Phân tích thành các lũy thừa bậc 2:
1) 8+2\(\sqrt{15}\)
2) 10- 2\(\sqrt{21}\)
3) 5+\(\sqrt{24}\)
4) 8-\(\sqrt{28}\)
5) 9-4\(\sqrt{2}\)
6) 51+10\(\sqrt{2}\)
Những câu hỏi liên quan
Phân tích các biểu thức sau thành các lũy thừa bậc hai:
a) \(8+2\sqrt{15}\) b) \(10-2\sqrt{21}\) c) \(12-\sqrt{140}\) d) \(5+\sqrt{24}\) e) \(14+6\sqrt{5}\) g) \(8-\sqrt{28}\)
Tínha.sqrt{8+2sqrt{5}} b.sqrt{10-2sqrt[]{5}} c.sqrt{5+sqrt{24}} d.sqrt{12-sqrt{140}} e.sqrt{14+2sqrt{5}} f. sqrt{8-sqrt{28}} g.sqrt{23-4sqrt{15}} h.sqrt{9+4sqrt{2}} giúp mik vs mai mik nộp rồi,cảm ơn mn nhiều
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Tính
a.\(\sqrt{8+2\sqrt{5}}\) b.\(\sqrt{10-2\sqrt[]{5}}\) c.\(\sqrt{5+\sqrt{24}}\) d.\(\sqrt{12-\sqrt{140}}\)
e.\(\sqrt{14+2\sqrt{5}}\) f. \(\sqrt{8-\sqrt{28}}\) g.\(\sqrt{23-4\sqrt{15}}\) h.\(\sqrt{9+4\sqrt{2}}\)
giúp mik vs mai mik nộp rồi,cảm ơn mn nhiều
c) \(\sqrt{5+\sqrt{24}}=\sqrt{5+2\sqrt{6}}=\sqrt{3}+\sqrt{2}\)
d) \(\sqrt{12-\sqrt{140}}=\sqrt{12-2\sqrt{35}}=\sqrt{7}-\sqrt{5}\)
f) \(\sqrt{8-\sqrt{28}}=\sqrt{8-2\sqrt{7}}=\sqrt{7}-1\)
g) \(\sqrt{23-4\sqrt{15}}=\sqrt{23-2\cdot\sqrt{60}}=2\sqrt{5}-\sqrt{3}\)
h) \(\sqrt{9+4\sqrt{2}}=\sqrt{\left(2\sqrt{2}+1\right)^2}=2\sqrt{2}+1\)
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Tính:
1) \(\sqrt{4-2\sqrt{3}}\)
2) \(\sqrt{5+2\sqrt{6}}\)
3) \(\sqrt{7-2\sqrt{10}}\)
4) \(\sqrt{14-6\sqrt{6}}\)
5) \(\sqrt{8+2\sqrt{15}}\)
6) \(\sqrt{10-2\sqrt{21}}\)
7) \(\sqrt{11+2\sqrt{18}}\)
LÀM CHI TIẾT GIÚP MK NHÉ!
