phân tích
\(14+6\sqrt{5}\)
\(12-\sqrt{140}\)
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) \(\sqrt{14-2\sqrt{33}}\)
2) \(\sqrt{12-2\sqrt{35}}\)
3) \(\sqrt{16-2\sqrt{55}}\)
4) \(\sqrt{14-6\sqrt{5}}\)
5) \(\sqrt{17-12\sqrt{2}}\)
6) \(\sqrt{27-12\sqrt{5}}\)
7) \(\sqrt{4+\sqrt{15}}\)
LÀM CHI TIẾT GIÚP MK NHÉ!
1)
\(=\sqrt{\left(\sqrt{11}\right)^2-2.\sqrt{11}.\sqrt{3}+\left(\sqrt{3}\right)^2}\)
\(=\sqrt{\left(\sqrt{11}-\sqrt{3}\right)^2}=\sqrt{11}-\sqrt{3}\)
2)
\(=\sqrt{\left(\sqrt{7}\right)^2-2.\sqrt{7}\sqrt{5}+\left(\sqrt{5}\right)^2}=\sqrt{\left(\sqrt{7}-\sqrt{5}\right)^2}=\sqrt{7}-\sqrt{5}\)
3)
\(=\sqrt{\left(\sqrt{11}\right)^2-2.\sqrt{11}\sqrt{5}+\left(\sqrt{5}\right)^2}=\sqrt{\left(\sqrt{11}-\sqrt{5}\right)}=\sqrt{11}-\sqrt{5}\)
4)
\(=\sqrt{3^2-2.3.\sqrt{5}+\left(\sqrt{5}\right)^2}=\sqrt{\left(3-\sqrt{5}\right)^2}=3-\sqrt{5}\)
5)
\(=\sqrt{3^2-2.3.2\sqrt{2}+\left(2\sqrt{2}\right)^2}=\sqrt{\left(3-2\sqrt{2}\right)^2}=3-2\sqrt{2}\)
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giúp em với ạ
\(\sqrt{5
+2\sqrt{ }6}\)
\(\sqrt{12+2\sqrt{ }35}-\sqrt{12-2\sqrt{ }35}\)
\(\sqrt{16+6\sqrt{ }7}\)
\(\sqrt{31-12\sqrt{ }3}\)
\(\sqrt{27+10\sqrt{ }2}\)
\(\sqrt{14+6\sqrt{ }5}\)
a: \(\sqrt{5+2\sqrt{6}}=\sqrt{3}+\sqrt{2}\)
b: \(\sqrt{12+2\sqrt{35}}-\sqrt{12-2\sqrt{35}}=\sqrt{7}+\sqrt{5}-\sqrt{7}+\sqrt{5}=2\sqrt{5}\)
c: \(\sqrt{16+6\sqrt{7}}=4+\sqrt{7}\)
d: \(\sqrt{31-12\sqrt{3}}=3\sqrt{3}-2\)
e: \(\sqrt{27+10\sqrt{2}}=5+\sqrt{2}\)
f: \(\sqrt{14+6\sqrt{5}}=3+\sqrt{5}\)
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Phân tích :1) sqrt{29+12sqrt{5}} - sqrt{29-12sqrt{5}}2) sqrt{8-2sqrt{15}}- sqrt{23-4sqrt{15}}3) sqrt{8-2sqrt{15}} + sqrt{48+6sqrt{15}}4) sqrt{49-5sqrt{96}}+sqrt{49+5sqrt{96}}5) sqrt{15-6sqrt{15}}+sqrt{33-12sqrt{6}}6) sqrt{16-6sqrt{7}}+sqrt{64-24sqrt{7}}7) sqrt{14-6sqrt{5}}+sqrt{14+6sqrt{5}}8) sqrt{1-6sqrt{2}}+sqrt{11-6sqrt{2}}9) sqrt{13+4sqrt{10}}+sqrt{13-4sqrt{10}}10) sqrt{46-6sqrt{5}}+sqrt{29-12sqrt{5}}
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Phân tích :
1) \(\sqrt{29+12\sqrt{5}}\) - \(\sqrt{29-12\sqrt{5}}\)
2) \(\sqrt{8-2\sqrt{15}}\)- \(\sqrt{23-4\sqrt{15}}\)
3) \(\sqrt{8-2\sqrt{15}}\) + \(\sqrt{48+6\sqrt{15}}\)
