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18 tháng 8 2021
Bài 1:
a: Ta có: \(P=\left(\dfrac{\sqrt{x}}{x\sqrt{x}-1}+\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}\)
\(=\dfrac{\sqrt{x}+x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{\sqrt{x}+1}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
23 tháng 10 2023
\(\dfrac{6}{\sqrt{2}-\sqrt{3}+3}\)
\(=\dfrac{6\left(\sqrt{2}-\sqrt{3}-3\right)}{\left(\sqrt{2}-\sqrt{3}\right)^2-9}\)
\(=\dfrac{6\left(\sqrt{2}-\sqrt{3}-3\right)}{-4-2\sqrt{6}}\)
\(=\dfrac{-3\left(\sqrt{2}-\sqrt{3}-3\right)}{2+\sqrt{6}}=\dfrac{-3\left(\sqrt{6}-2\right)\left(\sqrt{2}-\sqrt{3}-3\right)}{2}\)
LT
1
HP
5 tháng 1 2021
ĐK: \(x>0;x\ne1\)
\(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{1}{x-\sqrt{x}}\right):\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{x-1}\right)\)
\(=\left[\dfrac{\sqrt{x}.\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right]:\left[\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right]\)
\(=\dfrac{x\sqrt{x}-x-\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x\sqrt{x}-x-\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
19 tháng 8 2023
1:
\(A=\sqrt{x^2+\dfrac{2x^2}{3}}=\sqrt{\dfrac{5x^2}{3}}=\left|\sqrt{\dfrac{5}{3}}x\right|=-x\sqrt{\dfrac{5}{3}}\)
2: \(=\left(\dfrac{\sqrt{100}+\sqrt{40}}{\sqrt{5}+\sqrt{2}}+\sqrt{6}\right)\cdot\dfrac{2\sqrt{5}-\sqrt{6}}{2}\)
\(=\dfrac{\left(2\sqrt{5}+\sqrt{6}\right)\left(2\sqrt{5}-\sqrt{6}\right)}{2}\)
\(=\dfrac{20-6}{2}=7\)
LK
1
NT
0
31 tháng 8 2021
a: Ta có: \(\sqrt{2x-1}=4\)
\(\Leftrightarrow2x-1=16\)
\(\Leftrightarrow2x=17\)
hay \(x=\dfrac{17}{2}\)
b: Ta có: \(\sqrt{4x+4}-\sqrt{9x+9}=-6\)
\(\Leftrightarrow-\sqrt{x+1}=-6\)
\(\Leftrightarrow x+1=36\)
hay x=35
KK
0
11 tháng 7 2021
a, \(\sqrt{23-8\sqrt{7}}\)
\(=\sqrt{16-8\sqrt{7}+7}\)
\(=\sqrt{4^2-2.4.\sqrt{7}+\left(\sqrt{7}\right)^2}\)
\(=\sqrt{\left(4-\sqrt{7}\right)^2}\)
\(=\left|4-\sqrt{7}\right|\)
\(=4-\sqrt{7}\)
b, \(\sqrt{18-8\sqrt{2}}\)
\(=\sqrt{16-8\sqrt{2}+2}\)
\(=\sqrt{4^2-2.4.\sqrt{2}+\left(\sqrt{2}\right)^2}\)
\(=\sqrt{\left(4-\sqrt{2}\right)^2}\)
\(=\left|4-\sqrt{2}\right|\)
\(=4-\sqrt{2}\)
c, \(\sqrt{12+6\sqrt{3}}\)
\(=\sqrt{9+6\sqrt{3}+3}\)
\(=\sqrt{3^2+2.3.\sqrt{3}+\left(\sqrt{3}\right)^2}\)
\(=\sqrt{\left(3+\sqrt{3}\right)^2}\)
\(=\left|3+\sqrt{3}\right|\)
\(=3+\sqrt{3}\)
d, \(\sqrt{17+12\sqrt{2}}\)
\(=\sqrt{9+12\sqrt{2}+8}\)
\(=\sqrt{3^2+3.2.2\sqrt{2}+\left(2\sqrt{2}\right)^2}\)
\(=\sqrt{\left(3+2\sqrt{2}\right)^2}\)
\(=\left|3+2\sqrt{2}\right|\)
\(=3+2\sqrt{2}\)
12 tháng 7 2021
a) \(\sqrt{23-8\sqrt{7}}=4-\sqrt{7}\)
b) \(\sqrt{18-8\sqrt{2}}=4-\sqrt{2}\)
c) \(\sqrt{12+6\sqrt{3}}=3+\sqrt{3}\)
d) \(\sqrt{17+12\sqrt{2}}=3+2\sqrt{2}\)
a) \(\sqrt{6+2\sqrt{5}}=\sqrt{5+2\sqrt{5}+1}=\sqrt{\left(\sqrt{5}+1\right)}=\sqrt{5}+1\)
b) \(\sqrt{8-2\sqrt{7}}=\sqrt{7-2\sqrt{7}+1}=\sqrt{\left(\sqrt{7}-1\right)^2}=\sqrt{7}-1\left(do\sqrt{7}>1\right)\)
c) \(\sqrt{9+4\sqrt{5}}=\sqrt{5+2\cdot\sqrt{5}\cdot2+4}=\sqrt{\left(\sqrt{5}+2\right)^2}=2+\sqrt{5}\)