\(\frac{12}{\sqrt{7+\sqrt{24}}}\)
K
Khách
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AH
Akai Haruma
Giáo viên
27 tháng 6 2019
Lời giải:
\(\frac{\sqrt{12}-6}{\sqrt{8}-\sqrt{24}}-\frac{3+\sqrt{3}}{\sqrt{3}}+\frac{4}{1-\sqrt{7}}\)
\(=\frac{\sqrt{12}-\sqrt{36}}{\sqrt{8}-\sqrt{24}}-\frac{\sqrt{3}(\sqrt{3}+1)}{\sqrt{3}}+\frac{4(1+\sqrt{7})}{(1-\sqrt{7})(1+\sqrt{7})}\)
\(=\frac{\sqrt{12}(1-\sqrt{3})}{\sqrt{8}(1-\sqrt{3})}-(\sqrt{3}+1)+\frac{4(1+\sqrt{7})}{1-7}\)
\(=\frac{\sqrt{6}}{2}-(\sqrt{3}+1)-\frac{2(1+\sqrt{7})}{3}\)
4 tháng 8 2020
a/ \(\sqrt{5+\sqrt{24}}-\sqrt{2}=\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}-\sqrt{2}=\left|\sqrt{3}+\sqrt{2}\right|-\sqrt{2}=\sqrt{3}+\sqrt{2}-\sqrt{2}=\sqrt{3}\)
b/ \(\frac{3-2\sqrt{3}}{\sqrt{3}-2}=\frac{\sqrt{3}\left(\sqrt{3}-2\right)}{\sqrt{3}-2}=\sqrt{3}\)
c/ \(\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}=\frac{\sqrt{5}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}=-\sqrt{5}\)
d/ \(\frac{1}{1-\sqrt{2}}-\frac{1}{1+\sqrt{2}}=\frac{1+\sqrt{2}-1+\sqrt{2}}{\left(1-\sqrt{2}\right)\left(1+\sqrt{2}\right)}=\frac{2\sqrt{2}}{1-2}=-2\sqrt{2}\)
NV
Nguyễn Việt Lâm
Giáo viên
14 tháng 9 2020
\(a=\frac{1}{\sqrt{7-2\sqrt{6}}+1}+\frac{1}{\sqrt{7+2\sqrt{6}}-1}=\frac{1}{\sqrt{\left(\sqrt{6}-1\right)^2}+1}+\frac{1}{\sqrt{\left(\sqrt{6}+1\right)^2}-1}\)
\(=\frac{1}{\sqrt{6}}+\frac{1}{\sqrt{6}}=\frac{2}{\sqrt{6}}=\frac{\sqrt{6}}{3}\)
Coi lại đề câu b, quy luật ở số hạng cuối cùng sai (nhìn 2 số hạng đầu 2 số dưới căn hơn kém nhau 4 đơn vị, số cuối lại chỉ hơn kém nhau 1 đơn vị)
NA
5 tháng 8 2018
\(3x-7\sqrt{x}+4=0\)
\(3x-3\sqrt{x}-4\sqrt{x}+4=0\)
\(3\sqrt{x}\left(\sqrt{x}-1\right)-4\left(\sqrt{x}-1\right)=0\)
\(\left(\sqrt{x}-1\right)\left(3\sqrt{x}-4\right)=0\)
\(\orbr{\begin{cases}\sqrt{x}-1=0\\3\sqrt{x}-4=0\end{cases}}\Rightarrow\orbr{\begin{cases}\sqrt{x}=1\\3\sqrt{x}=4\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=\frac{16}{9}\end{cases}}\)
KT
6 tháng 8 2018
ĐK: \(x\ge1\)
\(\frac{1}{2}\sqrt{x-1}-\frac{3}{2}\sqrt{9x-9}+24\sqrt{\frac{x-1}{64}}=-17\)
<=> \(\frac{1}{2}\sqrt{x-1}-\frac{3}{2}\sqrt{9\left(x-1\right)}+24\sqrt{\frac{1}{64}\left(x-1\right)}=-17\)
<=> \(\frac{1}{2}\sqrt{x-1}-\frac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)
<=> \(-\sqrt{x-1}=-17\)
<=> \(x-1=17^2\)
<=> \(x=290\)
Vậy....
