rút gọn
A = \(\sqrt{3+\sqrt{5}}\)- \(\sqrt{3-\sqrt{5}}\)
Rút gọn biểu thức: \(A=\dfrac{\sqrt{3+\sqrt{5}}}{\sqrt{2}-\sqrt{3-\sqrt{5}}}-\dfrac{\sqrt{3-\sqrt{5}}}{\sqrt{2}+\sqrt{3+\sqrt{5}}}\)
\(A=\dfrac{\sqrt{6+2\sqrt{5}}}{2-\sqrt{6-2\sqrt{5}}}-\dfrac{\sqrt{6-2\sqrt{5}}}{2+\sqrt{6+2\sqrt{5}}}\)
\(=\dfrac{\sqrt{5}+1}{2-\sqrt{5}+1}-\dfrac{\sqrt{5}-1}{3+\sqrt{5}}\)
\(=\dfrac{\left(3+\sqrt{5}\right)\left(\sqrt{5}+1\right)-\left(\sqrt{5}-1\right)\left(3-\sqrt{5}\right)}{4}\)
\(=\dfrac{3\sqrt{5}+3+5+\sqrt{5}-3\sqrt{5}+5+3-\sqrt{5}}{4}\)
\(=4\)
Rút gọn A= \(\frac{3+\sqrt{5}}{\sqrt{2}+\sqrt{3+\sqrt{5}}}+\frac{3-\sqrt{5}}{\sqrt{2}-\sqrt{3-\sqrt{5}}}\)
\(A=\frac{3+\sqrt{5}}{\sqrt{2}+\sqrt{3+\sqrt{5}}}+\frac{3-\sqrt{5}}{\sqrt{2}-\sqrt{3-\sqrt{5}}}\)
\(=\frac{6+2\sqrt{5}}{2+\sqrt{6+2\sqrt{5}}}+\frac{6-2\sqrt{5}}{2-\sqrt{6-2\sqrt{5}}}\)
\(=\frac{6+2\sqrt{5}}{2+\sqrt{5+2\sqrt{5}+1}}+\frac{6-2\sqrt{5}}{2+\sqrt{5-2\sqrt{5}+1}}\)
\(=\frac{6+2\sqrt{5}}{2+\sqrt{\left(\sqrt{5}+1\right)^2}}+\frac{6-2\sqrt{5}}{2+\sqrt{\left(\sqrt{5}-1\right)^2}}\)
\(=\frac{6+2\sqrt{5}}{2+\left|\sqrt{5}+1\right|}+\frac{6-2\sqrt{5}}{2-\left|\sqrt{5}-1\right|}\)
\(=\frac{6+2\sqrt{5}}{2+\sqrt{5}+1}+\frac{6-2\sqrt{5}}{2-\sqrt{5}+1}\)( vì \(\sqrt{5}+1>0;\sqrt{5}-1>0\))
\(=\frac{6+2\sqrt{5}}{3+\sqrt{5}}+\frac{6-2\sqrt{5}}{3-\sqrt{5}}\)
\(=2+2\)
\(=4\)
Vậy A = 4
Tích cho mk nhoa !!!! ~~
Rút gọn: A = \(\frac{a+\sqrt{2+\sqrt{5}}.\sqrt{\sqrt{9-4\sqrt{5}}}}{\sqrt[3]{2-\sqrt{5}}.\sqrt[3]{\sqrt{9-4\sqrt{5}}}-\sqrt[3]{a^2}+\sqrt[3]{a}}\)
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: \(=\left(\dfrac{\sqrt{100}+\sqrt{40}}{\sqrt{5}+\sqrt{2}}+\sqrt{6}\right)\cdot\dfrac{2\sqrt{5}-\sqrt{6}}{2}\)
\(=\dfrac{\left(2\sqrt{5}+\sqrt{6}\right)\left(2\sqrt{5}-\sqrt{6}\right)}{2}\)
\(=\dfrac{20-6}{2}=7\)
Rút gọn biểu thức:
h) \(\sqrt{\dfrac{3+\sqrt{5}}{\sqrt{3-\sqrt{5}}}}+\sqrt{\dfrac{3-\sqrt{5}}{\sqrt{3+\sqrt{5}}}}\)
h) Ta có: \(\sqrt{\dfrac{3+\sqrt{5}}{\sqrt{3-\sqrt{5}}}}+\sqrt{\dfrac{3-\sqrt{5}}{\sqrt{3+\sqrt{5}}}}\)
\(=\sqrt{\dfrac{6+2\sqrt{5}}{\sqrt{2}\left(\sqrt{5}-1\right)}}+\sqrt{\dfrac{6-2\sqrt{5}}{\sqrt{2}\left(\sqrt{5}+1\right)}}\)
\(=\dfrac{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)\cdot\sqrt{2}}{\sqrt{2}\left(\sqrt{5}-1\right)}+\dfrac{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)\cdot\sqrt{2}}{\sqrt{2}\left(\sqrt{5}+1\right)}\)
\(=\dfrac{4\sqrt{2}}{\sqrt{2}\left(\sqrt{5}-1\right)}+\dfrac{4\sqrt{2}}{\sqrt{2}\left(\sqrt{5}+1\right)}\)
\(=\sqrt{5}+1+\sqrt{5}-1=2\sqrt{5}\)
Rút gọn biểu thức :
a) A=\(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\).
