Rút gọn biểu thức :
a, \(\frac{5\sqrt{3}}{\sqrt{3-\sqrt{5}}-\sqrt{3}}-\frac{5\sqrt{3}}{\sqrt{3-\sqrt{5}}+\sqrt{3}}\)
b, \(\frac{\sqrt{2}}{2\sqrt{2}+\sqrt{3+\sqrt{5}}}\)
a) Rút gọn biểu thức:\(\left(\frac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\frac{\sqrt{5}-5}{1-\sqrt{5}}\right):\frac{1}{\sqrt{2}-\sqrt{5}}\)
b) Tìm giá trị nhỏ nhất của biểu thức B=\(x^2-x\sqrt{3}+1\)
a) \(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{\sqrt{5}-5}{1-\sqrt{5}}\right):\dfrac{1}{\sqrt{2}-\sqrt{5}}\)
\(=\left[-\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}-\dfrac{\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}\right]\cdot\left(\sqrt{2}-\sqrt{5}\right)\)
\(=\left(-\sqrt{2}-\sqrt{5}\right)\left(\sqrt{2}-\sqrt{5}\right)\)
\(=-\left(\sqrt{2}+\sqrt{5}\right)\left(\sqrt{2}-\sqrt{5}\right)\)
\(=-\left(2-5\right)\)
\(=-\left(-3\right)\)
\(=3\)
b) Ta có:
\(x^2-x\sqrt{3}+1\)
\(=x^2-2\cdot\dfrac{\sqrt{3}}{2}\cdot x+\left(\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}\)
\(=\left(x-\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}\)
Mà: \(\left(x-\dfrac{\sqrt{3}}{2}\right)^2\ge0\forall x\) nên
\(\left(x-\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}\ge\dfrac{1}{4}\forall x\)
Dấu "=" xảy ra:
\(\left(x-\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}=\dfrac{1}{4}\)
\(\Leftrightarrow x=\dfrac{\sqrt{3}}{2}\)
Vậy: GTNN của biểu thức là \(\dfrac{1}{4}\) tại \(x=\dfrac{\sqrt{3}}{2}\)
a)
\(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{\sqrt{5}-5}{1-\sqrt{5}}\right):\dfrac{1}{\sqrt{2}-\sqrt{5}}\\ =\left(-\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}-\dfrac{\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}\right).\left(\sqrt{2}-\sqrt{5}\right)\\ =\left(-\sqrt{2}-\sqrt{5}\right).\left(\sqrt{2}-\sqrt{5}\right)\\ =-\left(\sqrt{2}+\sqrt{5}\right)\left(\sqrt{2}-\sqrt{5}\right)\\ =-\left(\sqrt{2}^2-\sqrt{5}^2\right)\\ =-\left(2-5\right)\\ =-\left(-3\right)\\ =3\)
Rút gọn biểu thức: \(P=\frac{3+\sqrt{5}}{\sqrt{2}+\sqrt{3}+\sqrt{5}}-\frac{3-\sqrt{5}}{\sqrt{2}-\sqrt{3}-\sqrt{5}}+\frac{\left(\sqrt{5}-1\right).\sqrt[3]{2+\sqrt{5}}}{\sqrt{28}-10\sqrt{3}+\sqrt{3}}\)
Giúp mk nha!
Xem kỹ lại đề nhé! loại này đề lệch một tý thôi -->Không rút được !
p/s: Tránh truongf hợp làm đến cuối mới biết đề sai.
