\(\left(\dfrac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\dfrac{\sqrt{30}-\sqrt{6}}{\sqrt{5}-1}\right)\) : \(\dfrac{1}{2\sqrt{5}-\sqrt{6}}\) =?
\(\dfrac{\sqrt{3-\sqrt{5}}\left(3+\sqrt{5}\right)}{\sqrt{10}+\sqrt{2}}\)
\(\dfrac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}\)+\(\dfrac{8}{1-\sqrt{5}}\)
\(\dfrac{5+\sqrt{7}}{9-\sqrt{23+8\sqrt{7}}}\)+\(\dfrac{5-\sqrt{7}}{2+\sqrt{16+6\sqrt{7}}}\)
\(\dfrac{1}{\sqrt{2}+\sqrt{2+\sqrt{3}}}\)+\(\dfrac{1}{\sqrt{2}-\sqrt{2+\sqrt{3}}}\)
đề là rút gọn các biểu thức sau
nhờ mọi người giải giúp mình. cảm ơn mn nhìu
a: \(=\dfrac{\sqrt{6-2\sqrt{5}}\left(3+\sqrt{5}\right)}{2\left(\sqrt{5}+1\right)}\)
\(=\dfrac{\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)}{2\left(\sqrt{5}+1\right)}=\dfrac{3\sqrt{5}+5-3-\sqrt{5}}{2\sqrt{5}+2}\)
\(=\dfrac{2\sqrt{5}+2}{2\sqrt{5}+2}=1\)
b: \(=\dfrac{2\sqrt{5}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}-2-2\sqrt{5}\)
=2căn 5-2-2căn 5
=-2
d: \(=\dfrac{\sqrt{2}}{2+\sqrt{3}+1}+\dfrac{\sqrt{2}}{2-\sqrt{3}+1}\)
\(=\dfrac{\sqrt{2}}{3+\sqrt{3}}+\dfrac{\sqrt{2}}{3-\sqrt{3}}\)
\(=\dfrac{3\sqrt{2}-\sqrt{6}+3\sqrt{2}+\sqrt{6}}{6}=\sqrt{2}\)
Bài 1.Rút gọn A = \(\sqrt{x^2+\dfrac{2x^2}{3}}\) với x<0
Bài 2.Rút gọn biểu thức \(\left(\dfrac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\dfrac{\sqrt{30}-\sqrt{6}}{\sqrt{5}-1}\right)\):\(\dfrac{2}{2\sqrt{5}-\sqrt{6}}\)
Bài 3.Cho ba biểu thức A = a\(\sqrt{b}\) + b\(\sqrt{a}\);B = \(a\sqrt{a}-b\sqrt{b}\) ;C = a-b.Trong ba biểu thức trên biểu thức bằng biểu thức \(\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)\) với a,b>0
Bài 7.Cho B = \(\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{4}}+...+\dfrac{1}{\sqrt{98}+\sqrt{99}}+\dfrac{1}{\sqrt{99}+\sqrt{100}}\).Giá trị của biểu thức B là
Bài 8.Gọi M là giá trị nhỏ nhất của \(\dfrac{\sqrt{x}+1}{\sqrt{x}+4}\) và N là giá trị lớn nhất của \(\dfrac{\sqrt{x}+5}{\sqrt{x}+2}\).Tìm M và N
Giúp mình với!Mình đang cần gấp
1:
\(A=\sqrt{x^2+\dfrac{2x^2}{3}}=\sqrt{\dfrac{5x^2}{3}}=\left|\sqrt{\dfrac{5}{3}}x\right|=-x\sqrt{\dfrac{5}{3}}\)
2: \(=\left(\dfrac{\sqrt{100}+\sqrt{40}}{\sqrt{5}+\sqrt{2}}+\sqrt{6}\right)\cdot\dfrac{2\sqrt{5}-\sqrt{6}}{2}\)
\(=\dfrac{\left(2\sqrt{5}+\sqrt{6}\right)\left(2\sqrt{5}-\sqrt{6}\right)}{2}\)
\(=\dfrac{20-6}{2}=7\)
1,Tính \(\dfrac{12}{4-\sqrt{10}}-6\sqrt{\dfrac{5}{2}}+\dfrac{5\sqrt{2}+\sqrt{10}}{\sqrt{5}+1}\)
