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Y
25 tháng 7 2019
a) \(=\sqrt{5-2\sqrt{5}+1}-2\sqrt{5}-1\)
\(=\sqrt{\left(\sqrt{5}-1\right)^2}-2\sqrt{5}-1\)
\(=\sqrt{5}-1-2\sqrt{5}-1=-\sqrt{5}-1\)
b) \(=3+4\sqrt{5}-\sqrt{5-4\sqrt{5}+4}\)
\(=3+4\sqrt{5}-\sqrt{\left(\sqrt{5}-2\right)^2}\)
\(=3+4\sqrt{5}-\left(\sqrt{5}-2\right)=5+3\sqrt{5}\)
c) \(=\left(\sqrt{\left(\sqrt{3}-1\right)^2}-1\right)\cdot\frac{1}{2\sqrt{3}-4}\)
\(=\left(\sqrt{3}-2\right)\cdot\frac{1}{2\left(\sqrt{3}-2\right)}=\frac{1}{2}\)
25 tháng 10 2020
a) Ta có: \(\frac{6}{\sqrt{2}-\sqrt{3}+3}\)
\(=\frac{6\left(\sqrt{2}-\sqrt{3}-3\right)}{\left(\sqrt{2}-\sqrt{3}+3\right)\left(\sqrt{2}-\sqrt{3}-3\right)}\)
\(=\frac{6\left(\sqrt{2}-\sqrt{3}-3\right)}{5-2\sqrt{6}-9}\)
\(=\frac{6\left(\sqrt{2}-\sqrt{3}-3\right)}{-4-2\sqrt{6}}\)
\(=\frac{6\left(\sqrt{2}-\sqrt{3}-3\right)}{-2\sqrt{2}\left(\sqrt{2}-\sqrt{3}\right)}\)
\(=\frac{3\left(\sqrt{2}-\sqrt{3}-3\right)\left(\sqrt{2}+\sqrt{3}\right)}{-\sqrt{2}\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}-\sqrt{3}\right)}\)
\(=\frac{3\sqrt{2}\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{2}-\sqrt{3}-3\right)}{2}\)
b) Ta có: \(\left(\frac{4}{\sqrt{5}+1}-\frac{4}{\sqrt{5}-1}\right):\sqrt{3+2\sqrt{2}}\)
\(=\left(\frac{4\left(\sqrt{5}-1\right)}{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}-\frac{4\left(\sqrt{5}+1\right)}{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}\right):\sqrt{2+2\cdot\sqrt{2}\cdot1+1}\)
\(=\left(\frac{4\left(\sqrt{5}-1\right)}{4}-\frac{4\left(\sqrt{5}+1\right)}{4}\right):\sqrt{\left(\sqrt{2}+1\right)^2}\)
\(=\left(\sqrt{5}-1-\sqrt{5}-1\right):\left|\sqrt{2}+1\right|\)
\(=-\frac{2}{\sqrt{2}+1}\)(Vì \(\sqrt{2}+1>0\))
\(=-\frac{2\left(\sqrt{2}-1\right)}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}\)
\(=-2\left(\sqrt{2}-1\right)\)
\(=-2\sqrt{2}+2\)
1 tháng 8 2018
\(a\text{) }\sqrt{10+\sqrt{9}}=\sqrt{10+3}=\sqrt{13}\)
\(b\text{) }\sqrt{21+6\sqrt{6}}-\sqrt{21-6\sqrt{6}}\\ =\sqrt{18+3+2\sqrt{54}}-\sqrt{18+3-2\sqrt{54}}\\ =\sqrt{\left(\sqrt{18}+\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{18}-\sqrt{3}\right)^2}\\ =\sqrt{18}+\sqrt{3}-\sqrt{18}+\sqrt{3}\\ =2\sqrt{3}\)
\(d\text{) }\sqrt{x+1+2\sqrt{x}}\left(x\ge0\right)\\ =\sqrt{\left(\sqrt{x}+1\right)^2}=\sqrt{x}+1\)
\(e\text{) }\sqrt{2x+3+2\sqrt{x^2+3x+2}}\left(x\le-2;x\ge-1\right)\\ =\sqrt{\left(x+2\right)+\left(x+1\right)+2\sqrt{\left(x+1\right)\left(x+2\right)}}=\sqrt{\left(\sqrt{x+1}+\sqrt{x+2}\right)^2}=\sqrt{x+1}+\sqrt{x+2}\)
Xem lại đề câu c nha.
