a)\(\sqrt{11-6\sqrt{2}}-\sqrt{27+10\sqrt{2}}\)
b) \(\sqrt{13-4\sqrt{3}}-\sqrt{16-8\sqrt{3}}\)
giải giúp mình với
1) Rút gọn:
a) A = \(\sqrt{5-2\sqrt{3-\sqrt{3}}}-\sqrt{3+\sqrt{3}}+\sqrt{2+\sqrt{3}}\)
b) B = \(\sqrt{13+\sqrt{2}+5\sqrt{1+2\sqrt{2}}}+\sqrt{13+\sqrt{2}+5\sqrt{1+2\sqrt{2}}}\)
c) C = \(\dfrac{\sqrt{21+3\sqrt{5}}+\sqrt{21-3\sqrt{5}}}{\sqrt{21}+6\sqrt{11}}+\sqrt{11-6\sqrt{2}}\)
d) D = \(\left(\sqrt{8+2\sqrt{10+2\sqrt{5}}}+\sqrt{8-2\sqrt{10+2\sqrt{5}}}\right).\sqrt{\dfrac{2+2\sqrt{5}}{2+\sqrt{5}}}\)
e) E = \(\dfrac{\left(27+10\sqrt{2}\right)\sqrt{27-10\sqrt{2}}-\left(27-10\sqrt{2}\right)\sqrt{27+10\sqrt{2}}}{\left(\sqrt{\sqrt{13}-3}+\sqrt{\sqrt{13}+3}\right):\sqrt{\sqrt{13}+2}}\)
rút gọn biểu thức :
A= \(\dfrac{\sqrt{4+\sqrt{3}}+\sqrt{4-\sqrt{3}}}{\sqrt{4+\sqrt{13}}}+\sqrt{27-10\sqrt{2}}\).
B= \(\dfrac{\sqrt{2-\sqrt{3}}+\sqrt{4-\sqrt{15}}+\sqrt{10}}{\sqrt{23-3\sqrt{5}}}\).
C= \(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\).
Ta có: \(C=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{4}+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\dfrac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)\left(1+\sqrt{2}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=1+\sqrt{2}\)
Ta có: \(B=\dfrac{\sqrt{2-\sqrt{3}}+\sqrt{4-\sqrt{15}}+\sqrt{10}}{\sqrt{23-3\sqrt{5}}}\)
\(=\dfrac{\sqrt{4-2\sqrt{3}}+\sqrt{8-2\sqrt{15}}+2\sqrt{5}}{3\sqrt{5}-1}\)
\(=\dfrac{\sqrt{3}-1+\sqrt{5}-\sqrt{3}+2\sqrt{5}}{3\sqrt{5}-1}\)
=1
Các bạn giúp mình nha
Bài 1: Giải phương trình
a,\(\sqrt[3]{x+1}+\sqrt[3]{7-x}=2\)
b, \(\sqrt[3]{x+3}-\sqrt[3]{6-x}=1\)
Bài 2: Tính giá trị của các biểu thức
a, \(A=\sqrt[3]{6\sqrt{3}+10}-\sqrt[3]{6\sqrt{3}-10}\)
b, \(B=\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}\)
c, \(D=\sqrt[3]{2+10\sqrt{\frac{1}{27}}}+\sqrt[3]{2-10\sqrt{\frac{1}{27}}}\)
Bài 3: Số m dưới đây có phải là nghiệm của phương trình \(x^3+12x-8=0\)không ?
