2}\Rightarrow M=\dfrac{\sqrt{3+2\sqrt{2}}-2}{\sqrt{3+2\sqrt{2}}}=\dfrac{\sqrt{2+2\sqrt{2}.1+1}-2}{\sqrt{2+2\sqrt{2}.1+1}}=\dfrac{\sqrt{2}+1-2}{\sqrt{2}+1}=\dfrac{\sqrt{2}-1}{\sqrt{2}+1}=\dfrac{\left(\sqrt{2}-1\right)^2}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}=\dfrac{2-2\sqrt{2}+1}{2-1}=3-2\sqrt{2}\)
c, \(M>0\Rightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}}>0\Rightarrow\sqrt{x}-2>0\Rightarrow\sqrt{x}>2\Rightarrow x>4\)
VA
0
B
21 tháng 8 2017
(14,78-a)/(2,87+a)=4/1
14,78+2,87=17,65
Tổng số phần bằng nhau là 4+1=5
Mỗi phần có giá trị bằng 17,65/5=3,53
=>2,87+a=3,53
=>a=0,66.
21 tháng 8 2017
a,\(\sqrt{x-4+4\sqrt{x-4}+4}\) +\(\sqrt{x-4-4\sqrt{x-4}+4}\)
=\(\sqrt{x-4}+2+\left|\sqrt{x-4}-2\right|\) (vi x>=8)
=\(\sqrt{x-4}+2+\sqrt{x-4}-2=2\sqrt{x-4}\)
b, \(\sqrt{x-1+2\sqrt{x\left(x-1\right)}+x}+\sqrt{x-1-2\sqrt{x\left(x-1\right)}+x}\)
=\(\sqrt{x-1}+\sqrt{x}+\left|\sqrt{x-1}-\sqrt{x}\right|\)
=\(\sqrt{x}+\sqrt{x-1}+\sqrt{x}-\sqrt{x-1}\) =\(2\sqrt{x}\)
c,d sai dau bai hay sao y
HL
1
23 tháng 9 2021
\(A=\dfrac{3x}{x-2}\cdot\sqrt{x^2-4x+4}\)
\(=\dfrac{3x}{x-2}\cdot\left(x-2\right)\)
=3x
\(B=\dfrac{-5y}{x+3}\cdot\sqrt{x^2+6x+9}\)
\(=\dfrac{-5y}{x+3}\cdot\left|x+3\right|\)
\(=\pm5y\)
22 tháng 6 2017
a, \(\sqrt{\left(\sqrt{2}\right)^2+2\times2\times\sqrt{2}+2^2}\)+ \(\sqrt{2^2+2\times2\times\sqrt{2}+\left(\sqrt{2}\right)^2}\)
= \(\sqrt{\left(\sqrt{2}+2\right)^2}\)+ \(\sqrt{\left(2-\sqrt{2}\right)^2}\)
= \(\sqrt{2}+2+2-\sqrt{2}\)
= 4
27 tháng 10 2023
a: Khi x=25 thì \(A=\dfrac{5-2}{5-3}=\dfrac{3}{2}\)
b: P=A*B
\(=\dfrac{\sqrt{x}-2}{\sqrt{x}-3}\left(\dfrac{6x+6\sqrt{x}-12}{x+5\sqrt{x}+4}-\dfrac{5\sqrt{x}}{\sqrt{x}+4}\right)\)
\(=\dfrac{\sqrt{x}-2}{\sqrt{x}-3}\cdot\left(\dfrac{6x+6\sqrt{x}-12}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+4\right)}-\dfrac{5\sqrt{x}}{\sqrt{x}+4}\right)\)
\(=\dfrac{\sqrt{x}-2}{\sqrt{x}-3}\cdot\dfrac{6x+6\sqrt{x}-12-5x-5\sqrt{x}}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x+\sqrt{x}-12}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}-2}{\sqrt{x}-1}=\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\)
c: \(\sqrt{P}< =\dfrac{1}{2}\)
=>0<=P<=1/4
