rút gọn biểu thức
P= \(\sqrt{\frac{4-\sqrt{7}}{4+\sqrt{7}}+\sqrt{\frac{4+\sqrt{7}}{4-\sqrt{-7}}}}\)
Giups mình với :< pls
Đề : Rút gọn biểu thức
B =\(\frac{1}{\sqrt{5}+\sqrt{7}}-\frac{1}{\sqrt{5}-\sqrt{7}}\)
C =\(\sqrt{\frac{4+\sqrt{7}}{4-\sqrt{7}}}+\sqrt{\frac{4-\sqrt{7}}{4+\sqrt{7}}}\)
\(B=\frac{1}{\sqrt{5}+\sqrt{7}}-\frac{1}{\sqrt{5}-\sqrt{7}}=\frac{\sqrt{5}-\sqrt{7}-\sqrt{5}-\sqrt{7}}{5-7}=\frac{-2\sqrt{7}}{-2}=\sqrt{7}\)
\(C=\sqrt{\frac{4+\sqrt{7}}{4-\sqrt{7}}}+\sqrt{\frac{4-\sqrt{7}}{4+\sqrt{7}}}=\sqrt{\left(\sqrt{\frac{4+\sqrt{7}}{4-\sqrt{7}}}+\sqrt{\frac{4-\sqrt{7}}{4+\sqrt{7}}}\right)^2}\)
\(C=\sqrt{\frac{4+\sqrt{7}}{4-\sqrt{7}}+2\sqrt{\frac{\left(4+\sqrt{7}\right)\left(4-\sqrt{7}\right)}{\left(4-\sqrt{7}\right)\left(4+\sqrt{7}\right)}}+\frac{4-\sqrt{7}}{4+\sqrt{7}}}\)
\(C=\sqrt{\frac{\left(4+\sqrt{7}\right)^2}{16-7}+\frac{\left(4-\sqrt{7}\right)^2}{16-7}+2}\)
\(C=\sqrt{\frac{\left(4+\sqrt{7}+4-\sqrt{7}\right)^2-2\left(4+\sqrt{7}\right)\left(4-\sqrt{7}\right)}{16-7}+2}\)
\(C=\sqrt{\frac{16^2-2\left(16-7\right)}{9}+2}=\sqrt{\frac{238}{9}+2}=\sqrt{\frac{256}{9}}=\frac{16}{3}\)
Chúc bạn học tốt ~
đoạn cuối sửa lại nhé -,- tính ngu
\(C=\sqrt{\frac{8^2-2\left(16-7\right)}{9}+2}=\sqrt{\frac{46}{9}+2}=\sqrt{\frac{64}{9}}=\frac{8}{3}\)
Chúc bạn học tốt ~
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 sau :
A = \(\frac{1}{4\sqrt{1}+1\sqrt{4}}+\frac{1}{7\sqrt{4}+4\sqrt{7}}+\frac{1}{10\sqrt{7}+7\sqrt{10}}...+\frac{1}{2007\sqrt{2004}+2004\sqrt{2007}}\)
1) Rút gọn biểu thức:
a) \(\frac{1}{7+4\sqrt{3}}+\frac{1}{7-4\sqrt{3}}\)
b) \(\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{6}-2}-\frac{12}{3-\sqrt{6}}-\sqrt{6}\)
c) \(\frac{5}{4-\sqrt{11}}+\frac{1}{3+\sqrt{7}}-\frac{6}{\sqrt{7}-2}-\frac{\sqrt{7}-5}{2}\)
d) \(\frac{4}{\sqrt{5}-\sqrt{2}}+\frac{3}{\sqrt{5}-2}-\frac{2}{\sqrt{3}-2}+\frac{\sqrt{3}-1}{6}\)
a) \(=\frac{7-4\sqrt{3}+7+4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}=\frac{14}{49-48}=14\)
b) \(=\frac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}-\frac{5\sqrt{6}}{5}+\frac{4\sqrt{3}-12\sqrt{2}}{\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)}\)
\(\frac{1-\frac{1}{\sqrt{49}}+\frac{1}{49}-\frac{1}{\left(7\sqrt{7}\right)^2}}{\frac{\sqrt{64}}{2}-\frac{4}{7}+\left(\frac{2}{7}\right)^2-\frac{4}{343}}\)
Rút gọn biểu thức
\(\frac{1-\frac{1}{\sqrt{49}}+\frac{1}{49}-\frac{1}{\left(7\sqrt{7}\right)^2}}{\frac{\sqrt{64}}{2}-\frac{4}{7}+\left(\frac{2}{7}\right)^2-\frac{4}{343}}\)
\(=\frac{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}{4-\frac{4}{7}+\frac{4}{49}-\frac{4}{343}}\)
\(=\frac{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}{4\left(1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}\right)}\)