1 tháng 8 2018
a)\(\sqrt{10+\sqrt{9}}=\sqrt{10+3}=\sqrt{13}\)
b)\(\sqrt{21+6\sqrt{6}}-\sqrt{21-6\sqrt{6}}\)
=\(\sqrt{\left(3\sqrt{2}\right)^2+2.3\sqrt{2}.\sqrt{3}+\sqrt{3^2}}-\sqrt{\left(3\sqrt{2}\right)^2-2.3.\sqrt{2}.\sqrt{3}+\sqrt{3^2}}\)
=\(\sqrt{\left(3\sqrt{2}+\sqrt{3}\right)^2}-\sqrt{\left(3\sqrt{2}-\sqrt{3}\right)^2}\)
=\(3\sqrt{2}+\sqrt{3}-3\sqrt{2}+\sqrt{3}\)
=\(2\sqrt{3}\)
c)\(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10-2\sqrt{5}}}\)
ÁP dụng HĐT \(\sqrt{a+b}\pm\sqrt{a-b}=\sqrt{2\left(a.\sqrt{a^2\pm b}\right)}\)ta có:
=\(\sqrt{2\left(4+\sqrt{4^2-10-2\sqrt{5}}\right)}\)
=\(\sqrt{2\left(4+\sqrt{16-10-2\sqrt{5}}\right)}\)
=\(\sqrt{2\left(4+\sqrt{6-2\sqrt{5}}\right)}\)
=\(\sqrt{2\left(4+\sqrt{\left(\sqrt{5}\right)^2-2\sqrt{5}.1+1^2}\right)}\)
=\(\sqrt{2\left(4+\sqrt{\left(\sqrt{5}-1\right)^2}\right)}\)
=\(\sqrt{2\left(4+\sqrt{5}-1\right)}\)
=\(\sqrt{2\left(3+\sqrt{5}\right)}\)
=\(\sqrt{6+\sqrt{5}}=\sqrt{5}+1\)
d)\(\sqrt{x+1+2\sqrt{x}}=\sqrt{\left(\sqrt{x}\right)^2+2\sqrt{x}.1+1^2}=\sqrt{x}+1\)
19 tháng 6 2019
1/\(\sqrt{\frac{4}{5}}\)+\(\sqrt{\frac{1}{2}}\)
=\(\sqrt{\frac{4.5}{5.5}}\)+\(\sqrt{\frac{1.2}{2.2}}\)
= \(5.2.\sqrt{5}\)+\(2\sqrt{2}\)
=\(10\sqrt{5}+2\sqrt{2}\)
19 tháng 6 2019
2.
\(\sqrt{\frac{1}{12}}\)+\(\sqrt{\frac{1}{3}}\)
=\(\sqrt{\frac{1.12}{12.12}}\)+\(\sqrt{\frac{1.3}{3.3}}\)
=\(12.2\sqrt{3}\)+\(3\sqrt{3}\)
=\(\sqrt{3}\left(24+3\right)\)
=\(27\sqrt{3}\)
26 tháng 10 2019
a/ \(A=\frac{1}{5+2\sqrt{6-x^2}}\)
Có: \(-x^2\le0\)với mọi x
=> \(6-x^2\le6\)
=> \(0\le\sqrt{6-x^2}\le\sqrt{6}\)
=> \(5\le5+2\sqrt{6-x^2}\le5+2\sqrt{6}\)
=> \(\frac{1}{5+2\sqrt{6}}\le\frac{1}{5+2\sqrt{6-x^2}}\le\frac{1}{5}\); với mọi x
=> \(\hept{\begin{cases}maxA=\frac{1}{5}\Leftrightarrow\sqrt{6-x^2}=0\Leftrightarrow x=\pm\sqrt{6}\\minA=\frac{1}{5+2\sqrt{6}}\Leftrightarrow\sqrt{6-x^2}=\sqrt{6}\Leftrightarrow x=0\end{cases}}\)
Vậy:...
b/ \(B=\sqrt{-x^2+2x+4}=\sqrt{-\left(x-1\right)^2+5}\)
Có: \(-\left(x-1\right)^2\le0\)với mọi x
=> \(-\left(x-1\right)^2+5\le5\)
=> \(0\le\sqrt{-\left(x-1\right)^2+5}\le\sqrt{5}\)
=> \(0\le B\le\sqrt{5}\)với mọi x
=> \(\hept{\begin{cases}maxB=\sqrt{5}\Leftrightarrow-\left(x-1\right)^2=0\Leftrightarrow x=1\\minB=0\Leftrightarrow\left(x-1\right)^2=5\Leftrightarrow x=\pm\sqrt{5}+1\end{cases}}\)
Vậy:...
26 tháng 10 2019
a)Ta có:
\(0\le2\sqrt{6-x^2}\le2\sqrt{6}\)
\(\Leftrightarrow\frac{1}{5}\ge\frac{1}{5+2\sqrt{6-x^2}}\ge\frac{1}{5+2\sqrt{6}}=5-2\sqrt{6}\)
\(\Rightarrow\hept{\begin{cases}MAX\left(A\right)=\frac{1}{5}\\MIN\left(A\right)=5-2\sqrt{6}\end{cases}}\)Dấu "=" xảy ra khi \(\hept{\begin{cases}x=0\left(MIN\right)\\x=\sqrt{6}\left(MAX\right)\end{cases}}\)
\(\sqrt{\frac{3a-4}{-5}}\)
\(\sqrt{\frac{3a-4}{-5}}\ge0\)
\(-5< 0< =>3a-4\le0\)
\(3a\le4< =>x\le\frac{4}{3}\)