b)B=\(\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}\)
c) C=\(\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}.\)
a) Ta có: \(A^3=\left(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\right)^3\)
\(=2+\sqrt{5}+2-\sqrt{5}+3\cdot\sqrt[3]{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}\left(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\right)\)
\(=4-3\cdot A\)
\(\Leftrightarrow A^3+3A-4=0\)
\(\Leftrightarrow A^3-A+4A-4=0\)
\(\Leftrightarrow A\left(A-1\right)\left(A+1\right)+4\left(A-1\right)=0\)
\(\Leftrightarrow\left(A-1\right)\left(A^2+A+4\right)=0\)
\(\Leftrightarrow A=1\)
Rút gọn :
a) \(\dfrac{3\sqrt{2-6}}{\sqrt{2-1}}\)
b) \(\dfrac{3\sqrt{5}+5\sqrt{3}}{\sqrt{3}+\sqrt{5}}\)
c) \(\dfrac{x-y}{\sqrt{x}-\sqrt{y}}\)
Câu a bạn xem lại đề nhé vì \(\sqrt{2-6}=\sqrt{-4}\left(VLý\right)\)
b) \(\dfrac{3\sqrt{5}+5\sqrt{3}}{\sqrt{3}+\sqrt{5}}=\dfrac{\sqrt{3.5}\left(\sqrt{3}+\sqrt{5}\right)}{\sqrt{3}+\sqrt{5}}=\sqrt{15}\)
c) \(\dfrac{x-y}{\sqrt{x}-\sqrt{y}}=\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}=\sqrt{x}+\sqrt{y}\)
Rút gọn biểu thức sau
\(a.\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}\)
\(b.\dfrac{\sqrt{\dfrac{5}{3}}+\sqrt{\dfrac{3}{5}}-2}{\sqrt{\dfrac{5}{3}}-\sqrt{\dfrac{3}{5}}}\)
a: \(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}\)
\(=\dfrac{\sqrt{2}+\sqrt{3}+2}{\sqrt{2}+\sqrt{3}+2+2+\sqrt{6}+\sqrt{8}}\)
\(=\dfrac{1}{\sqrt{2}+1}=\sqrt{2}-1\)
Rút gọn biểu thức sau
\(\dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}+\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}\)
\(=\dfrac{8+2\sqrt{15}+8-2\sqrt{15}}{2}\)
=8
nhân liên hợp lên là ra nha bạn! ('ω')
rút gọn A)\(\sqrt{\sqrt{5}-\sqrt{3-\left(2\sqrt{5-3}\right)^2}}\)
B) \(\sqrt{6+2\sqrt{5-\sqrt{\left(2\sqrt{3+1}\right)^2}}}\)
C) \(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}\)
\(c,\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\left(2+\sqrt{3}\right)}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-20-10\sqrt{3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}\)
\(=\sqrt{4+5\sqrt{3}+5\left(5-\sqrt{3}\right)}\)
\(=\sqrt{4+5\sqrt{3}+25-5\sqrt{3}}\)
\(=\sqrt{29}\)