\(\sqrt{13-2\sqrt{40}}-\sqrt{53+2\sqrt{360}}\)
\(\sqrt{74+40\sqrt{3}}-\sqrt{77+30\sqrt{6}}\)
Những câu hỏi liên quan
dfrac{2sqrt{30}}{sqrt{5}+sqrt{6}+sqrt{7}}
sqrt{24}+6sqrt{dfrac{2}{3}+dfrac{10}{sqrt{6}-1}}dfrac{2sqrt{15}+sqrt{16}}{sqrt{84}+sqrt{6}}2sqrt{40sqrt{2}}-2sqrt{sqrt{75}}-3sqrt{5sqrt{48}} dfrac{left(2+sqrt{3}right)^2-1}{left(sqrt{3}+1right)^2}:dfrac{left(3+sqrt{5}right)^2-4}{left(sqrt{5}+1right)^2}giúp em với ạ
Đọc tiếp
\(\dfrac{2\sqrt{30}}{\sqrt{5}+\sqrt{6}+\sqrt{7}} \)
\(\sqrt{24}+6\sqrt{\dfrac{2}{3}+\dfrac{10}{\sqrt{6}-1}}\)
\(\dfrac{2\sqrt{15}+\sqrt{16}}{\sqrt{84}+\sqrt{6}}\)
\(2\sqrt{40\sqrt{2}}-2\sqrt{\sqrt{75}}-3\sqrt{5\sqrt{48}}\)
\(\dfrac{\left(2+\sqrt{3}\right)^2-1}{\left(\sqrt{3}+1\right)^2}:\dfrac{\left(3+\sqrt{5}\right)^2-4}{\left(\sqrt{5}+1\right)^2}\)
giúp em với ạ
\(2\sqrt{40\sqrt{3}}-2\sqrt{\sqrt{75}}-3\sqrt{5\sqrt{48}}\)
\(=2\cdot\sqrt{40\sqrt{3}}-2\cdot\sqrt{5\sqrt{3}}-3\cdot\sqrt{20\sqrt{3}}\)
\(=2\cdot2\sqrt{10}\cdot\sqrt{\sqrt{3}}-2\cdot\sqrt{5}\cdot\sqrt{\sqrt{3}}-6\sqrt{5}\cdot\sqrt{\sqrt{3}}\)
\(=4\sqrt{10}\sqrt{\sqrt{3}}-4\cdot\sqrt{5}\cdot\sqrt{\sqrt{3}}\)
Đúng 0
Bình luận (0)
bài 1 rút gọn biểu thức sau:a)sqrt{16+6sqrt{7}}- sqrt{8-2sqrt{7}} b)Kdfrac{sqrt{4-2sqrt{3}}}{sqrt{6}-sqrt{2}}c)sqrt{60-24sqrt{6}}+sqrt{40-16sqrt{6}} d)B(3+sqrt{3})sqrt{12-6sqrt{13}}e)sqrt{6-4sqrt{2}}-sqrt{left(sqrt{2}-sqrt{6}right)^2}bài 2 cho biểu thức Aleft(dfrac{sqrt{x}}{sqrt{x}+3}+dfrac{3}{sqrt{x}-3}right).dfrac{sqrt{x}+3}{x+9}( với x≥0 và x≠ 9)a) rút gọn biểu thức Ab) tính giá trị biểu thứcx4+2sqrt{3}
Đọc tiếp
bài 1 rút gọn biểu thức sau:
a)\(\sqrt{16+6\sqrt{7}}\)- \(\sqrt{8-2\sqrt{7}}\) b)K=\(\dfrac{\sqrt{4-2\sqrt{3}}}{\sqrt{6}-\sqrt{2}}\)
c)\(\sqrt{60-24\sqrt{6}}\)+\(\sqrt{40-16\sqrt{6}}\) d)B=(3+\(\sqrt{3}\))\(\sqrt{12-6\sqrt{13}}\)
e)\(\sqrt{6-4\sqrt{2}}\)-\(\sqrt{\left(\sqrt{2}-\sqrt{6}\right)^2}\)
bài 2 cho biểu thức A=\(\left(\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{3}{\sqrt{x}-3}\right).\dfrac{\sqrt{x}+3}{x+9}\)( với x≥0 và x≠ 9)
a) rút gọn biểu thức A
b) tính giá trị biểu thức\(x=4+2\sqrt{3}\)