1) \(=\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{3}-1\)
2) \(=\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}=\sqrt{3}+\sqrt{2}\)
3) \(=\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}=\sqrt{5}-\sqrt{2}\)
5) \(=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}=\sqrt{5}+\sqrt{3}\)
6) \(=\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}=\sqrt{7}-\sqrt{3}\)
7) \(=\sqrt{\left(3+\sqrt{2}\right)^2}=3+\sqrt{2}\)
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RÚT GỌN BIỂU THỨC
A=\(4-\sqrt{21-8\sqrt{5}}\)
B=\(\sqrt{4-2\sqrt{3}+1}\)
C=\(\sqrt{8+2\sqrt{15}}-\sqrt{5-2\sqrt{6}}\)
D=\(\sqrt{28-10\sqrt{3}}+\sqrt{4+2\sqrt{3}}\)
E=\(\sqrt{14-6\sqrt{5}}-\sqrt{21-8\sqrt{5}}\)
F=\(\sqrt{19-2\sqrt{40}}-\sqrt{19+3\sqrt{40}}\)
\(A=4-\sqrt{21-8\sqrt{5}}=4-\sqrt{4^2-8\sqrt{5}+\left(\sqrt{5}\right)^2}.\)
\(A=4-\sqrt{\left(4-\sqrt{5}\right)^2}=4-\left(4-\sqrt{5}\right)\)
=> \(A=\sqrt{5}\)
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Dạng 3.Chứng minh đẳng thức
Bài 1: CM
a)\(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}=2\)
b)\(\left(5+\sqrt{21}\right)\left(\sqrt{14}-\sqrt{6}\right)\sqrt{5-\sqrt{21}}=8\)
Bài 2 :CM
\(\dfrac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{2}}=\sqrt{\sqrt{5}+1}\)
Bài 1
a) Đặt VT = A
<=> \(2\sqrt{2}A=\left(8+2\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{8-2\sqrt{15}}\)
<=> \(2\sqrt{2}A=\left(\sqrt{5}+\sqrt{3}\right)^2.\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right).\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
<=> \(2A=\left(\sqrt{5}+\sqrt{3}\right)^2.\left(\sqrt{5}-\sqrt{3}\right)^2\)
<=> 2A = \(\left(5-3\right)^2=4\)
<=> A = 2
b) Đặt VT = B
<=> \(2\sqrt{2}B=\left(10+2\sqrt{21}\right).\left(\sqrt{14}-\sqrt{6}\right)\sqrt{10-2\sqrt{21}}\)
<=> \(2\sqrt{2}B=\left(\sqrt{7}+\sqrt{3}\right)^2.\sqrt{2}\left(\sqrt{7}-\sqrt{3}\right).\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}\)
<=> \(2B=\left(\sqrt{7}+\sqrt{3}\right)^2.\left(\sqrt{7}-\sqrt{3}\right)^2=\left(7-3\right)^2=16\)
<=> B = 8
Bài 2
Đặt VT = A
<=> A2 = \(\dfrac{\sqrt{5}+2+\sqrt{5}-2+2\sqrt{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}}{2}\)
<=> A2 = \(\dfrac{2\sqrt{5}+2\sqrt{5-4}}{2}=\dfrac{2\sqrt{5}+2}{2}=\sqrt{5}+1\)
<=> \(A=\sqrt{\sqrt{5}+1}\)
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left(5+4sqrt{2}right)left(3+2sqrt{1+sqrt{2}}right)left(3-2sqrt{1+sqrt{2}}right) sqrt{frac{9}{4}-sqrt{2}} Sosanh2sqrt{27}vasqrt{147}
2sqrt{15}vasqrt{59}
2sqrt{2}-1va2
frac{sqrt{3}}{2}va1
-frac{sqrt{10}}{2}va-2sqrt{5}
sqrt{6}-1va3
2sqrt{5}-5sqrt{2}va1
frac{sqrt{8}}{3}vafrac{3}{4}
-2sqrt{6}va-sqrt{23}
2sqrt{6}-2va3
sqrt{111}-7va4
Xếp theo thứ tự tăng dần: 21,2sqrt{7},15sqrt{3},-sqrt{123} ; 28sqrt{2},sqrt{14},2sqrt{147},36sqrt{4}
giảm dần: 6sqrt{frac{1}{4}},4sqrt{frac{1}{2}...
