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4 tháng 8 2017
\(\frac{\sqrt{2}}{\sqrt{6}}\)=\(\frac{\sqrt{2}}{\sqrt{2}\sqrt{3}}\)=\(\frac{1}{\sqrt{3}}\)
4 tháng 8 2017
\(\sqrt{\frac{2}{6}=}\frac{\sqrt{3}}{3}\)
\(\sqrt{81}=9\)
\(\sqrt{\frac{81}{27}}=\sqrt{3}\)
\(\sqrt{\frac{2}{4}}=\frac{\sqrt{2}}{2}\)
\(\sqrt{32}=4\sqrt{2}\)
\(\sqrt{42}=\sqrt{42}\)
\(\sqrt{18}=3\sqrt{2}\)
Làm bừa chả biết có đúng không nữa
AT
13 tháng 6 2018
đkxđ: x≥0; x≠4
\(A=\dfrac{1}{2+\sqrt{x}}+\dfrac{1}{2-\sqrt{x}}-\dfrac{2\sqrt{x}}{4-x}\)
\(=\dfrac{2-\sqrt{x}}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}+\dfrac{2+\sqrt{x}}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}-\dfrac{2\sqrt{x}}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\)
\(=\dfrac{4-2\sqrt{x}}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}=\dfrac{2\left(2-\sqrt{x}\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}=\dfrac{2}{2+\sqrt{x}}\)
+) A = 1/4 <=> \(\dfrac{2}{2+\sqrt{x}}=\dfrac{1}{4}\Leftrightarrow2+\sqrt{x}=8\Leftrightarrow\sqrt{x}=6\Leftrightarrow x=36\)(tm)
Vậy x = 36
HC
13 tháng 6 2018
đkxđ \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)
\(A=\dfrac{2+\sqrt{x}+2-\sqrt{x}-2\sqrt{x}}{\left(\sqrt{x}+2\right)\left(2-\sqrt{x}\right)}\)
\(A=\dfrac{4-2\sqrt{x}}{\left(\sqrt{x}+2\right)\left(2-\sqrt{x}\right)}\)
\(A=\dfrac{2}{\sqrt{x}+2}\)
để \(A=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{2}{\sqrt{x}+2}=\dfrac{1}{4}\)
\(\Leftrightarrow\sqrt{x}+2=8\)
\(\Leftrightarrow x=36\left(tm\right)\)
vậy tại x=36 thì A=1/4
KL
1
3 tháng 8 2019
\(\sqrt{3}+\sqrt{8-2\sqrt{15}}\\ =\sqrt{3}+\sqrt{5-2\sqrt{5\cdot3}+3}\\ =\sqrt{3}+\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\\ =\sqrt{3}+\sqrt{5}-\sqrt{3}=\sqrt{5}\)
\(\sqrt{x-1-2\sqrt{x-2}}\left(x\ge2\right)\\ =\sqrt{x-2-2\sqrt{x-2}+1}\\ =\sqrt{\left(\sqrt{x-2}-1\right)^2}\\ =\left|\sqrt{x-2}-1\right|\\ =\left[{}\begin{matrix}\sqrt{x-2}-1\left(\sqrt{x-2}\ge1\Leftrightarrow x\ge3\right)\\1-\sqrt{x-2}\left(\sqrt{x-2}< 1\Leftrightarrow2\le x< 3\right)\end{matrix}\right.\)
Chúc bạn học tốt nha
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4 tháng 8 2018
\(a.D=\dfrac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\dfrac{2a+\sqrt{a}}{\sqrt{a}}+1=\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{a-\sqrt{a}+1}-\dfrac{\sqrt{a}\left(2\sqrt{a}+1\right)}{\sqrt{a}}+1=a+\sqrt{a}-2\sqrt{a}-1+1=a-\sqrt{a}\left(a>0\right)\)
\(b.D=2\Leftrightarrow a-\sqrt{a}-2=0\Leftrightarrow\left(\sqrt{a}+1\right)\left(\sqrt{a}-2\right)=0\Leftrightarrow a=4\left(TM\right)\)
\(c.D=a-\sqrt{a}=\sqrt{a}\left(\sqrt{a}-1\right)>0\left(a>1\right)\)\(\Rightarrow D=\left|D\right|\)
13 tháng 8 2020
Ta có: \(B=21\left(\sqrt{2+\sqrt{3}}+\sqrt{3-\sqrt{5}}\right)^2-6\left(\sqrt{2-\sqrt{3}}+\sqrt{3+\sqrt{5}}\right)^2-15\sqrt{15}\)
\(=21\cdot\left[2+\sqrt{3}+3-\sqrt{5}+2\sqrt{\left(2+\sqrt{3}\right)\left(3-\sqrt{5}\right)}\right]-6\cdot\left[2-\sqrt{3}+3+\sqrt{5}+2\cdot\sqrt{\left(2-\sqrt{3}\right)\left(3+\sqrt{5}\right)}\right]-15\sqrt{15}\)
\(=21\cdot\left(5+\sqrt{3}-\sqrt{5}+\sqrt{\left(4+2\sqrt{3}\right)\left(6-2\sqrt{5}\right)}\right)-6\cdot\left[5-\sqrt{3}+\sqrt{5}+\sqrt{\left(4-2\sqrt{3}\right)\left(6+2\sqrt{5}\right)}\right]-15\sqrt{15}\)
\(=21\cdot\left[5+\sqrt{3}-\sqrt{5}+\left(\sqrt{3}+1\right)\left(\sqrt{5}-1\right)\right]-6\cdot\left[5-\sqrt{3}+\sqrt{5}+\left(\sqrt{3}-1\right)\left(\sqrt{5}+1\right)\right]-15\sqrt{15}\)
\(=21\cdot\left(5+\sqrt{3}-\sqrt{5}+\sqrt{15}-\sqrt{3}+\sqrt{5}-1\right)-6\cdot\left(5-\sqrt{3}+\sqrt{5}+\sqrt{15}+\sqrt{3}-\sqrt{5}-1\right)-15\sqrt{15}\)
\(=21\cdot\left(4+\sqrt{15}\right)-6\left(4+\sqrt{15}\right)-15\sqrt{15}\)
\(=84+21\sqrt{15}-24-6\sqrt{15}-15\sqrt{15}\)
\(=60\)
2 tháng 7 2020
Bài 1:
\(A=\sqrt{8}-2\sqrt{2}+\sqrt{20}-2\sqrt{5}-2=2\sqrt{2}-2\sqrt{2}+2\sqrt{5}-2\sqrt{5}-2=-2\)\(B=\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{2\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{x-2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{\sqrt{x}-1}{\sqrt{x}}\)
NH
1
S
12 tháng 7 2019
\(\sqrt{227-30\sqrt{2}}+\sqrt{123+22\sqrt{2}}=\sqrt{225-30\sqrt{2}+2}+\sqrt{121+22\sqrt{2}+2}=\sqrt{15^2-15.2.\sqrt{2}+\left(\sqrt{2}\right)^2}+\sqrt{11^2+11.2\sqrt{2}+\left(\sqrt{2}\right)^2}=\sqrt{\left(15-\sqrt{2}\right)^2}+\sqrt{\left(11+\sqrt{2}\right)^2}=15-\sqrt{2}+11+\sqrt{2}\left(do:15-\sqrt{2}>0;11+\sqrt{2}>0\right)=26\)