2,Rút gọn:A=\(\left(\dfrac{\sqrt{x}}{\sqrt{x}-5}-\dfrac{5}{\sqrt{x}+5}+\dfrac{10\sqrt{x}}{25-x}\right):\dfrac{3}{\sqrt{x}+5}\)
1: \(=8+2\sqrt{10}-3\sqrt{10}+\sqrt{10}=8\)
\(\dfrac{6}{2-\sqrt{10}}-\dfrac{2\sqrt{5}-5\sqrt{2}}{\sqrt{2}-\sqrt{5}}+\sqrt{49+4\sqrt{10}}\)
\(\left(\dfrac{x-\sqrt{x}}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{x+\sqrt{x}}\right):\dfrac{\sqrt{x}+1}{x}\)
a: \(\dfrac{6}{2-\sqrt{10}}-\dfrac{2\sqrt{5}-5\sqrt{2}}{\sqrt{2}-\sqrt{5}}+\sqrt{49+4\sqrt{10}}\)
\(=\dfrac{6\left(2+\sqrt{10}\right)}{4-10}-\dfrac{\sqrt{10}\left(\sqrt{2}-\sqrt{5}\right)}{\sqrt{2}-\sqrt{5}}+\sqrt{49+2\cdot2\sqrt{10}}\)
\(=\dfrac{6\left(2+\sqrt{10}\right)}{-6}-\sqrt{10}+\sqrt{49+2\cdot\sqrt{40}}\)
\(=-2-\sqrt{10}-\sqrt{10}+\sqrt{49+4\sqrt{10}}\)
\(=-2-2\sqrt{10}+\sqrt{49+4\sqrt{10}}\)
b: ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x< >1\end{matrix}\right.\)
\(\left(\dfrac{x-\sqrt{x}}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{x+\sqrt{x}}\right):\dfrac{\sqrt{x}+1}{x}\)
\(=\left(\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\cdot\dfrac{x}{\sqrt{x}+1}\)
\(=\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right)\cdot\dfrac{x}{\sqrt{x}+1}\)
\(=\dfrac{x-1}{\sqrt{x}}\cdot\dfrac{x}{\sqrt{x}+1}=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\cdot\sqrt{x}\cdot\sqrt{x}}{\sqrt{x}\cdot\left(\sqrt{x}+1\right)}\)
\(=\sqrt{x}\left(\sqrt{x}-1\right)\)
B1: thực hiện phép tính
a )\(\dfrac{\sqrt{6}-\sqrt{15}}{\sqrt{35}-\sqrt{14}}\)
b ) \(\dfrac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\dfrac{8}{1-\sqrt{5}}\)
c )\(\dfrac{\sqrt{3-\sqrt{5}.}\left(3+\sqrt{5}\right)}{\sqrt{10}+\sqrt{2}}\)
d ) \(\dfrac{1}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\dfrac{1}{\sqrt{2}-\sqrt{2+\sqrt{3}}}\)
B2:chúng minh vế phải bằng vế trái
a) \(\dfrac{21+8\sqrt{5}}{4+\sqrt{5}}.\sqrt{9-4\sqrt{5}}=\sqrt{5}-2\)
b) \(\sqrt{8-2\sqrt{15}}-\sqrt{8+2\sqrt{15}}=-2\sqrt{3}\)
\(\dfrac{2\sqrt{30}}{\sqrt{5}+\sqrt{6}+\sqrt{7}} \)
\(\sqrt{24}+6\sqrt{\dfrac{2}{3}+\dfrac{10}{\sqrt{6}-1}}\)
\(\dfrac{2\sqrt{15}+\sqrt{16}}{\sqrt{84}+\sqrt{6}}\)
\(2\sqrt{40\sqrt{2}}-2\sqrt{\sqrt{75}}-3\sqrt{5\sqrt{48}}\)
\(\dfrac{\left(2+\sqrt{3}\right)^2-1}{\left(\sqrt{3}+1\right)^2}:\dfrac{\left(3+\sqrt{5}\right)^2-4}{\left(\sqrt{5}+1\right)^2}\)
giúp em với ạ
\(2\sqrt{40\sqrt{3}}-2\sqrt{\sqrt{75}}-3\sqrt{5\sqrt{48}}\)
\(=2\cdot\sqrt{40\sqrt{3}}-2\cdot\sqrt{5\sqrt{3}}-3\cdot\sqrt{20\sqrt{3}}\)
\(=2\cdot2\sqrt{10}\cdot\sqrt{\sqrt{3}}-2\cdot\sqrt{5}\cdot\sqrt{\sqrt{3}}-6\sqrt{5}\cdot\sqrt{\sqrt{3}}\)
