Câu1:\(\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}\)- \(\dfrac{\sqrt{6}-3}{\sqrt{2}-\sqrt{3}}\)
Câu2:\(\dfrac{\sqrt{2}}{\sqrt{3}-1}-\sqrt{\dfrac{3}{2}}\)
giải cụ thể giúp mk vớiiiii ạ
1.)\(\sqrt{\left(\sqrt{3}-3\right)^2}\)+\(\sqrt{4-2\sqrt{3}}\)
2)\(\dfrac{1}{\sqrt{5}-2}\) +\(\dfrac{\sqrt{10}-\sqrt{5}}{1-\sqrt{2}}\)
3)\(\dfrac{\sqrt{2}}{\sqrt{3}-1}\) -\(\sqrt{\dfrac{3}{2}}\)
giải chi tiết giúp mk vớiiiii ạ
1)\(=\left|\sqrt{3}-3\right|+\sqrt{\left(\sqrt{3}-1\right)^2}=3-\sqrt{3}+\left|\sqrt{3}-1\right|=3-\sqrt{3}+\sqrt{3}-1=2\)
2: \(=\sqrt{5}+2-\sqrt{5}=2\)
f)\(\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}\)- \(\dfrac{\sqrt{6}-3}{\sqrt{2}-\sqrt{3}}\)
g)\(\left(\sqrt{8}-3\sqrt{2}+\sqrt{10}\right).\left(\sqrt{2}-3\sqrt{0,4}\right)\)
giải chi tiết cụ thể giúp mk với ạ
a) √(√3 - 3)2 + √4 - 2√3
b)\(\dfrac{1}{\sqrt{5}-2}\)+ \(\dfrac{\sqrt{10}-\sqrt{5}}{1-\sqrt{2}}\)
giải giúp mk 2 câu này vớiiiii ạ
a) \(=\left|\sqrt{3}-3\right|+\sqrt{\left(\sqrt{3}-1\right)^2}=3-\sqrt{3}+\sqrt{3}-1=2\)
b) \(=\dfrac{\sqrt{5}+2}{5-4}-\dfrac{\sqrt{5}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}=\sqrt{5}+2-\sqrt{5}=2\)
M= \(\dfrac{3}{2}\sqrt{32x}-\dfrac{1}{3}\sqrt{18x}+\dfrac{2}{5}\sqrt{50x}-4\sqrt{2x}\) (x lớn hơn hoặc bằng 0)
giải chi tiết giúp mk vớiiiii ạ
\(M=\dfrac{3}{2}\cdot4\sqrt{2x}-\dfrac{1}{3}\cdot3\sqrt{2x}+\dfrac{2}{5}\cdot5\sqrt{2x}-4\sqrt{2x}=6\sqrt{2x}-\sqrt{2x}+2\sqrt{2x}-4\sqrt{2x}=3\sqrt{2x}\)
Rút gọn:
1) \(\dfrac{16-6\sqrt{7}}{\sqrt{7}-3}\)
2) \(\dfrac{\left(\sqrt{3}-\sqrt{2}\right)^2+4\sqrt{6}}{\sqrt{3}+\sqrt{2}}\)
3) \(\dfrac{\left(\sqrt{3}+2\sqrt{5}\right)^2-8\sqrt{15}}{\sqrt{6}-2\sqrt{10}}\)
Giúp em với ạ. Help mee !!!
Câu 1,2 bạn đã đăng và có lời giải rồi
Câu 3:
\(=\frac{(\sqrt{3})^2+(2\sqrt{5})^2-2.\sqrt{3}.2\sqrt{5}}{\sqrt{2}(\sqrt{3}-2\sqrt{5})}=\frac{(\sqrt{3}-2\sqrt{5})^2}{\sqrt{2}(\sqrt{3}-2\sqrt{5})}=\frac{\sqrt{3}-2\sqrt{5}}{\sqrt{2}}\)
\(\sqrt{\left(3-2\sqrt{5}\right)}^2+\sqrt{\left(5-2\sqrt{5}\right)^2}\)
\(\dfrac{2}{\sqrt{3}+1}-\dfrac{2}{\sqrt{3}-1}\)
\(\sqrt{7+2}\sqrt{10}-\sqrt{7-2\sqrt{10}}\)
giúp mk với ạ
1) \(=2\sqrt{5}-3+5-2\sqrt{5}=2\)
2) \(=\dfrac{2\sqrt{3}-2-2\sqrt{3}-2}{3-1}=\dfrac{-4}{2}=-2\)
3) \(=\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}=\sqrt{5}+\sqrt{2}-\sqrt{5}+\sqrt{2}=2\sqrt{2}\)
Giúp mk với!!!