Rút gọn biểu thức
A= \(\frac{\sqrt{2}\left(3+\sqrt{5}\right)}{2\sqrt{2}+\sqrt{3+\sqrt{5}}}+\frac{\sqrt{2}\left(3-\sqrt{5}\right)}{2\sqrt{2}-\sqrt{3-\sqrt{5}}}\)
Rút gọn biểu thức
\(\frac{2+\sqrt{5}}{\sqrt{2}+\sqrt{3+\sqrt{5}}}+\frac{2-\sqrt{5}}{\sqrt{2}-\sqrt{3-\sqrt{5}}}\)
bạn đặt A=biểu thức rồi tính \(\frac{1}{\sqrt{2}}A\) là ra
\(M=\frac{2+\sqrt{5}}{\sqrt{2}+\sqrt{3+\sqrt{5}}}+\frac{2-\sqrt{5}}{\sqrt{2}-\sqrt{3-\sqrt{5}}}\)
\(M.\frac{1}{\sqrt{2}}=\frac{2+\sqrt{5}}{2+\sqrt{6+2\sqrt{5}}}+\frac{2-\sqrt{5}}{2-\sqrt{6-2\sqrt{5}}}\)
\(M.\frac{1}{\sqrt{2}}=\frac{2+\sqrt{5}}{2+\sqrt{5}+1}+\frac{2-\sqrt{5}}{2-\sqrt{5}-1}\)
\(M.\frac{1}{\sqrt{2}}=\frac{2+\sqrt{5}}{3+\sqrt{5}}+\frac{2-\sqrt{5}}{1-\sqrt{5}}\)
P/s làm tiếp nha , hình như bạn ghi đề sai dấu
Rút gọn biểu thức:
a,\(\frac{3+\sqrt{5}}{\sqrt{10}+\sqrt{3+\sqrt{5}}}-\frac{3-\sqrt{5}}{\sqrt{10}+\sqrt{3-\sqrt{5}}}\)
\(b,\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}-\sqrt{2}\)
Rút gọn biểu thức
a) \(\sqrt{28-16\sqrt{3}}+\frac{12}{3-\sqrt{3}}\)
b) \(\frac{5+2\sqrt{5}}{\sqrt{5}}-\frac{20}{5+\sqrt{5}}-\sqrt{20}\)
Rút gọn các biểu thức sau:
A =\(\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}\)
B =\(\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{n-1}+\sqrt{n}}\)
Rút gọn các biểu thức sau :
a/ \(A=\frac{1}{\sqrt{5}+\sqrt{3}}-\frac{1}{\sqrt{5}-\sqrt{3}}\)
b/ \(\frac{\sqrt{4-2\sqrt{3}}}{\sqrt{6}-\sqrt{2}}\)
c/ \(\frac{1}{2+\sqrt{3}}+\frac{\sqrt{2}}{\sqrt{6}}-\frac{2}{3+\sqrt{3}}\)
cần gấp !!!!
a ) \(A=\frac{1}{\sqrt{5}+\sqrt{3}}-\frac{1}{\sqrt{5}-\sqrt{3}}\)
\(=\frac{\left(\sqrt{5}-\sqrt{3}\right)-\left(\sqrt{5}+\sqrt{3}\right)}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}\)
\(=\frac{\sqrt{5}-\sqrt{3}-\sqrt{5}-\sqrt{3}}{5-3}\)
\(=\frac{-2\sqrt{3}}{2}\)
\(=-\sqrt{3}\)
c ) \(\frac{1}{2+\sqrt{3}}+\frac{\sqrt{2}}{\sqrt{6}}-\frac{2}{3+\sqrt{3}}\)
\(=\frac{1}{2+\sqrt{3}}+\frac{1}{\sqrt{3}}-\frac{2}{\sqrt{3}\left(\sqrt{3}+1\right)}\)
\(=\frac{\sqrt{3}\left(\sqrt{3}+1\right)+\left(2+\sqrt{3}\right)\left(\sqrt{3}+1\right)-2\left(2+\sqrt{3}\right)}{\sqrt{3}\left(\sqrt{3}+1\right)\left(2+\sqrt{3}\right)}\)
\(=\frac{2\sqrt{3}+4}{\sqrt{3}\left(\sqrt{3}+1\right)\left(2+\sqrt{3}\right)}\)
\(=\frac{2\left(\sqrt{3}+2\right)}{\sqrt{3}\left(\sqrt{3}+1\right)\left(2+\sqrt{3}\right)}\)
\(=\frac{2.\sqrt{3}\left(\sqrt{3}-1\right)}{3\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\)
\(=\frac{2\sqrt{3}\left(\sqrt{3}-1\right)}{3.\left(3-1\right)}\)
\(=\frac{\sqrt{3}\left(\sqrt{3}-1\right)}{3}\)
\(=\frac{3-\sqrt{3}}{3}\)
\(=1-\frac{\sqrt{3}}{3}\)
b ) \(\frac{\sqrt{4-2\sqrt{3}}}{\sqrt{6}-\sqrt{2}}\)
\(=\frac{\sqrt{\left(\sqrt{3}\right)^2-2\sqrt{3}+1}}{\sqrt{2}\left(\sqrt{3}-1\right)}\)
\(=\frac{\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{2}\left(\sqrt{3}-1\right)}\)
\(=\frac{\left|\sqrt{3}-1\right|}{\sqrt{2}\left(\sqrt{3}-1\right)}=\frac{\sqrt{3}-1}{\sqrt{2}}=\frac{1}{\sqrt{2}}\)
\(=\frac{\sqrt{2}}{2}\)
Rút gọn biểu thức: \(\frac{\sqrt{20+8\sqrt{3}} +\sqrt{20-8\sqrt{3}}}{\sqrt{5+2\sqrt{3}}-\sqrt{5-2\sqrt{3}}}-\frac{\sqrt{4+\sqrt{3}}+\sqrt{4-\sqrt{3}}}{\sqrt{4+\sqrt{3}}-\sqrt{4-\sqrt{3}}}\)