=>\(\left\{{}\begin{matrix}P>=0\\P-\dfrac{1}{4}< =0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{\sqrt{x}-2}{\sqrt{x}-1}>=0\\\dfrac{\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{1}{4}< =0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left[{}\begin{matrix}x>=4\\0< =x< 1\end{matrix}\right.\\\dfrac{4\left(\sqrt{x}-2\right)-\sqrt{x}+1}{4\left(\sqrt{x}-1\right)}< =0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left[{}\begin{matrix}x>=4\\0< =x< 1\end{matrix}\right.\\\dfrac{3\sqrt{x}-7}{\sqrt{x}-1}< =0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x>=4\\0< =x< 1\end{matrix}\right.\\1< \sqrt{x}< =\dfrac{7}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x>=4\\0< =x< 1\end{matrix}\right.\\1< x< \dfrac{49}{9}\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}\left[{}\begin{matrix}x>=4\\0< =x< 1\end{matrix}\right.\\x=\dfrac{49}{9}\end{matrix}\right.\)
=>\(4< =x< =\dfrac{49}{9}\)
mà x nguyên
nên \(x\in\left\{4;5\right\}\)
DN
10 tháng 8 2015
\(a, A=2\sqrt{x-1}-3\sqrt{x-1}-4\sqrt{x-1}=\left(2-3-4\right)\sqrt{x-1}=-5\sqrt{x-1}\)
\(b, B=\frac{2}{x+y}.\left(x+y\right)\sqrt{\frac{3}{4}}=2\sqrt{\frac{3}{4}}=2.\frac{1}{2}.\sqrt{3}=\sqrt{3}\)
15 tháng 7 2019
\(E=\left(\frac{\sqrt{\sqrt{x}-1}}{\sqrt{\sqrt{x}+1}}+\frac{\sqrt{\sqrt{x}+1}}{\sqrt{\sqrt{x}-1}}\right):\sqrt{\frac{1}{x-1}}\) \(ĐKXĐ:x>1\)
\(E=\left(\frac{\left(\sqrt{\sqrt{x}-1}\right)^2}{\left(\sqrt{\sqrt{x}+1}\right)\left(\sqrt{\sqrt{x}-1}\right)}+\frac{\left(\sqrt{\sqrt{x}+1}\right)^2}{\left(\sqrt{\sqrt{x}-1}\right)\left(\sqrt{\sqrt{x}+1}\right)}\right)\cdot\sqrt{\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{1}}\)
\(E=\left(\frac{\sqrt{x}-1}{\left(\sqrt{\sqrt{x}+1}\right)\left(\sqrt{\sqrt{x}-1}\right)}+\frac{\sqrt{x}+1}{\left(\sqrt{\sqrt{x}-1}\right)\left(\sqrt{\sqrt{x}+1}\right)}\right)\cdot\sqrt{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(E=\frac{\sqrt{x}-1+\sqrt{x}+1}{\left(\sqrt{\sqrt{x}+1}\right)\left(\sqrt{\sqrt{x}-1}\right)}\cdot\sqrt{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(E=\frac{2\sqrt{x}}{\sqrt{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}}\cdot\sqrt{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=2\sqrt{x}\)
Ta có:\(x=19-8\sqrt{3}=16-2.4\sqrt{3}+3=\left(4-\sqrt{3}\right)^2\)
\(\Rightarrow2\sqrt{x}=2.\sqrt{\left(4-\sqrt{3}\right)^2}=2.\left(4-\sqrt{3}\right)=8-2\sqrt{3}\)
3 tháng 2 2022
b: \(=\dfrac{\sqrt{14+6\sqrt{5}}+\sqrt{14-6\sqrt{5}}}{\sqrt{2}}=\dfrac{3+\sqrt{5}+3-\sqrt{5}}{2}=\dfrac{6}{\sqrt{2}}=3\sqrt{2}\)
c: \(=\sqrt{x-1}+1+\sqrt{x-1}-1=2\sqrt{x-1}\)
\(\sqrt{x-4\sqrt{x-4}}\)
= \(\sqrt{\left(\sqrt{x-4}\right)^2-4\sqrt{x-4}+4}\)
= \(\sqrt{\left(\sqrt{x-4}-2\right)^2}\)
= \(\sqrt{x-4}\) - 2
\(\sqrt{x-4\sqrt{x-4}}=\sqrt{x-4-4\sqrt{x-4}+4}\)
\(=\sqrt{\left[\left(x-4\right)+2\right]^2}\)
\(=\left|\left(x-4\right)+2\right|\)
\(=x-4+2=x-2\)