\(\frac{\sqrt{20}-\sqrt{12}+2}{\sqrt{35}-\sqrt{21}+\sqrt{7}}+\frac{5}{\sqrt{7}}-\sqrt{8-2\sqrt{7}}\)
Những câu hỏi liên quan
So sánh
a)\(\sqrt{21}+\sqrt{5}\) và \(\sqrt{20}-\sqrt{6}\)
b)\(\frac{\sqrt{5^2}+\sqrt{35^2}}{\sqrt{7^2}+\sqrt{49^2}}\) và \(\frac{\sqrt{5^2}-\sqrt{35^2}}{\sqrt{7^2}-\sqrt{49^2}}\)
b) Ta có: \(\frac{\sqrt{5^2}+\sqrt{35^2}}{\sqrt{7^2}+\sqrt{49^2}}=\frac{5+35}{7+49}=\frac{40}{56}=\frac{5}{7}\) (1)
Lại có: \(\frac{\sqrt{5^2}-\sqrt{35^2}}{\sqrt{7^2}-\sqrt{49^2}}=\frac{5-35}{7-49}=\frac{-30}{-42}=\frac{5}{7}\) (2)
Từ biểu thức (1) và biểu thức (2)
=> \(\frac{\sqrt{5^2}+\sqrt{35^2}}{\sqrt{7^2}+\sqrt{49^2}}=\frac{\sqrt{5^2}-\sqrt{35^2}}{\sqrt{7^2}-\sqrt{49^2}}\)
Đúng 0
Bình luận (0)
Giải :
1) \(\frac{5\sqrt{7}-7\sqrt{5}+2\sqrt{70}}{\sqrt{35}}\)
2) \(\sqrt{\frac{4}{3}}+\sqrt{12}-\frac{4}{3}\sqrt{\frac{3}{4}}\)
3) \(\frac{1}{1+\sqrt{2}+\sqrt{3}}\)
4) \(\left(5\sqrt{\frac{1}{5}}+\frac{1}{2}\sqrt{20}-\frac{5}{4}\sqrt{\frac{4}{5}+\sqrt{5}}\right):2\sqrt{5}\)
\(\frac{5\sqrt{7}-7\sqrt{5}+2\sqrt{70}}{\sqrt{35}}\)
\(=\frac{\sqrt{35}.(5\sqrt{7}-7\sqrt{5}+2\sqrt{70})}{\sqrt{35}.\sqrt{35}}\)
\(=\frac{\sqrt{35}.(5\sqrt{7}-7\sqrt{5}+2\sqrt{70})}{35}\)
Đúng 0
Bình luận (0)
\(\sqrt{\frac{4}{3}}+\sqrt{12}-\frac{4}{3}\sqrt{\frac{3}{4}}\)
\(=\frac{\sqrt{4}}{\sqrt{3}}+\sqrt{12}-\frac{4}{3}\cdot\frac{\sqrt{3}}{\sqrt{4}}\)
\(=\frac{2\sqrt{3}}{\sqrt{3}.\sqrt{3}}+\sqrt{12}-\frac{4}{3}\cdot\frac{\sqrt{3}}{2}\)
\(=\frac{2\sqrt{3}}{3}+2\sqrt{3}-\frac{2\sqrt{3}}{3}\)
\(=2\sqrt{3}\left(\frac{1}{3}+1-\frac{1}{3}\right)\)
\(=2\sqrt{3}\)
Đúng 0
Bình luận (0)
\(\left(5\sqrt{\frac{1}{5}}+\frac{1}{2}\sqrt{20}-\frac{5}{4}\sqrt{\frac{4}{5}+5}\right):2\sqrt{5}\)
\(=\left(5\cdot\frac{\sqrt{1}}{\sqrt{5}}+\frac{1}{2}\sqrt{4.5}-\frac{5}{4}\sqrt{\frac{4+25}{5}}\right)\cdot\frac{1}{2\sqrt{5}}\)
\(=\left(\frac{5\sqrt{5}}{\sqrt{5}.\sqrt{5}}+\frac{1}{2}.2\sqrt{5}-\frac{5}{4}\sqrt{\frac{29}{5}}\right)\cdot\frac{\sqrt{5}}{2\cdot\sqrt{5}\cdot\sqrt{5}}\)