\(=4\sqrt{10}\sqrt{\sqrt{3}}-4\cdot\sqrt{5}\cdot\sqrt{\sqrt{3}}\)
Tính:
\(A=3\sqrt{20}-\sqrt{45}+2\sqrt{18}+\sqrt{72}\)
\(B=\dfrac{12}{3-\sqrt{5}}-\dfrac{16}{\sqrt{5}+1}\)
\(C=10\sqrt{\dfrac{1}{5}}+\dfrac{1}{5}\sqrt{125}-2\sqrt{20}\)
\(E=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-2\right)^2}\)
\(F=\sqrt{6+2\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
a) Ta có: \(A=3\sqrt{20}-\sqrt{45}+2\sqrt{18}+\sqrt{72}\)
\(=6\sqrt{5}-3\sqrt{5}+6\sqrt{2}+6\sqrt{2}\)
\(=3\sqrt{5}+12\sqrt{2}\)
b) Ta có: \(B=\dfrac{12}{3-\sqrt{5}}-\dfrac{16}{\sqrt{5}+1}\)
\(=\dfrac{12\left(3+\sqrt{5}\right)}{4}-\dfrac{16\left(\sqrt{5}-1\right)}{4}\)
\(=3\left(3+\sqrt{5}\right)-4\left(\sqrt{5}-1\right)\)
\(=9+3\sqrt{5}-4\sqrt{5}+4\)
\(=13-\sqrt{5}\)
c) Ta có: \(C=10\sqrt{\dfrac{1}{5}}+\dfrac{1}{5}\sqrt{125}-2\sqrt{20}\)
\(=\dfrac{10}{\sqrt{5}}+\dfrac{1}{5}\cdot5\sqrt{5}-2\cdot2\sqrt{5}\)
\(=2\sqrt{5}+\sqrt{5}-4\sqrt{5}\)
\(=-\sqrt{5}\)
e) Ta có: \(E=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-2\right)^2}\)
\(=\sqrt{3}+1-2+\sqrt{3}\)
\(=2\sqrt{3}-1\)
f) Ta có: \(F=\sqrt{6+2\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(=\sqrt{5}+1-\sqrt{5}+2\)
=3
Tính:
\(A=3\sqrt{20}-\sqrt{45}+2\sqrt{18}+\sqrt{72}\)
\(B=\dfrac{12}{3-\sqrt{5}}-\dfrac{16}{\sqrt{5}+1}\)
\(C=10\sqrt{\dfrac{1}{5}}+\dfrac{1}{5}\sqrt{125}-2\sqrt{20}\)
\(E=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-2\right)^2}\)
\(F=\sqrt{6+2\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
e) Ta có: \(E=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-2\right)^2}\)
\(=\sqrt{3}+1-2+\sqrt{3}\)
\(=2\sqrt{3}-1\)
f) Ta có: \(F=\sqrt{6+2\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(=\sqrt{5}+1-\sqrt{5}+2\)
=3
a) Ta có: \(A=3\sqrt{20}-\sqrt{45}+2\sqrt{18}+\sqrt{72}\)
\(=6\sqrt{5}-3\sqrt{5}+6\sqrt{2}+6\sqrt{2}\)
\(=3\sqrt{5}+12\sqrt{2}\)
b) Ta có: \(B=\dfrac{12}{3-\sqrt{5}}-\dfrac{16}{\sqrt{5}+1}\)
\(=\dfrac{12\left(3+\sqrt{5}\right)}{4}-\dfrac{16\left(\sqrt{5}-1\right)}{4}\)
\(=3\left(3+\sqrt{5}\right)-4\left(\sqrt{5}-1\right)\)
\(=9+3\sqrt{5}-4\sqrt{5}+4\)
\(=13-\sqrt{5}\)
Rút gọn:
a) \(\dfrac{\left(5\sqrt{2}+2\sqrt{5}\right)\left(\sqrt{3}-3\sqrt{2}\right)}{\sqrt{30}}\)
b) \(\dfrac{5\sqrt{7}-4\sqrt{35}+7\sqrt{5}}{\sqrt{35}}\)
c) \(\dfrac{6\sqrt{6}-2\sqrt{12}+3-\sqrt{2}}{2\sqrt{6}+1}\)
d) \(\dfrac{10\sqrt{18}+5\sqrt{3}-15\sqrt{27}}{\sqrt{3\left(\sqrt{6}-4\right)}}\)