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\(a.11+2\sqrt{30}=6+2.\sqrt{6}.\sqrt{5}+5=\left(\sqrt{6}+\sqrt{5}\right)^2\)
\(b.10+2\sqrt{21}=7+2\sqrt{7}.\sqrt{3}+3=\left(\sqrt{7}+\sqrt{3}\right)^2\)
\(c.6x-\sqrt{x}-1=6x+2\sqrt{x}-3\sqrt{x}-1=2\sqrt{x}\left(3\sqrt{x}+1\right)-\left(3\sqrt{x}+1\right)=\left(3\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)\) \(d.14-2\sqrt{45}=14-6\sqrt{5}=9-2.3\sqrt{5}+5=\left(3-\sqrt{5}\right)^2\)
\(e.4x-3\sqrt{x}-1=4x-4\sqrt{x}+\sqrt{x}-1=4\sqrt{x}\left(\sqrt{x}-1\right)+\left(\sqrt{x}-1\right)=\left(\sqrt{x}-1\right)\left(4\sqrt{x}+1\right)\) \(j.12-2\sqrt{27}=9-2.3.\sqrt{3}+3=\left(3-\sqrt{3}\right)^2\)
a) 4x-3√x-1=4x-4√x+√x-1=4√x(√x-1)+(√x+1)
=(4√x+1).(√x-1)
c) 12-2√27=9-2.3√3+3=32-2.3√3+(√3)2=(3-√3)2
d) 6x-√x-1=6x-3√x+2√x-1=3√x(2√x-1)+(2√x-1)
=(3√x+1).(2√x-1)
e) 11+2√30=6+2.√6.√5+5=(√6)2+2.√6.√5+(√5)2=(√6+√5)2
f) 10+2√21=7+2.√7.√3+3=(√7)2+2.√7.√3+(√3)2=(√7+√3)2
úi sao bạn cũng là quản lý giống mình à, mình trả lời câu hỏi của bạn có được không nhỉ
a. \(11+2\sqrt{10}=\left(\sqrt{10}+1\right)^2\)
b. \(12-2\sqrt{11}=\left(\sqrt{11}-1\right)^2\)
c.\(23+2\sqrt{22}=\left(\sqrt{22}+1\right)^2\)
\(x\sqrt{x}+4x-12\sqrt{x}-27\)
\(=\left(x\sqrt{x}-27\right)+\left(4x-12\sqrt{x}\right)\)
\(=\left(\sqrt{x}-3\right)\left(x+3\sqrt{x}+9\right)+4\sqrt{x}\left(\sqrt{x}-3\right)\)
\(=\left(\sqrt{x}-3\right)\left(x+3\sqrt{x}+9+4\sqrt{x}\right)\)
\(=\left(\sqrt{x}-3\right)\left(x+7\sqrt{x}+9\right)\)
a, \(\sqrt{a^2-b^2}-\sqrt{a^3+b^3}\)
\(=\sqrt{\left(a+b\right)\left(a-b\right)}-\sqrt{\left(a+b\right)\left(a^2-ab+b^2\right)}\)
\(=\sqrt{a+b}\left(\sqrt{a-b}-\sqrt{a^2-ab+b^2}\right)\)
Bài 3:
a) Ta có: \(4+2\sqrt{3}\)
\(=3+2\cdot\sqrt{3}\cdot1+1\)
\(=\left(\sqrt{3}+1\right)^2\)
b) Ta có: \(7+4\sqrt{3}\)
\(=4+2\cdot2\cdot\sqrt{3}+3\)
\(=\left(2+\sqrt{3}\right)^2\)
c) Ta có: \(9+4\sqrt{5}\)
\(=5+2\cdot\sqrt{5}\cdot2+4\)
\(=\left(\sqrt{5}+2\right)^2\)
d) Ta có: \(31+10\sqrt{6}\)
\(=25+2\cdot5\cdot\sqrt{6}+6\)
\(=\left(5+\sqrt{6}\right)^2\)
e) Ta có: \(13+4\sqrt{3}\)
\(=12+2\cdot2\sqrt{3}\cdot1+1\)
\(=\left(2\sqrt{3}+1\right)^2\)
g) Ta có: \(21+12\sqrt{3}\)
\(=12+2\cdot2\sqrt{3}\cdot3+9\)
\(=\left(2\sqrt{3}+3\right)^2\)