\(=\left(\frac{5\sqrt{5}}{5}+\sqrt{5}-\frac{5}{4}\cdot\frac{\sqrt{29}}{\sqrt{5}}\right)\cdot\frac{\sqrt{5}}{10}\)
\(=\left(\sqrt{5}+\sqrt{5}-\frac{5}{4}\cdot\frac{\sqrt{29}\sqrt{5}}{\sqrt{5}\sqrt{5}}\right)\cdot\frac{\sqrt{5}}{10}\)
\(=\left(2\sqrt{5}-\frac{5}{4}\cdot\frac{\sqrt{145}}{5}\right)\cdot\frac{\sqrt{5}}{10}\)
\(=\left(2\sqrt{5}-\frac{\sqrt{145}}{4}\right)\cdot\frac{\sqrt{5}}{10}\)
Đúng 0
Bình luận (0)
Xem thêm câu trả lời
a. P (3+sqrt{2}+sqrt{6})(sqrt{6-3sqrt{3}})
b. A(frac{15}{sqrt{6}+1}+frac{4}{sqrt{6}-2}-frac{12}{3-sqrt{6}}): (sqrt{6}+11)
c. B frac{sqrt{8-2sqrt{12}}}{sqrt{3}-1}-sqrt{8}
d. C sqrt{4+sqrt{7}}-sqrt{4-sqrt{7}}-sqrt{2}
đ. Dfrac{1}{sqrt{2}-sqrt{3}}sqrt{frac{3sqrt{2}-2sqrt{3}}{3sqrt{2}+2sqrt{3}}}
e. E sqrt{8+2sqrt{10+2sqrt{5}}}+sqrt{8-2sqrt{10+2sqrt{5}}}
ê. G sqrt{4+5sqrt{3}+5sqrt{48-10sqrt{7+4sqrt{3}}}}
g. Hfrac{2sqrt{4+sqrt{5+21+sqrt{80}}}}{sqrt{10}-sqrt{2}}
i. Isqrt{frac{4-sqrt{7}}{4+sqrt{7...
Đọc tiếp
a. P= (\(3+\sqrt{2}+\sqrt{6}\))(\(\sqrt{6-3\sqrt{3}}\))
b. A=(\(\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{6}-2}-\frac{12}{3-\sqrt{6}}\)): (\(\sqrt{6}+11\))
c. B= \(\frac{\sqrt{8-2\sqrt{12}}}{\sqrt{3}-1}\)-\(\sqrt{8}\)
d. C= \(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}-\sqrt{2}\)
đ. D=\(\frac{1}{\sqrt{2}-\sqrt{3}}\sqrt{\frac{3\sqrt{2}-2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}}}\)
e. E= \(\sqrt{8+2\sqrt{10+2\sqrt{5}}}+\sqrt{8-2\sqrt{10+2\sqrt{5}}}\)
ê. G= \(\sqrt{4+5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}\)
g. H=\(\frac{2\sqrt{4+\sqrt{5+21+\sqrt{80}}}}{\sqrt{10}-\sqrt{2}}\)
i. I=\(\sqrt{\frac{4-\sqrt{7}}{4+\sqrt{7}}}+\sqrt{\frac{4+\sqrt{7}}{4-\sqrt{7}}}\)
k. K=\(\frac{3+\sqrt{5}}{\sqrt{2}+\sqrt{3+\sqrt{5}}}+\frac{3-\sqrt{5}}{\sqrt{2}-\sqrt{3-\sqrt{5}}}\)
1/ Rut gon bieu thuc sau:
a) \(\sqrt{12-2\sqrt{35}}+\sqrt{7-2\sqrt{10}}-\sqrt{\sqrt{49}}\)
b) \(\frac{\sqrt{7}-5}{2}-\frac{6}{\sqrt{7}-2}+\frac{1}{3+\sqrt{7}}+\frac{3}{5+2\sqrt{7}}\)
Tính
A=\(\left(\frac{15}{\sqrt{7}+2}+\frac{12}{\sqrt{7}-1}-\frac{8}{3-\sqrt{7}}\right)\cdot\left(3\sqrt{7}+20\right)\)