\(1,\\ a,=\sqrt{\left(3+\sqrt{7}\right)^2}-\sqrt{\left(\sqrt{7}-1\right)^2}=3+\sqrt{7}-\sqrt{7}+1=4\\ b,K=\dfrac{\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{2}\left(\sqrt{3}-1\right)}=\dfrac{\sqrt{3}-1}{\sqrt{2}\left(\sqrt{3}-1\right)}=\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}\\ c,=\sqrt{\left(6-2\sqrt{6}\right)^2}+\sqrt{\left(2\sqrt{6}-4\right)^2}=6-2\sqrt{6}+2\sqrt{6}-4=2\\ e,=\sqrt{\left(2-\sqrt{2}\right)^2}-\left(\sqrt{6}-\sqrt{2}\right)=2-\sqrt{2}-\sqrt{6}+\sqrt{2}=2-\sqrt{6}\)
\(2,\\ a,A=\dfrac{x-3\sqrt{x}+3\sqrt{x}+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}+3}{x+9}\\ A=\dfrac{x+9}{\left(\sqrt{x}-3\right)\left(x+9\right)}=\dfrac{1}{\sqrt{x}-3}\\ b,x=4+2\sqrt{3}\Leftrightarrow\sqrt{x}=\sqrt{3}+1\\ \Leftrightarrow A=\dfrac{1}{\sqrt{3}+1-3}=\dfrac{1}{\sqrt{3}+2}=2-\sqrt{3}\)
Đúng 2
Bình luận (1)
Tính giá trị của biểu thức
a. \(A=\sqrt[3]{6\sqrt{3}+10}-\sqrt[3]{6\sqrt{3}-10}\)
b. \(\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}\)
c. \(\sqrt[3]{45+29\sqrt{2}}+\sqrt[3]{45-29\sqrt{2}}\)
cảm ơn các bạn trước nhaa 
ta có: A3=\(6\sqrt{3}+10-6\sqrt{3}+10-3\sqrt[3]{\left(6\sqrt{3}+10\right)\left(6\sqrt{3}-10\right)}.\left(\sqrt[3]{6\sqrt{3}+10}-\sqrt[3]{6\sqrt{3}-10}\right)\)
=\(20-3.\sqrt[3]{8}.A\)=\(20-6A\)
do đó A3=20-6A↔A3+6A-20=0↔(A2+2A+10)(A-2)=0
dễ thấy A2+2A+10>0→A=2
b) giống a)
c)giống b)
Đúng 0
Bình luận (0)
1,Rút gọn B=\(\dfrac{3\sqrt{8}-2\sqrt{12}+\sqrt{20}}{3\sqrt{18}-2\sqrt{27}+\sqrt{40}}\)
2,Tính
a, A=\(\dfrac{2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}}{\sqrt{6}+\sqrt{2}}\)
Bài 2:
\(A=\dfrac{2\sqrt{3+\sqrt{5-2\sqrt{3}-1}}}{\sqrt{2}\left(\sqrt{3}+1\right)}\)
\(=\dfrac{2\sqrt{3+\sqrt{3}-1}}{\sqrt{2}\left(\sqrt{3}+1\right)}=1\)
Bài 1:
\(=\dfrac{2\left(3\sqrt{2}-2\sqrt{3}+\sqrt{5}\right)}{3\left(3\sqrt{2}-2\sqrt{3}+\sqrt{5}\right)}=\dfrac{2}{3}\)
Đúng 0
Bình luận (0)
Bài 1: Tính
\(\sqrt{3+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}\\ \sqrt{12+6\sqrt{3}+\sqrt{12-6\sqrt{3}}}\\ \sqrt{9-4\sqrt{2}+\sqrt{9+4\sqrt{2}}}\)
\(\sqrt{\sqrt{2}+2+\sqrt{4+\sqrt{9-\sqrt{32}}}}\\ \sqrt{6+2\sqrt{5}-\sqrt{29+12\sqrt{5}}}\\ \sqrt{8+\sqrt{8}+\sqrt{20}+\sqrt{40}}-\sqrt{\sqrt{49}+\sqrt{40}}\\ \sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
1.
$\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}=\sqrt{3+1+2\sqrt{3}}-\sqrt{3+1-2\sqrt{3}}$
$=\sqrt{(\sqrt{3}+1)^2}-\sqrt{(\sqrt{3}-1)^2}$
$=|\sqrt{3}+1|-|\sqrt{3}-1|=2$
2.