h) Ta có: \(29+12\sqrt{5}\)
\(=20+2\cdot2\sqrt{5}\cdot3+3\)
\(=\left(2\sqrt{5}+3\right)^2\)
i) Ta có: \(49+8\sqrt{3}\)
\(=48+2\cdot4\sqrt{3}\cdot1\)
\(=\left(4\sqrt{3}+1\right)^2\)
k) Sửa đề: \(14-6\sqrt{5}\)
Ta có: \(14-6\sqrt{5}\)
\(=9-2\cdot3\cdot\sqrt{5}+5\)
\(=\left(3-\sqrt{5}\right)^2\)
l) Ta có: \(23-8\sqrt{7}\)
\(=16-2\cdot4\cdot\sqrt{7}+7\)
\(=\left(4-\sqrt{7}\right)^2\)
m) Ta có: \(15-4\sqrt{11}\)
\(=11-2\cdot\sqrt{11}\cdot2+4\)
\(=\left(\sqrt{11}-2\right)^2\)
n) Sửa đề: \(28-10\sqrt{3}\)
Ta có: \(28-10\sqrt{3}\)
\(=25-2\cdot5\cdot\sqrt{3}+3\)
\(=\left(5-\sqrt{3}\right)^2\)
o) Ta có: \(17-12\sqrt{2}\)
\(=9-2\cdot3\cdot2\sqrt{2}+8\)
\(=\left(3-2\sqrt{2}\right)^2\)
p) Ta có: \(43-30\sqrt{2}\)
\(=25-2\cdot5\cdot3\sqrt{2}+18\)
\(=\left(5-3\sqrt{2}\right)^2\)
q) Ta có: \(51-10\sqrt{2}\)
\(=50-2\cdot5\sqrt{2}\cdot1\)
\(=\left(5\sqrt{2}-1\right)^2\)
r) Ta có: \(49-12\sqrt{5}\)
\(=45-2\cdot3\sqrt{5}\cdot2+4\)
\(=\left(3\sqrt{5}-2\right)^2\)
a) 2a−4b=2(a−2b)2a−4b=2(a−2b)
c) 2ax−2ay+2a=2a(x−y+1)2ax−2ay+2a=2a(x−y+1)
e) 3xy(x−4)−9x(4−x)=3x(x−4)(y+3)3xy(x−4)−9x(4−x)=3x(x−4)(y+3)
b,d xem lại đề
a) \(21-8\sqrt{5}=16-2\times4\times\sqrt{5}+5=\left(4-\sqrt{5}\right)^2\)
b) \(47-12\sqrt{11}=36-2\times6\times\sqrt{11}+11=\left(6-\sqrt{11}\right)^2\)
c) \(13-4\sqrt{3}=12-2\times1\times\sqrt{3}+1=\left(2\sqrt{3}-1\right)^2\)
d) \(43+30\sqrt{2}=25+2\times5\times3\sqrt{2}+18=\left(5+3\sqrt{2}\right)^2\)
e) \(41+24\sqrt{2}=9+2\times3\times4\sqrt{2}+32=\left(3+4\sqrt{2}\right)^2\)
g) \(29-12\sqrt{5}=9+2\times3\times2\sqrt{5}+20=\left(3+2\sqrt{5}\right)^2\)
h) \(49-8\sqrt{3}=48-2\times4\sqrt{3}\times1+1=\left(4\sqrt{3}-1\right)^2\)
i) \(37-12\sqrt{7}=28-2\times3\times2\sqrt{7}+9=\left(2\sqrt{7}-3\right)^2\)
ta có :
Hay quá, Minh Quang không bị lừa :)))
a) 10+2\(\sqrt{21}\)= 3+7 +2\(\sqrt{3.7}\) =3+2\(\sqrt{3.7}\) +7= (\(\sqrt{3}\) +\(\sqrt{7}\))2
b) 12-2\(\sqrt{27}\) = 9+3-2\(\sqrt{9.3}\) =9-2\(\sqrt{9.3}\) +3=(\(\sqrt{9}\)-\(\sqrt{3}\))2
c) 11+2\(\sqrt{30}\) =6+5+2\(\sqrt{6.5}\) = 6+2\(\sqrt{6.5}\) +5=(\(\sqrt{6}\) +\(\sqrt{5}\) )2
d) 14- 2\(\sqrt{45}\)= 9+5-2\(\sqrt{9.5}\) =9-2\(\sqrt{9.5}\) +5= (\(\sqrt{9}\) -\(\sqrt{5}\) )2
a) \(10+2\sqrt{21}=2\left(5+\sqrt{21}\right)\)
