1 Tính
\(P=\sqrt{1+99999...9^2+0,99999...9^2}\)( n chứ số 9)
\(F=\sqrt{13-\sqrt{160}}-\sqrt{53-4\sqrt{9}}\)
\(H=\sqrt{0.25\sqrt{961}+2\sqrt{10}+\sqrt{15}+\sqrt{6}}\)
Những câu hỏi liên quan
Tính
\(a,F=\sqrt{13-\sqrt{160}}-\sqrt{53-4\sqrt{9}}\)
\(b,A=\sqrt{13-30\sqrt{2+\sqrt{9+4\sqrt{2}}}}\)
\(c,D=\frac{1}{\sqrt{8}+\sqrt{7}}+\sqrt{175}-2\sqrt{2}\)
\(d,M=\sqrt{0,25\sqrt{961}+2\sqrt{10}+\sqrt{15}+\sqrt{6}}\)
Rút gọn biểu thức
A. (2-√3)sqrt{7+4sqrt{3}}
B. sqrt{13+4sqrt{10}}:+:sqrt[]{13-4sqrt{10}}
C.(3 - √2) sqrt{11+6sqrt{2}}
D. (√5+√7) sqrt{12-2sqrt{35}}
E. (√2-√9)sqrt{11+2sqrt{18}}
F. sqrt{46-6sqrt{5}}:+:sqrt{29-12sqrt{5}}
G.sqrt{49-5sqrt{96}}:+:sqrt{49+5sqrt{96}}
H.sqrt{13-sqrt{160::::}}:+:sqrt{53+4sqrt{90}}
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Rút gọn biểu thức
A. (2-√3)\(\sqrt{7+4\sqrt{3}}\)
B. \(\sqrt{13+4\sqrt{10}}\:+\:\sqrt[]{13-4\sqrt{10}}\)
C.(3 - √2) \(\sqrt{11+6\sqrt{2}}\)
D. (√5+√7) \(\sqrt{12-2\sqrt{35}}\)
E. (√2-√9)\(\sqrt{11+2\sqrt{18}}\)
F. \(\sqrt{46-6\sqrt{5}}\:+\:\sqrt{29-12\sqrt{5}}\)
G.\(\sqrt{49-5\sqrt{96}}\:+\:\sqrt{49+5\sqrt{96}}\)
H.\(\sqrt{13-\sqrt{160\:\:\:\:}}\:+\:\sqrt{53+4\sqrt{90}}\)
\(A=\left(2-\sqrt{3}\right)\sqrt{4+2.2.\sqrt{3}+3}=\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)=1\)
các câu còn lại làm tương tự nhé bạn !
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Tính: \(N=\frac{\sqrt{15-10\sqrt{2}}+\sqrt{13+4\sqrt{10}}-\sqrt{11+2\sqrt{10}}}{2\sqrt{3+2\sqrt{2}}+\sqrt{9-4\sqrt{2}}+\sqrt{12+8\sqrt{2}}}\)
Tính \(E=\frac{\sqrt{15-10\sqrt{2}}+\sqrt{13+4\sqrt{10}}-\sqrt{11+2\sqrt{10}}}{2\sqrt{3+2\sqrt{2}}+\sqrt{9-4\sqrt{2}}+\sqrt{12+8\sqrt{2}}}\)
a,Tính giá trị của biểu thức:
N=\(\frac{\sqrt{15-10\sqrt{2}}+\sqrt{13+4\sqrt{10}}-\sqrt{11+2\sqrt{10}}}{2\sqrt{3+2\sqrt{2}}+\sqrt{9-4\sqrt{2}}+\sqrt{12+8\sqrt{2}}}\)
Rút gọn biểu thức:
a) dfrac{sqrt{9-2sqrt{6}}-sqrt{6}}{sqrt{3}} b)dfrac{5+sqrt{5}}{5-sqrt{5}}+dfrac{5-sqrt{5}}{5+sqrt{5}}
c) sqrt{13+30sqrt{2+sqrt{9+4sqrt{2}}}} d) sqrt{8+2sqrt{10+2sqrt{5}}}+sqrt{8-2sqrt{10+2sqrt{5}}}
e) dfrac{2sqrt{3+sqrt{5-sqrt{13+sqrt{48}}}}}{sqrt{6}-sqrt{2}} f) sqrt{9-sqrt{5sqrt{3}+5sqrt{8+10sqrt{7-4sqrt{3}}}}}
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Rút gọn biểu thức:
a) \(\dfrac{\sqrt{9-2\sqrt{6}}-\sqrt{6}}{\sqrt{3}}\) b)\(\dfrac{5+\sqrt{5}}{5-\sqrt{5}}+\dfrac{5-\sqrt{5}}{5+\sqrt{5}}\)