\(m=\sqrt[3]{4+\sqrt{80}}-\sqrt[3]{\sqrt{80}-4}\)
\(\sqrt{5-\sqrt{6}+\sqrt{10}-\sqrt{15}}-\frac{\sqrt{16-4\sqrt{15}}}{2}\)
\(\frac{\sqrt{1+2\sqrt{27\sqrt{2}-38}}-\sqrt{5-3\sqrt{2}}}{\sqrt{3\sqrt{2}}-4}\)
\(\sqrt{3+\sqrt{2}+\sqrt{3}+\sqrt{6}}-\sqrt{2+\sqrt{3}}\)
\(\frac{3-\sqrt{6+\sqrt{3+\sqrt{6+\sqrt{3}}}}}{3-\sqrt{3+\sqrt{6+\sqrt{3}}}}+\frac{2+\sqrt{6+\sqrt{3+\sqrt{6+\sqrt{3}}}}}{3+\sqrt{6+\sqrt{3+\sqrt{6+\sqrt{3}}}}}\)
\(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
Giải dùm mình nhé!!!!!!! Không cần giải hết đâu!!!!! Mọi người thấy bài nào dạng giống nhau thì giải 1 lần rồi nói mình đó là bài nào lun nha!!!!!!!!!! Mình cần rất gấp ạ!!!!!! Mong các bạn giúp mình
giải hộ mik
a)\(\sqrt{11+6\sqrt{2}}-\left(3+\sqrt{2}\right)\)
b)\(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
a) Ta có: \(\sqrt{11+6\sqrt{2}}-\left(3+\sqrt{2}\right)\)
\(=3+\sqrt{2}-3-\sqrt{2}\)
=0
b) Ta có: \(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{6}+\sqrt{8}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+2}\)
\(=\sqrt{2}+1\)
a) \(A=\sqrt{8+2\sqrt{10+2\sqrt{5}}}+\sqrt{8-2\sqrt{10+2\sqrt{5}}}\)
b) \(B=\left(\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{6}-2}+\frac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)
Giúp mình với dang cần gấp
có ai biết giải ko giải hộ mình mấy bài này với ( giải chi tiết hộ mình nhé)
1, \(2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}\)
2, \(\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}\)
3, \(\sqrt{4+\sqrt{5\sqrt{3+}5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
4, \(\sqrt{30-2\sqrt{16+6\sqrt{11+4\sqrt{4-2\sqrt{3}}}}}\)
5, \(\dfrac{\left(5\sqrt{3}+\sqrt{50}\right)\left(5-\sqrt{24}\right)}{\sqrt{75}-5\sqrt{2}}\)
6, \(\sqrt{4+\sqrt{8}.\sqrt{2+\sqrt{2}}}.\sqrt{2-\sqrt{2+\sqrt{2}}}\)
7, \(\sqrt{8\sqrt{3}-2\sqrt{25\sqrt{12}+4\sqrt{192}}}\)
\(\sqrt{13+\sqrt{48}}=\sqrt{13+\sqrt{4.12}}=\sqrt{13+2\sqrt{12}}=\sqrt{\left(\sqrt{12}+1\right)^2}\)
\(=\sqrt{12}+1=2\sqrt{3}+1\)
\(\Rightarrow\sqrt{5-\sqrt{13+\sqrt{48}}}=\sqrt{5-2\sqrt{3}-1}=\sqrt{4-2\sqrt{3}}=\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(=\sqrt{3}-1\)
\(\Rightarrow\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}=\sqrt{3+\sqrt{3}-1}=\sqrt{2+\sqrt{3}}\)
\(\Rightarrow\sqrt{\dfrac{4+2\sqrt{3}}{2}}=\sqrt{\dfrac{\left(\sqrt{3}+1\right)^2}{2}}=\dfrac{\sqrt{3}+1}{\sqrt{2}}\)
\(\Rightarrow2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}==2.\dfrac{\sqrt{3}+1}{\sqrt{2}}=\sqrt{6}+\sqrt{2}\)
2) biến đổi khúc sau như câu 1:
\(\Rightarrow\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}=\sqrt{6+2\left(\sqrt{3}-1\right)}=\sqrt{4+2\sqrt{3}}\)
\(=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)
4) Ta có: \(\sqrt{30-2\sqrt{16+6\sqrt{11+4\sqrt{4-2\sqrt{3}}}}}\)
\(=\sqrt{30-2\sqrt{16+6\sqrt{11+4\left(\sqrt{3}-1\right)}}}\)