b)\(12-2\sqrt{27}=12-2.3\sqrt{3}=12-6\sqrt{3}=6\left(6-\sqrt{3}\right)\)
c)\(11+2\sqrt{30}=10+2\sqrt{30}+1=\left(\sqrt{10}+1\right)^2\)
d)\(14-2\sqrt{45}=9-2\sqrt{9.5}+5=\left(3-\sqrt{5}\right)^2\)
a) (\(\sqrt{7}+\sqrt{3}\))2
b) ( \(\sqrt{9}-\sqrt{3}\))2
c) (\(\sqrt{6}+\sqrt{5}\))2
d) (\(\sqrt{9}-\sqrt{5}\))2
a/ 2(5+\(\sqrt{21}\))
b/ 2(6-\(\sqrt{27}\))
c/(\(\sqrt{5}\)+\(\sqrt{6}\))^2
d/2(7-\(45\))
\(\left(\sqrt{ }7+\sqrt{ }3\right)^2\)
\(\left(\sqrt{ }9-\sqrt{ }3\right)^2\)
\(\left(\sqrt{ }5+\sqrt{ }6\right)^2\)
\(\left(\sqrt{ }9-\sqrt{ }5\right)^2\)
a)\(\left(\sqrt{7}+\sqrt{3}\right)^2\)
b)\(\left(\sqrt{9}-\sqrt{3}\right)^2\)
c)\(\left(\sqrt{5}+\sqrt{6}\right)^2\)
d)\(\left(\sqrt{9}-\sqrt{5}\right)^2\)
a) \(\left(\sqrt{3}+\sqrt{7}\right)^2\)
b) \(\left(\sqrt{3}-\sqrt{9}\right)^2\)
c) \(\left(\sqrt{5}+\sqrt{6}\right)^2\)
d) \(\left(\sqrt{9}-\sqrt{5}\right)^2\)
a) (\(\sqrt{7}+\sqrt{3}\))2
b) (\(\sqrt{9}-\sqrt{3}\))2
c) (\(\sqrt{6}+\sqrt{5}\))2
d) (\(\sqrt{9}-\sqrt{5}\))2
a) \(\left(\sqrt{3}+\sqrt{7}\right)^2\)
b) \(\left(\sqrt{3}+\sqrt{9}\right)^2\)
c) \(\left(\sqrt{5}+\sqrt{6}\right)^2\)
d)\(\left(\sqrt{9}+\sqrt{5}\right)^2\)
a, 2(5+\(\sqrt{21}\))
a) \(\left(\sqrt{3}+\sqrt{7}\right)^2\)
b) \(\left(\sqrt{9}-\sqrt{3}\right)^2\)
c) \(\left(\sqrt{5}+\sqrt{6}\right)^2\)
d) \(\left(\sqrt{9}-\sqrt{5}\right)^2\)
a. 10+ 2\(\sqrt{21}\)= 3+ \(2\sqrt{3}.\sqrt{7}\)+7 =(\(\sqrt{3}+\sqrt{7}\))\(^2\)
b. 12- 2\(\sqrt{27}\)= 9- \(2.\sqrt{9}.\sqrt{3}\)= (3-\(\sqrt{3}\))\(^2\)
c. 11+ 2\(\sqrt{30}\)= 5+ 2\(\sqrt{5}.\sqrt{6}\)+ 6= (\(\sqrt{5}+\sqrt{6}\))\(^2\)
d. 14- 2\(\sqrt{45}\)= 9- 2\(\sqrt{9}.\sqrt{5}\)+ 5= (3- \(\sqrt{5}\))\(^2\)
a) 10 + 2\(\sqrt{21}\)=\(\left(\sqrt{7}+\sqrt{3}\right)^2\)
b) 12 - 2\(\sqrt{27}\)=\(3\times\left(\sqrt{3}-1\right)^2\)
c) 11 + 2\(\sqrt{30}\)=\(\left(\sqrt{5}+\sqrt{6}\right)^2\)
d) 14 - 2\(\sqrt{45}\)=\(\left(3-\sqrt{5}\right)^2\)
a)\(10+2\sqrt{21}=\left(\sqrt{3}+\sqrt{7}\right)^2\)
b)\(12-2\sqrt{27}=\left(\sqrt{3}-3\right)^2\)
c)\(11+2\sqrt{30}=\left(\sqrt{5}+\sqrt{6}\right)^2\)
d)\(14-2\sqrt{45}=\left(3-\sqrt{5}\right)^2\)
a) 10+2\sqrt{21}=\left(\sqrt{3}+\sqrt{7}\right)^210+221=(3+7)2