c) \(\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}\) d) \(\sqrt{8+2\sqrt{10+2\sqrt{5}}}+\sqrt{8-2\sqrt{10+2\sqrt{5}}}\)
e) \(\dfrac{2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}}{\sqrt{6}-\sqrt{2}}\) f) \(\sqrt{9-\sqrt{5\sqrt{3}+5\sqrt{8+10\sqrt{7-4\sqrt{3}}}}}\)
b: \(=\dfrac{\sqrt{5}+1}{\sqrt{5}-1}+\dfrac{\sqrt{5}-1}{\sqrt{5}+1}\)
\(=\dfrac{6+2\sqrt{5}+6-2\sqrt{5}}{4}=\dfrac{12}{4}=3\)
c: \(=\sqrt{13+30\sqrt{2+2\sqrt{2}+1}}\)
\(=\sqrt{13+30\left(\sqrt{2}+1\right)}=\sqrt{43+30\sqrt{2}}\)
e: \(=\dfrac{2\sqrt{3+\sqrt{5-2\sqrt{3}-1}}}{\sqrt{6}-\sqrt{2}}\)
\(=\dfrac{\sqrt{2}\cdot\sqrt{3+\sqrt{3}-1}}{\sqrt{3}-1}=\dfrac{\sqrt{4+2\sqrt{3}}}{\sqrt{3}-1}=\dfrac{\sqrt{3}+1}{\sqrt{3}-1}\)
\(=\dfrac{4-2\sqrt{3}}{2}=2-\sqrt{3}\)
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Tính \(\sqrt{13-\sqrt{160}}+\sqrt{53-4\sqrt{9}}\)
Rút gọn
Hleft(sqrt{3}-1right)sqrt{6+2sqrt{2}.sqrt{3-sqrt{sqrt{2}+sqrt{12}+sqrt{18-sqrt{128}}}}}
Ffrac{left(5+2sqrt{6}right)left(49-20sqrt{6}right)sqrt{5-2sqrt{6}}}{9sqrt{3}-11sqrt{2}}
Gsqrt{4+sqrt{5sqrt{3}+5sqrt{48-10sqrt{7+4sqrt{3}}}}}
Efrac{2sqrt{3+sqrt{5-13+sqrt{48}}}}{sqrt{6}+sqrt{2}}
Dsqrt{4+sqrt{15}}+sqrt{4-sqrt{15}}-2sqrt{3-sqrt{5}}
Zsqrt{4+sqrt{10+2sqrt{5}}}+sqrt{4-sqrt{10-2sqrt{5}}}
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Rút gọn
H=\(\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}}\)
F=\(\frac{\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right)\sqrt{5-2\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}\)
G=\(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
E=\(\frac{2\sqrt{3+\sqrt{5-13+\sqrt{48}}}}{\sqrt{6}+\sqrt{2}}\)
D=\(\sqrt{4+\sqrt{15}}+\sqrt{4-\sqrt{15}}-2\sqrt{3-\sqrt{5}}\)
Z=\(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10-2\sqrt{5}}}\)
Đề thiếu nha:
\(E=\frac{2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}}{\sqrt{6}+\sqrt{2}}\)
\(=\frac{2\sqrt{3+\sqrt{5-\sqrt{12+4\sqrt{3}+1}}}}{\sqrt{6}+\sqrt{2}}\)
\(=\frac{2\sqrt{3+\sqrt{5-\sqrt{\left(2\sqrt{3}+1\right)^2}}}}{\sqrt{6}+\sqrt{2}}\)
\(=\frac{2\sqrt{3+\sqrt{5-2\sqrt{3}-1}}}{\sqrt{6}+\sqrt{2}}\)
\(=\frac{2\sqrt{3+\sqrt{3-2\sqrt{3}+1}}}{\sqrt{6}+\sqrt{2}}\)
\(=\frac{2\sqrt{3+\sqrt{\left(\sqrt{3}-1\right)^2}}}{\sqrt{6}+\sqrt{2}}\)
\(=\frac{2\sqrt{3+\sqrt{3}-1}}{\sqrt{2}\left(\sqrt{3}+1\right)}\)(vì \(\sqrt{3}>1\))
\(=\frac{\sqrt{2}.\sqrt{2+\sqrt{3}}}{\sqrt{3}+1}\)
\(=\frac{\sqrt{4+2\sqrt{3}}}{\sqrt{3}+1}\)