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\(\left(5+4\sqrt{2}\right)\left(3+2\sqrt{1+\sqrt{2}}\right)\left(3-2\sqrt{1+\sqrt{2}}\right)\\ \\ \\ \sqrt{\frac{9}{4}-\sqrt{2}}\\ \\ \\ Sosanh2\sqrt{27}va\sqrt{147}\\ \\ \\ 2\sqrt{15}va\sqrt{59}\\ \\ \\ 2\sqrt{2}-1va2\\ \\ \\ \frac{\sqrt{3}}{2}va1\\ \\ \\ -\frac{\sqrt{10}}{2}va-2\sqrt{5}\\ \\ \\ \sqrt{6}-1va3\\ \\ \\ 2\sqrt{5}-5\sqrt{2}va1\\ \\ \\ \frac{\sqrt{8}}{3}va\frac{3}{4}\\ \\ \\ -2\sqrt{6}va-\sqrt{23}\\ \\ \\ 2\sqrt{6}-2va3\\ \\ \\ \sqrt{111}-7va4\)
Xếp theo thứ tự tăng dần: \(21,2\sqrt{7},15\sqrt{3},-\sqrt{123}\) ; \(28\sqrt{2},\sqrt{14},2\sqrt{147},36\sqrt{4}\)
giảm dần: \(6\sqrt{\frac{1}{4}},4\sqrt{\frac{1}{2}},-\sqrt{132},2\sqrt{3},\sqrt{\frac{15}{5}}\); \(-27,4\sqrt{3},16\sqrt{5},21\sqrt{2}\)
a,\(\left(5+4\sqrt{2}\right)\left(3+2\sqrt{1+\sqrt{2}}\right)\left(3-2\sqrt{1+\sqrt{2}}\right)\)
=\(\left(5+4\sqrt{2}\right)\left(9-4\left(1+\sqrt{2}\right)\right)\)
=\(\left(5+4\sqrt{2}\right)\left(9-4-4\sqrt{2}\right)\)
=\(\left(5+4\sqrt{2}\right)\left(5-4\sqrt{2}\right)=25-\left(4\sqrt{2}\right)^2\)
=-7
b, \(\sqrt{\frac{9}{4}-\sqrt{2}}=\sqrt{\frac{9-4\sqrt{2}}{4}}=\frac{\sqrt{9-4\sqrt{2}}}{2}=\frac{\sqrt{9-2\sqrt{8}}}{2}=\frac{\sqrt{\left(\sqrt{8}-1\right)^2}}{2}=\frac{\left|\sqrt{8}-1\right|}{2}=\frac{\sqrt{8}-1}{2}\)
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So sánh:
1) \(2\sqrt{27}\) và \(\sqrt{147}\)
+ \(2\sqrt{27}\) = \(6\sqrt{3}\)
+ \(\sqrt{147}\) = \(7\sqrt{3}\)
⇒ \(6\sqrt{3}\) < \(7\sqrt{3}\)
Vậy: \(2\sqrt{27}\)< \(\sqrt{147}\)
2) \(2\sqrt{15}\) và \(\sqrt{59}\)
+ \(2\sqrt{15}\) = \(\sqrt{60}\)
⇒ \(\sqrt{60}\) > \(\sqrt{59}\)
Vậy: \(2\sqrt{15}\) > \(\sqrt{59}\)
3) \(2\sqrt{2}-1\) và 2
\(giống\left(-1\right)\left\{{}\begin{matrix}3-1\\2\sqrt{2}-1\end{matrix}\right.\)
So sánh: 3 và \(2\sqrt{2}\)
+ 3 = \(\sqrt{9}\)
+ \(2\sqrt{2}=\sqrt{8}\)
⇒ \(\sqrt{8}\) < \(\sqrt{9}\)
⇒ \(\sqrt{8}\) -1 < \(\sqrt{9}\) -1
⇒ \(2\sqrt{2}\) - 1 < 3 - 1
Vậy: \(2\sqrt{2}-1< 2\)