\(C=\dfrac{\sqrt{2+\sqrt{3}}}{2}:\left(\dfrac{\sqrt{2+\sqrt{3}}}{2}-\dfrac{2}{\sqrt{6}}+\dfrac{\sqrt{2+\sqrt{3}}}{2\sqrt{3}}\right)\)
Ta có: \(C=\dfrac{\sqrt{2+\sqrt{3}}}{2}:\left(\dfrac{\sqrt{2+\sqrt{3}}}{2}-\dfrac{2}{\sqrt{6}}+\dfrac{\sqrt{2+\sqrt{3}}}{2\sqrt{3}}\right)\)
\(=\dfrac{\sqrt{4+2\sqrt{3}}}{2\sqrt{2}}:\left(\dfrac{\sqrt{6+3\sqrt{3}}}{2\sqrt{3}}-\dfrac{2\sqrt{2}}{2\sqrt{3}}+\dfrac{\sqrt{2+\sqrt{3}}}{2\sqrt{3}}\right)\)
\(=\dfrac{\sqrt{3}+1}{2\sqrt{2}}:\left(\dfrac{\sqrt{12+6\sqrt{3}}-4+\sqrt{4+2\sqrt{3}}}{2\sqrt{6}}\right)\)
\(=\dfrac{\sqrt{3}+1}{2\sqrt{2}}:\dfrac{3+\sqrt{3}-4+\sqrt{3}+1}{2\sqrt{6}}\)
\(=\dfrac{\sqrt{3}+1}{2\sqrt{2}}\cdot\dfrac{2\sqrt{6}}{2\sqrt{3}}\)
\(=\dfrac{\sqrt{3}+1}{2}\)
(1) thực hiện phép tính:
a) \(\sqrt{5}.\left(\sqrt{20}-3\right)+\sqrt{45}\)
b) \(\sqrt{\left(5-\sqrt{3}\right)^2}-\sqrt{\left(2-\sqrt{3}\right)^2}\)
c) \(\dfrac{2}{\sqrt{5}+1}-\dfrac{2}{3-\sqrt{5}}\)
giúp mk vs ạ mai mk học rồi
\(a,=\sqrt{5}\left(2\sqrt{5}-3\right)+3\sqrt{5}=10-3\sqrt{5}+3\sqrt{5}=10\\ b,=5-\sqrt{3}-\left(2-\sqrt{3}\right)=3\\ c,=\dfrac{2\left(\sqrt{5}-1\right)}{4}-\dfrac{2\left(3+\sqrt{5}\right)}{4}=\dfrac{2\sqrt{5}-2-6-2\sqrt{5}}{4}=\dfrac{-8}{4}=-2\)
mn cs thể giải chi tiết đc ko ạ
\(A=\left(\dfrac{x-3\sqrt{x}}{x-9}-1\right):\left(\dfrac{9-x}{x+\sqrt{x}-6}-\dfrac{\sqrt{x}-3}{2-\sqrt{x}}-\dfrac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)
Lời giải:
ĐKXĐ: $x\geq 0; x\neq 4$
\(A=\left[\frac{\sqrt{x}(\sqrt{x}-3)}{(\sqrt{x}-3)(\sqrt{x}+3)}-1\right]:\left[\frac{(3-\sqrt{x})(3+\sqrt{x})}{(\sqrt{x}-2)(\sqrt{x}+3)}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}+3}\right]\)
\(=\left(\frac{\sqrt{x}}{\sqrt{x}+3}-1\right):\left(\frac{3-\sqrt{x}}{\sqrt{x}-2}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)
\(=\frac{-3}{\sqrt{x}+3}:\frac{-(\sqrt{x}-2)}{\sqrt{x}+3}=\frac{-3}{\sqrt{x}+3}.\frac{\sqrt{x}+3}{-(\sqrt{x}-2)}=\frac{3}{\sqrt{x}-2}\)