4) \(\sqrt{49-5\sqrt{96}}\)+\(\sqrt{49+5\sqrt{96}}\)
5) \(\sqrt{15-6\sqrt{15}}\)+\(\sqrt{33-12\sqrt{6}}\)
6) \(\sqrt{16-6\sqrt{7}}\)+\(\sqrt{64-24\sqrt{7}}\)
7) \(\sqrt{14-6\sqrt{5}}\)+\(\sqrt{14+6\sqrt{5}}\)
8) \(\sqrt{1-6\sqrt{2}}\)+\(\sqrt{11-6\sqrt{2}}\)
9) \(\sqrt{13+4\sqrt{10}}\)+\(\sqrt{13-4\sqrt{10}}\)
10) \(\sqrt{46-6\sqrt{5}}\)+\(\sqrt{29-12\sqrt{5}}\)
\(\sqrt{29+12\sqrt{5}}-\sqrt{29-12\sqrt{5}}=\left(2\sqrt{5}+3\right)-\left(2\sqrt{5}-3\right)=6\)
\(\sqrt{8-2\sqrt{15}}-\sqrt{23-4\sqrt{15}}=\left(\sqrt{5}-\sqrt{3}\right)-\left(2\sqrt{5}-\sqrt{3}\right)=-\sqrt{5}\)
\(\sqrt{8-12\sqrt{5}}+\sqrt{48+6\sqrt{15}}=\left(\sqrt{5}-\sqrt{3}\right)+\left(3\sqrt{5}+\sqrt{3}\right)=4\sqrt{5}\)
\(\sqrt{49-5\sqrt{96}}+\sqrt{49+5\sqrt{96}}=\left(5-2\sqrt{6}\right)+\left(5+2\sqrt{6}\right)=10\)
\(\sqrt{15-6\sqrt{15}}+\sqrt{33-12\sqrt{6}}\) đề này sai ạ
\(\sqrt{16-6\sqrt{7}}+\sqrt{64-24\sqrt{7}}=\left(3-\sqrt{7}\right)+\left(6-2\sqrt{7}\right)=9-3\sqrt{7}\)
\(\sqrt{14-6\sqrt{5}}+\sqrt{14+6\sqrt{5}}=\left(3-\sqrt{5}\right)+\left(3+\sqrt{5}\right)=6\)
\(\sqrt{1-6\sqrt{2}}+\sqrt{11-6\sqrt{2}}\)
\(\sqrt{13+4\sqrt{10}}+\sqrt{13-4\sqrt{10}}=\left(2\sqrt{2}+5\right)+\left(2\sqrt{2}-5\right)=4\sqrt{2}\)
\(\sqrt{46-6\sqrt{5}}+\sqrt{29-12\sqrt{5}}=\left(3\sqrt{5}-1\right)+\left(2\sqrt{5}-3\right)=5\sqrt{5}-4\)
#Học tốt ạ
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\(6-12\sqrt{15}+3\sqrt{5}-30\sqrt{3}\) phân tích thành nhân tử hộ e ạ
Phân tích: \(\sqrt{12-2\sqrt{14+2.\sqrt{13}}}\)
\(\sqrt{12-2\sqrt{14+2\sqrt{13}}}\)
\(=\sqrt{12-2\sqrt{\left(\sqrt{13}\right)^2+2\sqrt{13}+1}}\)
\(=\sqrt{12-2\sqrt{\left(\sqrt{13}+1\right)^2}}\)
\(=\sqrt{12-2.\left|\sqrt{13}+1\right|}\)
\(=\sqrt{12-2\left(\sqrt{13}+1\right)}\)
\(=\sqrt{12-2\sqrt{13}+2}\)
\(=\sqrt{14+2\sqrt{13}}\)
\(=\sqrt{\left(\sqrt{13}\right)^2+2\sqrt{13}+1}\)
\(=\sqrt{\left(\sqrt{13}+1\right)^2}\)
\(=\left|\sqrt{13}+1\right|\)
\(=\sqrt{13}+1\) ( vi \(\sqrt{13}+1>0\))
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2 tháng 7 2021 lúc 17:05
Phân tích ra thừa số :
a) \(3\sqrt{2}-2\sqrt{3}\)
b) \(\sqrt{2}+\sqrt{6}+\sqrt{14}+\sqrt{42}\)
c)\(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}\)
a) \(3\sqrt{2}-2\sqrt{3}=\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)\)