\(=\frac{\sqrt{3+2\sqrt{3}+1}}{\sqrt{3}+1}\)
\(=\frac{\sqrt{\left(\sqrt{3}+1\right)^2}}{\sqrt{3}+1}=\frac{\sqrt{3}+1}{\sqrt{3}+1}=1\)
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\(D=\sqrt{4+\sqrt{15}}+\sqrt{4-\sqrt{15}}-2\sqrt{3-\sqrt{5}}\)
\(\Rightarrow D\sqrt{2}=\sqrt{8+2\sqrt{15}}+\sqrt{8-2\sqrt{15}}-2\sqrt{6-2\sqrt{5}}\)
\(=\sqrt{5+2\sqrt{15}+3}+\sqrt{5-2\sqrt{15}+3}-2\sqrt{5-2\sqrt{5}+1}\)
\(=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}-2\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=\sqrt{5}+\sqrt{3}+\sqrt{5}-\sqrt{3}-2\left(\sqrt{5}-1\right)\)
\(=2\sqrt{5}-2\sqrt{5}+2=2\)
\(\Rightarrow D=\frac{2}{\sqrt{2}}=\sqrt{2}\)
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\(H=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{\sqrt{2}+2\sqrt{3}+\sqrt{16-8\sqrt{2}+2}}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{\sqrt{2}+2\sqrt{3}+\sqrt{\left(4-\sqrt{2}\right)^2}}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}.\sqrt{\sqrt{2}+2\sqrt{3}+4-\sqrt{2}}}\)(vì \(4>\sqrt{2}\))
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}.\sqrt{3+2\sqrt{3}+1}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}.\sqrt{\left(\sqrt{3}+1\right)^2}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}.\left(\sqrt{3}+1\right)}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{6}+2\sqrt{2}}\)
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Xem thêm câu trả lời
Biến đổi biểu thức trong căn thành bình phương 1 tổng hay 1 hiệu rồi từ đó phá bớt 1 lớp căn:a)sqrt{4-sqrt{15}} b) sqrt{7-sqrt{33}} c) sqrt{10-2sqrt{6}+2sqrt{10}-2sqrt{15}}d) sqrt{25+4sqrt{10}+4sqrt{15}+4sqrt{6}}e) sqrt{49-5sqrt{96}}+sqrt{49+5sqrt{96}}f) sqrt{13-sqrt{160}}-sqrt{53+4sqrt{90}}g) left(sqrt{2}-3right)sqrt{11+6sqrt{2}}h) left(3sqrt{2}+sqrt{10}right)sqrt{38-12sqrt{5}}Giúp tớ với ạ! Càng nhanh càng tốt nha! :))
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Biến đổi biểu thức trong căn thành bình phương 1 tổng hay 1 hiệu rồi từ đó phá bớt 1 lớp căn:
a)\(\sqrt{4-\sqrt{15}}\)
b) \(\sqrt{7-\sqrt{33}}\)
c) \(\sqrt{10-2\sqrt{6}+2\sqrt{10}-2\sqrt{15}}\)
d) \(\sqrt{25+4\sqrt{10}+4\sqrt{15}+4\sqrt{6}}\)
e) \(\sqrt{49-5\sqrt{96}}+\sqrt{49+5\sqrt{96}}\)
f) \(\sqrt{13-\sqrt{160}}-\sqrt{53+4\sqrt{90}}\)
g) \(\left(\sqrt{2}-3\right)\sqrt{11+6\sqrt{2}}\)
h) \(\left(3\sqrt{2}+\sqrt{10}\right)\sqrt{38-12\sqrt{5}}\)
Giúp tớ với ạ! Càng nhanh càng tốt nha! :))