tim x bt
x2 -(căn3 +căn5)*x+căn3 * căn5=0
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Những câu hỏi liên quan
NL
Nguyễn Lê Phước Thịnh
CTVHS
12 tháng 8 2021
a: \(\dfrac{5+2\sqrt{5}}{\sqrt{5}+\sqrt{2}}=\dfrac{\left(5+2\sqrt{5}\right)\left(\sqrt{5}-\sqrt{2}\right)}{3}=\dfrac{5\sqrt{5}-5\sqrt{2}+10-2\sqrt{10}}{3}\)
b: \(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}=\sqrt{\left(2-\sqrt{3}\right)^2}=2-\sqrt{3}\)
NM
1
27 tháng 3 2016
\(<=>x^2-\sqrt{3}x-\sqrt{5}x+\sqrt{15}=0<=>x\left(x-\sqrt{3}\right)-\sqrt{5}\left(x-\sqrt{3}\right)=0<=>\left(x-\sqrt{3}\right)\left(x-\sqrt{5}\right)=0\)
<=>Tự làm
NY
0
NL
Nguyễn Lê Phước Thịnh
CTVHS
2 tháng 7
a: \(\left(2\sqrt6-4\sqrt3+5\sqrt2-\frac14\cdot\sqrt8\right)\cdot3\sqrt6\)
\(=\left(2\sqrt6-4\sqrt3+5\sqrt2-\frac12\sqrt2\right)\cdot3\sqrt6\)
\(=\left(2\sqrt6-4\sqrt3+\frac92\cdot\sqrt2\right)\cdot3\sqrt6\)
\(=2\sqrt6\cdot3\sqrt6-4\sqrt3\cdot3\sqrt6+\frac92\cdot\sqrt2\cdot3\sqrt6\)
\(=36-12\sqrt{18}+\frac{27}{2}\sqrt{12}=36-36\sqrt2+27\sqrt3\)
b: \(\left(\sqrt{\frac17}-\sqrt{\frac{16}{7}}+\sqrt7\right):\sqrt7=\left(\frac{\sqrt7}{7}-\frac{4\sqrt7}{7}+\sqrt7\right):\sqrt7\)
\(=\frac17-\frac47+1=\frac87-\frac47=\frac47\)
c: \(\left(\sqrt{3-\sqrt5}+\sqrt{3+\sqrt5}\right)^2\)
\(=3-\sqrt5+3+\sqrt5+2\cdot\sqrt{\left(3-\sqrt5\right)\left(3+\sqrt5\right)}\)
\(=6+2\cdot\sqrt{9-5}=6+2\cdot2=10\)
21 tháng 9 2021
\(a,9\sqrt{5}+3\sqrt{20}-7\sqrt{45}=9\sqrt{5}+6\sqrt{5}-21\sqrt{5}=-6\sqrt{5}\\ b,\dfrac{2\sqrt{6}+\sqrt{40}}{\sqrt{3}+\sqrt{5}}=\dfrac{2\sqrt{6}+2\sqrt{10}}{\sqrt{3}+\sqrt{5}}\\ =\dfrac{2\sqrt{2}\left(\sqrt{3}+\sqrt{5}\right)\left(\sqrt{5}-\sqrt{3}\right)}{\left(\sqrt{3}+\sqrt{5}\right)\left(\sqrt{5}-\sqrt{3}\right)}=\dfrac{2\sqrt{2}\left(5-3\right)}{5-3}=2\sqrt{2}\)
TM
0
NL
Nguyễn Lê Phước Thịnh
CTVHS
23 giờ trước (16:44)
a: \(\frac{3}{4+\sqrt{9+4\sqrt5}}\)
\(=\frac{3}{4+\sqrt{\left(\sqrt5+2\right)^2}}\)
\(=\frac{3}{4+\sqrt5+2}=\frac{3}{6+\sqrt5}=\frac{3\left(6-\sqrt5\right)}{36-5}=\frac{3\left(6-\sqrt5\right)}{31}\)
b: \(\frac{\sqrt3}{\sqrt2+\sqrt{5+2\sqrt6}}\)
\(=\frac{\sqrt3}{\sqrt2+\sqrt{\left(\sqrt3+\sqrt2\right)^2}}=\frac{\sqrt3}{\sqrt2+\sqrt3+\sqrt2}\)
\(=\frac{\sqrt3}{2\sqrt2+\sqrt3}=\frac{\sqrt3\left(2\sqrt2-\sqrt3\right)}{8-3}=\frac{2\sqrt6-3}{5}\)
c: \(\frac{3}{\sqrt5+\sqrt7-\sqrt2}=\frac{3\left(\sqrt5+\sqrt7+\sqrt2\right)}{\left(\sqrt5+\sqrt7\right)^2-2}\)
\(=\frac{3\left(\sqrt5+\sqrt7+\sqrt2\right)}{10+2\sqrt{35}}=\frac{3\left(\sqrt5+\sqrt7+\sqrt2\right)\left(\sqrt{35}-5\right)}{2\left(\sqrt{35}+5\right)\left(\sqrt{35}-5\right)}\)
\(=\frac{3\left(\sqrt5+\sqrt7+\sqrt2\right)\left(\sqrt{35}-5\right)}{2\left(35-25\right)}=\frac{3\left(\sqrt5+\sqrt7+\sqrt2\right)\left(\sqrt{35}-5\right)}{20}\)
N
2
\(x^2-\left(\sqrt{3}+\sqrt{5}\right).x+\sqrt{3}.\sqrt{5}=0\)
\(\Leftrightarrow x^2-\sqrt{3}.x-\sqrt{5}.x+\sqrt{3}.\sqrt{5}=0\)
\(\Leftrightarrow x^2-\sqrt{3}.x-\sqrt{5}.x+\sqrt{3}.\sqrt{5}=0\)
\(\Leftrightarrow x\left(x-\sqrt{3}\right)-\sqrt{5}\left(x-\sqrt{3}\right)=0\)
\(\Leftrightarrow\left(x-\sqrt{5}\right)\left(x-\sqrt{3}\right)=0\)
\(\Leftrightarrow\int^{x-\sqrt{5}=0}_{x-\sqrt{3}=0}\Leftrightarrow\int^{x=\sqrt{5}}_{x=\sqrt{3}}\)
Vậy x \(\in\left\{\sqrt{3};\sqrt{5}\right\}\)