4) \(\frac{\sqrt{3}}{2}\) và 1
+ 1 = \(\frac{2}{2}\)
⇒ \(\frac{\sqrt{3}}{2}\) < \(\frac{2}{2}\)
Vậy: \(\frac{\sqrt{3}}{2}\) < 1
5) \(\frac{-\sqrt{10}}{2}\) và \(-2\sqrt{5}\)
+ \(-2\sqrt{5}\) = \(\frac{-4\sqrt{5}}{2}\) = \(\frac{-\sqrt{80}}{2}\)
⇒ \(\frac{-\sqrt{10}}{2}\) > \(\frac{-\sqrt{80}}{2}\)
Vậy: \(\frac{-\sqrt{10}}{2}\) > \(-2\sqrt{5}\)
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Bài 4: Rút gọn căn bậc 2 theo hằng đẳng thức
a sqrt{8+2sqrt{15}}
b sqrt{23+4sqrt{15}}
c sqrt{11+4sqrt{6}}
d sqrt{14-6sqrt{5}}
e sqrt{22-8sqrt{6}}
f sqrt{16-6sqrt{7}}
g sqrt{9-4sqrt{2}}
h sqrt{13-4sqrt{3}}
i sqrt{7-4sqrt{3}}
j sqrt{21-8sqrt{5}}
k sqrt{4-2sqrt{3}}
p sqrt{5-2sqrt{6}}
ssqrt{10-2sqrt{21}}
x sqrt{28-10sqrt{3}}
y sqrt{9-4sqrt{5}}
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Bài 4: Rút gọn căn bậc 2 theo hằng đẳng thức
a> \(\sqrt{8+2\sqrt{15}}\)
b> \(\sqrt{23+4\sqrt{15}}\)
c> \(\sqrt{11+4\sqrt{6}}\)
d> \(\sqrt{14-6\sqrt{5}}\)
e> \(\sqrt{22-8\sqrt{6}}\)
f> \(\sqrt{16-6\sqrt{7}}\)
g> \(\sqrt{9-4\sqrt{2}}\)
h> \(\sqrt{13-4\sqrt{3}}\)
i> \(\sqrt{7-4\sqrt{3}}\)
j> \(\sqrt{21-8\sqrt{5}}\)
k> \(\sqrt{4-2\sqrt{3}}\)
p> \(\sqrt{5-2\sqrt{6}}\)
s>\(\sqrt{10-2\sqrt{21}}\)
x> \(\sqrt{28-10\sqrt{3}}\)
y> \(\sqrt{9-4\sqrt{5}}\)
a, \(\sqrt{8+2\sqrt{15}}=\left(\sqrt{5}\right)^2-2\sqrt{3}.\sqrt{5}-\left(\sqrt{3}\right)^2\)
\(=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
\(=\sqrt{5}-\sqrt{3}\)
b,
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Tính
1) sqrt{18}.sqrt{2}
2) sqrt{15^2-9^2}
3) sqrt{46-6sqrt{5}}-sqrt{46+6sqrt{5}}
4)sqrt{21+6sqrt{6}}-sqrt{21-6sqrt{6}}
5) left(2+sqrt{5}right).sqrt{9-4sqrt{5}}
6)left(3-sqrt{2}right).sqrt{7+4sqrt{3}}
7)left(sqrt{3}+sqrt{5}right).sqrt{7-2sqrt{10}}
8)left(sqrt{6}+sqrt{10}right).sqrt{4-sqrt{15}}
9) sqrt{2}.left(sqrt{8}-sqrt{32}+3sqrt{18}right)
10) sqrt{2}left(sqrt{2}-sqrt{3-sqrt{5}}right)
11) sqrt{3}-sqrt{2}-sqrt{left(sqrt{3}+sqrt{2}right)^2}
12) left(sqrt{2}-sqrt{3+sqrt{5}}right).sqrt...