\(=\sqrt{30-2\sqrt{16+6\sqrt{7+4\sqrt{3}}}}\)
\(=\sqrt{30-2\sqrt{16+6\left(2+\sqrt{3}\right)}}\)
\(=\sqrt{30-2\sqrt{28+6\sqrt{3}}}\)
\(=\sqrt{30-2\left(3\sqrt{3}+1\right)}\)
\(=\sqrt{28-6\sqrt{3}}=3\sqrt{3}-1\)
5) Ta có: \(\dfrac{\left(5\sqrt{3}+\sqrt{50}\right)\left(5-\sqrt{24}\right)}{\sqrt{75}-5\sqrt{2}}\)
\(=\dfrac{5\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)^2}{\sqrt{75}-5\sqrt{2}}\)
\(=\dfrac{5\left(\sqrt{3}-\sqrt{2}\right)}{5\left(\sqrt{3}-\sqrt{2}\right)}=1\)
bài 1: thực hiện phép tính
1) 2\(\sqrt{5}\)\(-\sqrt{125}-\sqrt{80}+\sqrt{605}\)
2) \(\sqrt{15-\sqrt{216}}+\sqrt{33-12\sqrt{6}}\)
3) \(\frac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\frac{8}{1-\sqrt{5}}\)
4) \(\frac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\frac{\sqrt{5}-\sqrt{27}}{\sqrt{30}-\sqrt{162}}\)
5) \(2\sqrt{\frac{16}{3}}-3\sqrt{\frac{1}{27}}-6\sqrt{\frac{4}{75}}\)
6) \(\sqrt{\frac{2-\sqrt{3}}{2+\sqrt{3}}-\sqrt{\frac{2+\sqrt{3}}{2-\sqrt{3}}}}\)
GIÚP MÌNH VỚI M.N~~~!!!!!!!!! MÌNH ĐANG CẦN GẤP LẮM (T_T)
1) \(2\sqrt{5}-\sqrt{125}-\sqrt{80}+\sqrt{605}\)
\(=2\sqrt{5}-\sqrt{5^2.5}-\sqrt{4^2.5}+\sqrt{11^2.5}\)
\(=2\sqrt{5}-5\sqrt{5}-4\sqrt{5}+11\sqrt{5}\)
\(=4\sqrt{5}\)
2) \(\sqrt{15-\sqrt{216}}+\sqrt{33-12\sqrt{6}}\)
\(=\sqrt{15-\sqrt{6^2.6}}+\sqrt{33-12\sqrt{6}}\)
\(=\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)
\(=\sqrt{\left(\sqrt{6}\right)^2-6\sqrt{6}+3^2}+\sqrt{\left(2\sqrt{6}\right)^2-12\sqrt{6}+3^2}\)
\(=\sqrt{\left(\sqrt{6}-3\right)^2}+\sqrt{\left(2\sqrt{6}-3\right)^2}\)
\(=\left|\sqrt{6}-3\right|+\left|2\sqrt{6}-3\right|\)
\(=3-\sqrt{6}+2\sqrt{6}-3\) ( vi \(\sqrt{6}-3< 0\))
\(=\sqrt{6}\)
5) \(2\sqrt{\frac{16}{3}}-3\sqrt{\frac{1}{27}}-6\sqrt{\frac{4}{75}}\)
\(=2\frac{4}{\sqrt{3}}-3.\frac{1}{3}-6\sqrt{\frac{2^2}{3.5^2}}\)
\(=\frac{8\sqrt{3}}{3}-1-6.\frac{2}{5}.\sqrt{\frac{1}{3}}\)
\(=8\frac{\sqrt{3}}{3}-1-\frac{12}{5}.\frac{\sqrt{3}}{3}\)
\(=\frac{28}{5}.\frac{\sqrt{3}}{3}-1\)
Báo cáo sai phạm
1) 2√5−√125−√80+√605
=2√5−√52.5−√42.5+√112.5
=2√5−5√5−4√5+11√5
=4√5
2) √15−√216+√33−12√6
=√15−√62.6+√33−12√6
=√15−6√6+√33−12√6
=√(√6)2−6√6+32+√(2√6)2−12√6+32
=√(√6−3)2+√(2√6−3)2
=|√6−3|+|2√6−3|
=3−√6+2√6−3 ( vi √6−3<0)
=√6
5) 2√163 −3√127 −6√475
=24√3 −3.13 −6√223.52
=8√33 −1−6.25 .√13
=8√33 −1−125 .√33
=285 .√33 −1
1.\(\frac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}\) + \(\frac{8}{1-\sqrt{5}}\)
2.\(\frac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}\)- \(\frac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}\)
3.\(\sqrt{\frac{2-\sqrt{3}}{2+\sqrt{3}}}\)+\(\sqrt{\frac{2+\sqrt{3}}{2-\sqrt{3}}}\)
4.\(2\sqrt{\frac{16}{3}}\)- \(3\sqrt{\frac{1}{27}}\)-\(6\sqrt{\frac{4}{75}}\)
5.\(2\sqrt{27}\)- \(6\sqrt{\frac{4}{3}}\)+\(\frac{3}{5}\sqrt{75}\)
6.\(\frac{\sqrt{3-\sqrt{5}}\times\left(3+\sqrt{5}\right)}{\sqrt{10}+\sqrt{2}}\)
Mn giúp mình với ạ mình cần gấp