\(=\frac{1}{4}\)
\(A=\left(\sqrt{3}+1\right)^2+\frac{5}{4}\sqrt{48}-\frac{2}{\sqrt{3}+1}\)
\(B=\frac{4}{3-\sqrt{5}}-\frac{3}{\sqrt{5}+\sqrt{2}}-\frac{1}{\sqrt{2}-1}\)
\(C=\sqrt{4-2\sqrt{3}}-\sqrt{7+4\sqrt{3}}\)
Rút gọn biểu thức
Giúp mình với mình cần gấp
\(C=\sqrt{4-2\sqrt{3}}-\sqrt{7+4\sqrt{3}}\)
\(\Leftrightarrow C=\sqrt{3-2\sqrt{3}+1}-\sqrt{4+4\sqrt{3}+3}\)
\(\Leftrightarrow C=\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{\left(2+\sqrt{3}\right)^2}\)
\(\Leftrightarrow C=\left|\sqrt{3}-1\right|-\left|2+\sqrt{3}\right|\)
\(\Leftrightarrow C=\sqrt{3}-1-2-\sqrt{3}\)
\(\Leftrightarrow C=-3\)
Rút gọn các biểu thức sau
a)\(\left(\sqrt{14}+\sqrt{16}\right)\sqrt{5-\sqrt{21}}\)
b)\(\frac{5}{\sqrt{21}-4}+\frac{3\sqrt{7}-7\sqrt{3}}{\sqrt{7}-\sqrt{3}}\)
a. \(=\sqrt{2}.\left(\sqrt{7}+\sqrt{8}\right)\sqrt{5-\sqrt{3}\sqrt{7}}\)
\(=\left(\sqrt{7}+\sqrt{8}\right)\sqrt{3-2\sqrt{3}.\sqrt{7}+7}\)
\(=\left(\sqrt{7}+\sqrt{8}\right)\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}\)
\(=\left(\sqrt{7}+\sqrt{8}\right)\left(\sqrt{7}-\sqrt{3}\right)\)
Rồi nhân ra. bạn làm tiếp nhé. Tuy nhiên minh nghĩ bạn bị nhầm đề. là \(\sqrt{6}\) chứ không phải căn 16
b. \(=\frac{5\left(\sqrt{21}+1\right)}{21-16}+\frac{\sqrt{3}.\sqrt{7}\left(\sqrt{3}-\sqrt{7}\right)}{-\left(\sqrt{3}-\sqrt{7}\right)}\)
\(=\sqrt{21}+4-\sqrt{21}=4\)
Rút gọn biểu thức sau:
\(\dfrac{\left(4+\sqrt{7}\right).\sqrt{4-\sqrt{7}}}{\sqrt{4+\sqrt{7}}}\)
\(\left(4+\sqrt{7}\right)\cdot\dfrac{\sqrt{4-\sqrt{7}}}{\sqrt{4+\sqrt{7}}}\)
\(=\left(4+\sqrt{7}\right)\cdot\dfrac{\sqrt{7}-1}{\sqrt{7}+1}\)
\(=\dfrac{\left(\sqrt{7}+1\right)^2\cdot\left(\sqrt{7}-1\right)}{\sqrt{7}+1}\cdot\dfrac{1}{2}\)
\(=\dfrac{6}{2}=3\)
\(=\dfrac{\left(8+2\sqrt{7}\right)\sqrt{8-2\sqrt{7}}}{2\sqrt{8+2\sqrt{7}}}=\dfrac{\left(\sqrt{7}+1\right)^2\sqrt{\left(\sqrt{7}-1\right)^2}}{2\sqrt{\left(\sqrt{7}+1\right)^2}}\)
\(=\dfrac{\left(\sqrt{7}+1\right)^2\left(\sqrt{7}-1\right)}{2\left(\sqrt{7}+1\right)}=\dfrac{\left(\sqrt{7}+1\right)\left(\sqrt{7}-1\right)}{2}\)
\(=\dfrac{7-1}{2}=3\)
Rút gọn biểu thức
\(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\)
√4−√7−√4+√7+√7=√2(√4−√7−√4+√7+√7)√2=√8−2√7−√8+2√7+√14√2=√7−2√7+1−√7+2√7+1+√14√2=√(√7−1)2−√(√7+1)2+√14√2=∣∣√7−1∣∣−∣∣√7+1∣∣+√14√2=√7−1−√7−1+√14√2=√14−2√2=√2(√7−√2)√2=√7−√2
Lời giải:
\(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}=\sqrt{\frac{8-2\sqrt{7}}{2}}-\sqrt{\frac{8+2\sqrt{7}}{2}}=\sqrt{\frac{(\sqrt{7}-1)^2}{2}}-\sqrt{\frac{(\sqrt{7}+1)^2}{2}}\)
\(=\frac{|\sqrt{7}-1|}{\sqrt{2}}-\frac{|\sqrt{7}+1|}{\sqrt{2}}=\frac{\sqrt{7}-1-(\sqrt{7}+1)}{\sqrt{2}}=\frac{-2}{\sqrt{2}}=-\sqrt{2}\)