B=\(\left(9+4\sqrt{5}\right):\left(\frac{\sqrt{5}+2}{\sqrt{5}-2}\right)\)
a) frac{2sqrt{5}-4sqrt{10}}{3sqrt{10}}b)frac{6sqrt{6}-2sqrt{12}+3-sqrt{2}}{2sqrt{6}+1}c)frac{sqrt{3left(3-sqrt{11}right)^2}}{sqrt{6}left(3-sqrt{11}right)}d)frac{5sqrt{7}-4sqrt{35}+7sqrt{5}}{sqrt{35}}e) left(sqrt{3}-1right)sqrt{2sqrt{19+8sqrt{3}-4}}g) sqrt{47+sqrt{5}}.sqrt{7-sqrt{2+sqrt{5}}.}sqrt{7+sqrt{2+sqrt{5}}}
Đọc tiếp
a) \(\frac{2\sqrt{5}-4\sqrt{10}}{3\sqrt{10}}\)
b)\(\frac{6\sqrt{6}-2\sqrt{12}+3-\sqrt{2}}{2\sqrt{6}+1}\)
c)\(\frac{\sqrt{3\left(3-\sqrt{11}\right)^2}}{\sqrt{6}\left(3-\sqrt{11}\right)}\)
d)\(\frac{5\sqrt{7}-4\sqrt{35}+7\sqrt{5}}{\sqrt{35}}\)
e) \(\left(\sqrt{3}-1\right)\sqrt{2\sqrt{19+8\sqrt{3}-4}}\)
g) \(\sqrt{47+\sqrt{5}}.\sqrt{7-\sqrt{2+\sqrt{5}}.}\sqrt{7+\sqrt{2+\sqrt{5}}}\)
Rút gọn
a)\(\frac{2\sqrt{5\:\:\:}-4\sqrt{10}}{3\sqrt{10}}\)
b)\(\frac{6\sqrt{6}-2\sqrt{12}+3-\sqrt{2}}{2\sqrt{6}+1}\)
c)\(\frac{5\sqrt{7}-4\sqrt{35}+7\sqrt{5}}{\sqrt{35}}\)
d)\(\left(\sqrt{3}-1\right)\sqrt{2\sqrt{19+8\sqrt{3}}-4}\)
\(\frac{8+2\sqrt{15}+\sqrt{21}+\sqrt{35}}{\sqrt{3}+\sqrt{5}+\sqrt{7}}\)
rút gọn
Tính:
a) Asqrt{8-2sqrt{15}}left(sqrt{3}+sqrt{5}right)-left(sqrt{45}-sqrt{20}right)
b) Bleft(frac{sqrt{21}-sqrt{3}}{sqrt{7}-1}-frac{sqrt{15}-sqrt{3}}{1-sqrt{5}}right)left(frac{1}{2}sqrt{6}-sqrt{frac{3}{2}}+3sqrt{frac{2}{3}}right)
c) C2sqrt{3}+sqrt{7-4sqrt{3}}+left(sqrt{frac{1}{3}}-sqrt{frac{4}{3}+}sqrt{3}right):sqrt{3}
d) Dleft(frac{5+sqrt{5}}{5-sqrt{5}}+frac{5-sqrt{5}}{5+sqrt{5}}right):frac{1}{sqrt{7-4sqrt{3}}}
Đọc tiếp
Tính:
a) \(A=\sqrt{8-2\sqrt{15}}\left(\sqrt{3}+\sqrt{5}\right)-\left(\sqrt{45}-\sqrt{20}\right)\)
b) \(B=\left(\frac{\sqrt{21}-\sqrt{3}}{\sqrt{7}-1}-\frac{\sqrt{15}-\sqrt{3}}{1-\sqrt{5}}\right)\left(\frac{1}{2}\sqrt{6}-\sqrt{\frac{3}{2}}+3\sqrt{\frac{2}{3}}\right)\)
c) \(C=2\sqrt{3}+\sqrt{7-4\sqrt{3}}+\left(\sqrt{\frac{1}{3}}-\sqrt{\frac{4}{3}+}\sqrt{3}\right):\sqrt{3}\)
d) \(D=\left(\frac{5+\sqrt{5}}{5-\sqrt{5}}+\frac{5-\sqrt{5}}{5+\sqrt{5}}\right):\frac{1}{\sqrt{7-4\sqrt{3}}}\)