11 tháng 12 2019
a) = \(5\sqrt{2}-3\sqrt{6}+3\sqrt{2}+5\sqrt{6}\)
= \(8\sqrt{2}+2\sqrt{6}\)
b) = \(2\sqrt{3}-4\sqrt{2}-5\sqrt{3}-\sqrt{2}\)
= \(-3\sqrt{3}-5\sqrt{2}\)
c) = \(\frac{\left(\sqrt{2}-1\right)\left(2+\sqrt{2}\right)}{\left(2-\sqrt{2}\right)\left(2+\sqrt{2}\right)}\)
=\(\frac{2\sqrt{2}+2-2-\sqrt{2}}{2^2-\sqrt{2^2}}\)
=\(\frac{\sqrt{2}}{4-2}\) = \(\frac{\sqrt{2}}{2}\)
d) = \(2\sqrt{6}-5\sqrt{6}+2\sqrt{2}\)
=\(-3\sqrt{6}+2\sqrt{2}\)
e) = \(8\sqrt{6}+3\sqrt{6}-6\sqrt{6}=5\sqrt{6}\)
f) = \(4\sqrt{3}+9\sqrt{3}-4\sqrt{3}=9\sqrt{3}\)
g) = \(10+5\sqrt{10}-5\sqrt{10}=10\)
h) = \(\frac{\left(3+\sqrt{3}\right)\left(3+\sqrt{3}\right)}{\left(3-\sqrt{3}\right)\left(3+\sqrt{3}\right)}+\frac{\left(3-\sqrt{3}\right)\left(3-\sqrt{3}\right)}{\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)}\)
= \(\frac{9+3\sqrt{3}+3\sqrt{3}+3}{3^2-\sqrt{3^2}}+\frac{9-3\sqrt{3}-3\sqrt{3}+3}{3^2-\sqrt{3^2}}\)
= \(\frac{12+6\sqrt{3}}{9-3}+\frac{12-6\sqrt{3}}{9-3}\)
= \(\frac{12+6\sqrt{3}+12-6\sqrt{3}}{6}\)
= \(\frac{24}{6}=4\)
k) = \(\left(2\sqrt{7}-2\sqrt{3}+\sqrt{7}\right).\sqrt{7}+2\sqrt{21}\)
= \(\left(3\sqrt{7}-2\sqrt{3}\right).\sqrt{7}+2\sqrt{21}\)
= \(21-2\sqrt{21}+2\sqrt{21}=21\)
l) = \(\frac{\left(2\sqrt{3}-\sqrt{6}\right)\left(\sqrt{8}+2\right)}{\left(\sqrt{8}-2\right)\left(\sqrt{8}+2\right)}\)
= \(\frac{4\sqrt{6}+4\sqrt{3}-4\sqrt{3}-2\sqrt{6}}{\sqrt{8^2}-2^2}\)
= \(\frac{2\sqrt{6}}{8-4}=\frac{2\sqrt{6}}{4}=\frac{\sqrt{6}}{2}\)
L
0
\(\frac{12}{\sqrt{7+\sqrt{24}}}=\frac{12}{\sqrt{6+2\sqrt{6}+1}}\)
\(=\frac{12}{\sqrt{\left(\sqrt{6}+1\right)^2}}\)
\(=\frac{12}{\sqrt{6}+1}\)
\(=\frac{12\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}\)
\(=\frac{12\left(\sqrt{6}-1\right)}{6-1}\)
\(=\frac{12\left(\sqrt{6}-1\right)}{5}\)