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Tính
1) \(\sqrt{18}.\sqrt{2}\)
2) \(\sqrt{15^2-9^2}\)
3) \(\sqrt{46-6\sqrt{5}}-\sqrt{46+6\sqrt{5}}\)
4)\(\sqrt{21+6\sqrt{6}}-\sqrt{21-6\sqrt{6}}\)
5) \(\left(2+\sqrt{5}\right).\sqrt{9-4\sqrt{5}}\)
6)\(\left(3-\sqrt{2}\right).\sqrt{7+4\sqrt{3}}\)
7)\(\left(\sqrt{3}+\sqrt{5}\right).\sqrt{7-2\sqrt{10}}\)
8)\(\left(\sqrt{6}+\sqrt{10}\right).\sqrt{4-\sqrt{15}}\)
9) \(\sqrt{2}.\left(\sqrt{8}-\sqrt{32}+3\sqrt{18}\right)\)
10) \(\sqrt{2}\left(\sqrt{2}-\sqrt{3-\sqrt{5}}\right)\)
11) \(\sqrt{3}-\sqrt{2}-\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}\)
12) \(\left(\sqrt{2}-\sqrt{3+\sqrt{5}}\right).\sqrt{2}+2\sqrt{5}\)
1: \(=\sqrt{36}=6\)
2: \(=\sqrt{\left(15-9\right)\left(15+9\right)}=\sqrt{24\cdot6}=12\)
3: \(=3\sqrt{5}-1-3\sqrt{5}-1=-2\)
4: \(=3\sqrt{2}+\sqrt{3}-3\sqrt{2}+\sqrt{3}=2\sqrt{3}\)
5: \(=\left(2+\sqrt{5}\right)\left(\sqrt{5}-2\right)=5-4=1\)
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Bài 1: Tính
1, Aleft(1-frac{5+sqrt{5}}{1+sqrt{5}}right).left(frac{5-sqrt{5}}{1-sqrt{5}}-1right)
2, Bleft(frac{3sqrt{125}}{15}-frac{10-4sqrt{6}}{sqrt{5}-2}right).frac{1}{sqrt{5}}
3, Cleft(frac{sqrt{1000}}{100}-frac{5sqrt{2}-2sqrt{5}}{2sqrt{5}-8}right).frac{sqrt{10}}{10}
4, Dfrac{1}{sqrt{49+20sqrt{6}}}-frac{1}{sqrt{49-20sqrt{6}}}+frac{1}{sqrt{7-4sqrt{3}}}
5, Efrac{1}{sqrt{4-2sqrt{3}}}-frac{1}{sqrt{7-sqrt{48}}}+frac{3}{sqrt{14-6sqrt{5}}}
6, Ffrac{1}{sqrt{2}-sqrt{3}}sqrt{frac{3sqrt{2}-2sqrt{3}...
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Bài 1: Tính
1, \(A=\left(1-\frac{5+\sqrt{5}}{1+\sqrt{5}}\right).\left(\frac{5-\sqrt{5}}{1-\sqrt{5}}-1\right)\)
2, \(B=\left(\frac{3\sqrt{125}}{15}-\frac{10-4\sqrt{6}}{\sqrt{5}-2}\right).\frac{1}{\sqrt{5}}\)
3, \(C=\left(\frac{\sqrt{1000}}{100}-\frac{5\sqrt{2}-2\sqrt{5}}{2\sqrt{5}-8}\right).\frac{\sqrt{10}}{10}\)
4, \(D=\frac{1}{\sqrt{49+20\sqrt{6}}}-\frac{1}{\sqrt{49-20\sqrt{6}}}+\frac{1}{\sqrt{7-4\sqrt{3}}}\)
5, \(E=\frac{1}{\sqrt{4-2\sqrt{3}}}-\frac{1}{\sqrt{7-\sqrt{48}}}+\frac{3}{\sqrt{14-6\sqrt{5}}}\)
6, \(F=\frac{1}{\sqrt{2}-\sqrt{3}}\sqrt{\frac{3\sqrt{2}-2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}}}\)
7, \(G=\frac{\sqrt{15-10\sqrt{2}}+\sqrt{13+4\sqrt{10}-\sqrt{11-2\sqrt{10}}}}{2\sqrt{3+2\sqrt{2}}+\sqrt{9-4\sqrt{2}+\sqrt{12+8\sqrt{2}}}}\)