a) Ta có: \(A=\sqrt{8-2\sqrt{15}}\cdot\left(\sqrt{3}+\sqrt{5}\right)-\left(\sqrt{45}-\sqrt{20}\right)\)
\(=\sqrt{5-2\cdot\sqrt{5}\cdot\sqrt{3}+3}\cdot\left(\sqrt{5}+\sqrt{3}\right)-\sqrt{5}\left(\sqrt{9}-\sqrt{4}\right)\)
\(=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\cdot\left(\sqrt{5}+\sqrt{3}\right)-\sqrt{5}\)
\(=\left|\sqrt{5}-\sqrt{3}\right|\cdot\left(\sqrt{5}+\sqrt{3}\right)-\sqrt{5}\)
\(=\left(\sqrt{5}-\sqrt{3}\right)\cdot\left(\sqrt{5}+\sqrt{3}\right)-\sqrt{5}\)(Vì \(\sqrt{5}>\sqrt{3}\))
\(=5-3-\sqrt{5}\)
\(=2-\sqrt{5}\)
b) Ta có: \(B=\left(\frac{\sqrt{21}-\sqrt{3}}{\sqrt{7}-1}-\frac{\sqrt{15}-\sqrt{3}}{1-\sqrt{5}}\right)\left(\frac{1}{2}\sqrt{6}-\sqrt{\frac{3}{2}}+3\sqrt{\frac{2}{3}}\right)\)
\(=\left(\frac{\sqrt{3}\left(\sqrt{7}-1\right)}{\sqrt{7}-1}+\frac{\sqrt{3}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}\right)\left(\sqrt{\frac{3}{2}}-\sqrt{\frac{3}{2}}+\sqrt{6}\right)\)
\(=\sqrt{3}+\sqrt{3}+\sqrt{6}\)
\(=2\sqrt{3}+\sqrt{6}\)
c) Ta có: \(C=2\sqrt{3}+\sqrt{7-4\sqrt{3}}+\left(\sqrt{\frac{1}{3}}-\sqrt{\frac{4}{3}}+\sqrt{3}\right):\sqrt{3}\)
\(=2\sqrt{3}+\sqrt{4-2\cdot2\cdot\sqrt{3}+3}+\sqrt{\frac{1}{3}:3}-\sqrt{\frac{4}{3}:3}+\sqrt{3:3}\)
\(=2\sqrt{3}+\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\frac{1}{9}}-\sqrt{\frac{4}{9}}+\sqrt{1}\)
\(=2\sqrt{3}+\left|2-\sqrt{3}\right|+\frac{1}{3}-\frac{2}{3}+1\)
\(=2\sqrt{3}+2-\sqrt{3}+\frac{2}{3}\)(Vì \(2>\sqrt{3}\))
\(=\sqrt{3}+\frac{8}{3}\)
d) Ta có: \(D=\left(\frac{5+\sqrt{5}}{5-\sqrt{5}}+\frac{5-\sqrt{5}}{5+\sqrt{5}}\right):\frac{1}{\sqrt{7-4\sqrt{3}}}\)
\(=\left(\frac{\left(5+\sqrt{5}\right)^2+\left(5-\sqrt{5}\right)^2}{\left(5-\sqrt{5}\right)\left(5+\sqrt{5}\right)}\right)\cdot\sqrt{4-2\cdot2\cdot\sqrt{3}+3}\)
\(=\frac{25+10\sqrt{5}+5+25-10\sqrt{5}+5}{25-5}\cdot\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=\frac{60}{20}\cdot\left|2-\sqrt{3}\right|\)
\(=3\cdot\left(2-\sqrt{3}\right)\)(Vì \(2>\sqrt{3}\))
\(=6-3\sqrt{3}\)
Đúng 0
Bình luận (0)