\(\sqrt{12+6\sqrt{3}+\sqrt{12-6\sqrt{3}}}=\sqrt{12+6\sqrt{3}+\sqrt{9+3-2\sqrt{9.3}}}=\sqrt{12+6\sqrt{3}+\sqrt{(3-\sqrt{3})^2}}\)
\(=\sqrt{12+6\sqrt{3}+3-\sqrt{3}}=\sqrt{15+5\sqrt{3}}\)
Đúng 0
Bình luận (0)
3.
\(\sqrt{9-4\sqrt{2}+\sqrt{9+4\sqrt{2}}}=\sqrt{9-4\sqrt{2}+\sqrt{8+1+2\sqrt{8.1}}}\)
\(=\sqrt{9-4\sqrt{2}+\sqrt{2\sqrt{2}+1)^2}}=\sqrt{9-4\sqrt{2}+2\sqrt{2}+1}=\sqrt{10-2\sqrt{2}}\)
4.
\(\sqrt{\sqrt{2}+2+\sqrt{4+\sqrt{9-\sqrt{32}}}}=\sqrt{\sqrt{2}+2+\sqrt{4+\sqrt{8+1-2\sqrt{8.1}}}}\)
\(=\sqrt{\sqrt{2}+2+\sqrt{4+\sqrt{(\sqrt{8}-1)^2}}}\) \(=\sqrt{\sqrt{2}+2+\sqrt{4+\sqrt{8}-1}}=\sqrt{\sqrt{2}+2+\sqrt{3+2\sqrt{2}}}\)
\(=\sqrt{\sqrt{2}+2+\sqrt{(2+1+2\sqrt{2}}}=\sqrt{\sqrt{2}+2+\sqrt{(\sqrt{2}+1)^2}}=\sqrt{\sqrt{2}+2+\sqrt{2}+1}\)
\(=\sqrt{3+2\sqrt{2}}=\sqrt{(\sqrt{2}+1)^2}=\sqrt{2}+1\)
Đúng 0
Bình luận (0)
5.
\(\sqrt{6+2\sqrt{5}-\sqrt{29+12\sqrt{5}}}=\sqrt{6+2\sqrt{5}-\sqrt{20+9+2\sqrt{20.9}}}\)
\(=\sqrt{6+2\sqrt{5}-\sqrt{(\sqrt{20}+3)^2}}=\sqrt{6+2\sqrt{5}-(\sqrt{20}+3)}=\sqrt{3}\)
6.
\(\sqrt{8+\sqrt{8}+\sqrt{20}+\sqrt{40}}-\sqrt{\sqrt{49}+\sqrt{40}}\)
\(=\sqrt{8+2\sqrt{2}+2\sqrt{5}+2\sqrt{10}}-\sqrt{7+2\sqrt{10}}\)
\(=\sqrt{(2+5+2\sqrt{2.5})+2(\sqrt{2}+\sqrt{5})+1}-\sqrt{2+5+2\sqrt{2.5}}\)
\(=\sqrt{(\sqrt{2}+\sqrt{5})^2+2(\sqrt{2}+\sqrt{5})+1}-\sqrt{(\sqrt{2}+\sqrt{5})^2}\)
\(=\sqrt{(\sqrt{2}+\sqrt{5}+1)^2}-\sqrt{(\sqrt{2}+\sqrt{5})^2}=|\sqrt{2}+\sqrt{5}+1|-|\sqrt{2}+\sqrt{5}|=1\)
Đúng 0
Bình luận (0)
Xem thêm câu trả lời
Tính
a, \(\sqrt{8-\sqrt{28}}\) + \(\sqrt{21+12\sqrt{3}}\)
b, \(\sqrt{5+\sqrt{24}}\) - \(\sqrt{57-40\sqrt{2}}\)
c, \(\sqrt{13-\sqrt{160}}\) + \(\sqrt{53+4\sqrt{90}}\)
\(a.\sqrt{8-\sqrt{28}}+\sqrt{21+12\sqrt{3}}=\sqrt{7-2\sqrt{7}+1}+\sqrt{12+2.2\sqrt{3}.3+9}=\sqrt{7}-1+2\sqrt{3}+3=2\sqrt{3}+\sqrt{7}+2\) \(b.\sqrt{5+\sqrt{24}}-\sqrt{57-40\sqrt{2}}=\sqrt{3+2.\sqrt{3}.\sqrt{2}+2}-\sqrt{32-2.4\sqrt{2}.5+25}=\sqrt{3}+\sqrt{2}-4\sqrt{2}+5=\sqrt{3}-3\sqrt{2}+5\) \(c.\sqrt{13-\sqrt{160}}+\sqrt{53+4\sqrt{90}}=\sqrt{8-2.2\sqrt{2}.\sqrt{5}+5}+\sqrt{45+2.3\sqrt{5}.2\sqrt{2}+8}=2\sqrt{2}-\sqrt{5}+3\sqrt{5}+2\sqrt{2}=4\sqrt{2}+2\sqrt{5}\)