b) \(\sqrt{2}+\sqrt{6}+\sqrt{14}+\sqrt{42}=\sqrt{2}\left(1+\sqrt{3}+\sqrt{7}+\sqrt{21}\right)\)
\(=\sqrt{2}\left(1+\sqrt{3}\right)\left(1+\sqrt{7}\right)\)
c) \(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}=\dfrac{\sqrt{3}\left(2-\sqrt{2}\right)}{\sqrt{2}\left(2-\sqrt{2}\right)}=\dfrac{\sqrt{6}}{2}\)
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a) \(3\sqrt{2}-2\sqrt{3}=\sqrt{3}.\sqrt{3}.\sqrt{2}-\sqrt{2}.\sqrt{2}.\sqrt{3}=\left(\sqrt{3}-\sqrt{2}\right).\sqrt{6}\)
b) \(\sqrt{2}+\sqrt{6}+\sqrt{14}+\sqrt{42}=\left(\sqrt{3}+1\right)\sqrt{2}+\sqrt{14}\left(\sqrt{3}+1\right)=\sqrt{2}\left(\sqrt{7}+1\right)\left(\sqrt{3}+1\right)\)
c) \(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}=\dfrac{\sqrt{3}\left(2-\sqrt{2}\right)}{\sqrt{2}\left(2-\sqrt{2}\right)}=\dfrac{\sqrt{3}}{\sqrt{2}}=\sqrt{\dfrac{9}{4}}\)
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\(\sqrt{12-6\sqrt{3}}\)
\(\sqrt{19+8\sqrt{3}}\)
\(\sqrt{14-6\sqrt{5}}\)
tính giải chi tiết hộ mình nha
\(\sqrt{12-6\sqrt{3}}=\sqrt{9-6\sqrt{3}+3}=\sqrt{3^2-2.3.\sqrt{3}+\left(\sqrt{3}\right)^2}=\sqrt{\left(3-\sqrt{3}\right)^2}\)
\(=\left|3-\sqrt{3}\right|=3-\sqrt{3}\)
\(\sqrt{19+8\sqrt{3}}=\sqrt{16+8\sqrt{3}+3}=\sqrt{4^2+2.4.\sqrt{3}+\left(\sqrt{3}\right)^2}=\sqrt{\left(4+\sqrt{3}\right)^2}\)
\(=\left|4+\sqrt{3}\right|=4+\sqrt{3}\)
\(\sqrt{14-6\sqrt{5}}=\sqrt{9-6\sqrt{5}+5}=\sqrt{3^2-2.3.\sqrt{5}+\left(\sqrt{5}\right)^2}=\sqrt{\left(3-\sqrt{5}\right)^2}\)
\(=\left|3-\sqrt{5}\right|=3-\sqrt{5}\)
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\(\sqrt{12-6\sqrt{3}}=\sqrt{3^2-2.3.\sqrt{3}+\left(\sqrt{3}\right)^2}=\sqrt{\left(3-\sqrt{3}\right)^2}=\left|3-\sqrt{3}\right|=3-\sqrt{3}\)
\(\sqrt{19+8\sqrt{3}}=\sqrt{4^2+2.4.\sqrt{3}+\left(\sqrt{3}\right)^2}=\sqrt{\left(4+\sqrt{3}\right)^2}=\left|4+\sqrt{3}\right|=4+\sqrt{3}\)
\(\sqrt{14-6\sqrt{5}}=\sqrt{3^2-2.3.\sqrt{5}+\left(\sqrt{5}\right)^2}=\sqrt{\left(3-\sqrt{5}\right)^2}=\left|3-\sqrt{5}\right|=3-\sqrt{5}\)
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\(\sqrt{12-6\sqrt{3}}=3-\sqrt{3}\)
\(\sqrt{19+8\sqrt{3}}=4+\sqrt{3}\)
\(\sqrt{14-6\sqrt{5}}=3-\sqrt{5}\)
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