b) 12-2\sqrt{27}=\left(3-\sqrt{3}\right)^212−227=(3−3)2
c) 11+2\sqrt{30}=\left(\sqrt{5}+\sqrt{6}\right)^211+230=(5+6)2
d) 14-2\sqrt{45}=\left(3-\sqrt{5}\right)^214−245=(3−5)2
a) \(\left(\sqrt{7}+\sqrt{3}\right)^2\)
b) \(\left(3-\sqrt{3}\right)^2\)
c) \(\left(\sqrt{6}+\sqrt{5}\right)^2\)
d) \(\left(3-\sqrt{5}\right)^{^{ }2}\)
a)\(\left(\sqrt{3}+\sqrt{7}\right)^2\)
b)\(\left(3-\sqrt{3}\right)^2\)
c)\(\left(\sqrt{5}+\sqrt{6}\right)^2\)
d)\(\left(3-\sqrt{5}\right)^2\)
a) \(\left(\sqrt{3}+\sqrt{7}\right)^2\)
b) \(\left(3-\sqrt{3}\right)^2\)
c) \(\left(\sqrt{5}+\sqrt{6}\right)^2\)
d) \(\left(3-\sqrt{5}\right)^2\)
a) \(10+2\sqrt{21}=3+2\sqrt{3.7}+7=\left(\sqrt{3}\right)^2+2\sqrt{3}.\sqrt{7}+\left(\sqrt{7}\right)^2=\left(\sqrt{3}+\sqrt{7}\right)^2\)
b) \(12-2\sqrt{27}=9-2\sqrt{3.9}+3=3^2-2.3.\sqrt{3}+\left(\sqrt{3}\right)^2=\left(3-\sqrt{3}\right)^2\)
c) \(11+2\sqrt{30}=6+2\sqrt{6.5}+5=\left(\sqrt{6}\right)^2+2\sqrt{6}.\sqrt{5}+\left(\sqrt{5}\right)^2=\left(\sqrt{6}+\sqrt{5}\right)^2\)
d) \(14-2\sqrt{45}=9-2\sqrt[]{9.5}+5=3^2-2.3\sqrt{5}+\left(\sqrt{5}\right)^2=\left(3-\sqrt{5}\right)^2\)
a) 10+2\(\sqrt{21}\) = (\(\sqrt{7}\))2+(\(\sqrt{3}\))2+2\(\sqrt{7}\).\(\sqrt{3}\) = (\(\sqrt{7}\))2+ 2\(\sqrt{7}\).\(\sqrt{3}\)+(\(\sqrt{3}\)) 2 = (\(\sqrt{7}\)+\(\sqrt{3}\))2
b) 12-2\(\sqrt{27}\) = 32+(\(\sqrt{3}\))2-2\(\sqrt{9}\).\(\sqrt{3}\) = 32-2\(\sqrt{9}\).\(\sqrt{3}\)+(\(\sqrt{3}\))2 = (\(\sqrt{9}\)-\(\sqrt{3}\))2
c) 11+2\(\sqrt{30}\) = (\(\sqrt{6}\))2+(\(\sqrt{5}\))2+2\(\sqrt{6}\).\(\sqrt{5}\) = (\(\sqrt{6}\)+\(\sqrt{5}\))2
d) 14-2\(\sqrt{45}\) = 32+(\(\sqrt{5}\))2-2\(\sqrt{9}\).\(\sqrt{5}\) = (\(\sqrt{9}\)-\(\sqrt{5}\))2
a) \(10+2\sqrt{21}\)
\(=3+2\sqrt{3\times7}+7\)
\(=\left(\sqrt{3}\right)^2+2\sqrt{3\times7}+\left(\sqrt{9}\right)^2\)
\(=\left(\sqrt{3}+\sqrt{7}\right)^2\)
b) \(12-2\sqrt{27}\)
\(=9-2\sqrt{9\times3}+3\)
\(=3^2-2\sqrt{9\times3}+\left(\sqrt{3}\right)^2\)
\(=\left(3-\sqrt{3}\right)^2\)
c) \(11+2\sqrt{30}\)
\(=5+2\sqrt{5\times6}+6\)
\(=\left(\sqrt{5}\right)^2+2\sqrt{5\times6}+\left(\sqrt{6}\right)^2\)
\(=\left(\sqrt{5}+\sqrt{6}\right)^2\)
d) \(14-2\sqrt{45}\)
\(=9-2\sqrt{9\times5}+5\)
\(=3^2-2\times3\times\sqrt{5}+\left(\sqrt{5}\right)^2\)
\(=\left(3-\sqrt{5}\right)^2\)