Đúng 0
Bình luận (0)
❤ Tính:
a) sqrt{5-sqrt{21}}-sqrt{5+sqrt{21}}
b)sqrt{13-sqrt{160}}-sqrt{53+4sqrt{90}}
c)sqrt{7+sqrt{24}}+sqrt{31-sqrt{600}}
d)sqrt{28-sqrt{300}}+sqrt{4-sqrt{12}}
e)sqrt{7-sqrt{40}}-sqrt{5-sqrt{24}}-sqrt{6-sqrt{20}}
f)sqrt{48-10sqrt{7+sqrt{48}}}
g) frac{1}{1-sqrt{2}}-frac{1}{sqrt{2}-sqrt{3}}+frac{1}{sqrt{3}-sqrt{4}}-frac{1}{sqrt{4}-sqrt{5}}+...+frac{1}{sqrt{15}-sqrt{16}}
Đọc tiếp
❤ Tính:
a) \(\sqrt{5-\sqrt{21}}-\sqrt{5+\sqrt{21}}\)
b)\(\sqrt{13-\sqrt{160}}-\sqrt{53+4\sqrt{90}}\)
c)\(\sqrt{7+\sqrt{24}}+\sqrt{31-\sqrt{600}}\)
d)\(\sqrt{28-\sqrt{300}}+\sqrt{4-\sqrt{12}}\)
e)\(\sqrt{7-\sqrt{40}}-\sqrt{5-\sqrt{24}}-\sqrt{6-\sqrt{20}}\)
f)\(\sqrt{48-10\sqrt{7+\sqrt{48}}}\)
g) \(\frac{1}{1-\sqrt{2}}-\frac{1}{\sqrt{2}-\sqrt{3}}+\frac{1}{\sqrt{3}-\sqrt{4}}-\frac{1}{\sqrt{4}-\sqrt{5}}+...+\frac{1}{\sqrt{15}-\sqrt{16}}\)
Rút gọn các biểu thức sau:a.2sqrt{3+sqrt{5-sqrt{13+sqrt{48}}}}b.sqrt{2+sqrt{3}}.sqrt{2+sqrt{2+sqrt{3}}.sqrt{2+sqrt{2+sqrt{2+sqrt{3}}}}}.sqrt{2-sqrt{2+sqrt{2+sqrt{3}}}}c.sqrt{8+sqrt{40}+sqrt{20}+sqrt{8}}d.sqrt{10+sqrt{24}+sqrt{20}+sqrt{8}}d.sqrt{10+sqrt{24}-sqrt{40}-sqrt{60}}
Đọc tiếp
Rút gọn các biểu thức sau:
a.\(2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}\)
b.\(\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}}.\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}\)
c.\(\sqrt{8+\sqrt{40}+\sqrt{20}+\sqrt{8}}\)
d.\(\sqrt{10+\sqrt{24}+\sqrt{20}+\sqrt{8}}\)
d.\(\sqrt{10+\sqrt{24}-\sqrt{40}-\sqrt{60}}\)
a/ \(\sqrt{2}+\sqrt{6}\)
b/ Sửa đề:
\(\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}.\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}=1\)
c/ \(1+\sqrt{2}+\sqrt{5}\)
Đúng 0
Bình luận (0)
a/ \(2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}\)
\(=2\sqrt{3+\sqrt{5-\sqrt{12+2.2\sqrt{3}+1}}}\)
\(=2\sqrt{3+\sqrt{5-\sqrt{\left(2\sqrt{3}+1\right)^2}}}\)
\(=2\sqrt{3+\sqrt{5-\left(2\sqrt{3}+1\right)}}\)
\(=2\sqrt{3+\sqrt{4-2\sqrt{3}}}\)
\(=2\sqrt{3+\sqrt{3-2\sqrt{3}+1}}\)
\(=2\sqrt{3+\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(=2\sqrt{3+\left(\sqrt{3}-1\right)}\)
\(=\sqrt{2}\sqrt{4+2\sqrt{3}}\)
\(=\sqrt{2}\sqrt{3+2\sqrt{3}+1}\)
\(=\sqrt{2}\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=\sqrt{2}\left(\sqrt{3}+1\right)\)
\(=\sqrt{2}+\sqrt{6}\)
Đúng 0
Bình luận (0)
Xem thêm câu trả lời
bài 5 Tính:
a) \(\sqrt{6-2\sqrt{5}}\)
b)\(\sqrt{7-4\sqrt{3}}\)
c)\(\sqrt{3-2\sqrt{2}}\) -\(\sqrt{6-4\sqrt{2}}\)
d)\(\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}\)
Lời giải:
a. \(\sqrt{6-2\sqrt{5}}=\sqrt{5-2\sqrt{5}.\sqrt{1}+1}=\sqrt{(\sqrt{5}-1)^2}=\sqrt{5}-1\)
b. \(\sqrt{7-4\sqrt{3}}=\sqrt{4-2\sqrt{4}.\sqrt{3}+3}=\sqrt{(\sqrt{4}-\sqrt{3})^2}=\sqrt{4}-\sqrt{3}=2-\sqrt{3}\)
c.
\(\sqrt{3-2\sqrt{2}}-\sqrt{6-4\sqrt{2}}=\sqrt{2-2\sqrt{2}+1}-\sqrt{4-4\sqrt{2}+2}\)
\(=\sqrt{(\sqrt{2}-1)^2}-\sqrt{(\sqrt{4}-\sqrt{2})^2}\)
\(=|\sqrt{2}-1|-|\sqrt{4}-\sqrt{2}|=\sqrt{2}-1-(2-\sqrt{2})=2\sqrt{2}-3\)
d.
\(=\sqrt{13+30\sqrt{2+\sqrt{(\sqrt{8}+1)^2}}}=\sqrt{13+30\sqrt{2+\sqrt{8}+1}}\)
\(=\sqrt{13+30\sqrt{3+2\sqrt{2}}}=\sqrt{13+30\sqrt{(\sqrt{2}+1)^2}}\)
\(=\sqrt{13+30(\sqrt{2}+1)}=\sqrt{43+30\sqrt{2}}=\sqrt{18+2\sqrt{18.25}+25}\)
\(=\sqrt{(\sqrt{18}+\sqrt{25})^2}=\sqrt{18}+\sqrt{25}=5+3\sqrt{2}\)
Đúng 3
Bình luận (0)
a) \(\sqrt{6-2\sqrt{5}}=\sqrt{5}-1\)
b) \(\sqrt{7-4\sqrt{3}}=2-\sqrt{3}\)
c) \(\sqrt{3-2\sqrt{2}}-\sqrt{6-4\sqrt{2}}=\sqrt{2}-1-2+\sqrt{2}=-3+2\sqrt{2}\)
d) Ta có: \(\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}\)
\(=\sqrt{13+30\sqrt{2+1+2\sqrt{2}}}\)
\(=\sqrt{13+30\left(\sqrt{2}+1\right)}\)
\(=\sqrt{43+30\sqrt{2}}\)
\(=5+3\sqrt{2}\)
Đúng 0
Bình luận (0)
a.\(\sqrt{19-6\sqrt{2}}\) b.\(\sqrt{11-6\sqrt{2}}\) c.\(\sqrt{9-6\sqrt{2}}\)
d.\(\sqrt{21+12\sqrt{3}}\) e.\(\sqrt{57-40\sqrt{2}}\)
a) \(\sqrt{19-6\sqrt{2}}=3\sqrt{2}-1\)
b) \(\sqrt{11-6\sqrt{2}}=3-\sqrt{2}\)
d) \(\sqrt{21+12\sqrt{3}}=2\sqrt{3}+3\)
e) \(\sqrt{57-40\sqrt{2}}=4\sqrt{2}-5\)
Đúng 0
Bình luận (0)
