Viết các biểu thức sau thành bình phương của 1 biểu thức:
a) 7 + 2√10
b) 11 - 2√28
c) 4 - 2√3
d) 7 + 4√3
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a: \(4-2\sqrt{3}=\left(\sqrt{3}-1\right)^2\)
b; \(7+4\sqrt{3}=\left(2+\sqrt{3}\right)^2\)
c: \(13-4\sqrt{3}=\left(2\sqrt{3}-1\right)^2\)
\(a,\)
\(3+2\sqrt{2}=2+2\sqrt{2}+1=\sqrt{2}^2+2\sqrt{2}+1=\left(\sqrt{2}+1\right)^2\)
\(3-2\sqrt{2}=2-2\sqrt{2}+1=\left(\sqrt{2}\right)^2-2\sqrt{2}+1=\left(\sqrt{2}-1\right)^2\)
\(b,\)
\(6+2\sqrt{5}=5+2\sqrt{5}+1=\left(\sqrt{5}\right)^2+2\sqrt{5}+1=\left(\sqrt{5}+1\right)^2\)
\(6-2\sqrt{5}=5-2\sqrt{5}+1=\left(\sqrt{5}\right)^2-2\sqrt{5}+1=\left(\sqrt{5}-1\right)^2\)
\(c,\)
\(7+4\sqrt{3}=4+2.2\sqrt{3}+3=2^2+2.2.\sqrt{3}+\left(\sqrt{3}\right)^2=\left(2+\sqrt{3}\right)^2\)
\(7-4\sqrt{3}=2^2-2.2.\sqrt{3}+\left(\sqrt{3}\right)^2=\left(2-\sqrt{3}\right)^2\)
`a)3+-2sqrt2`
`=2+-2sqrt2+1`
`=(sqrt2+-1)^2`
`b)6+-2sqrt5`
`=5+-2sqrt5+1`
`=(sqrt5+-1)^2`
`7)7+-4sqrt3`
`=4+-2.2.sqrt3+3`
`=(2+-sqrt3)^2`
1/ \(7-2\sqrt{6}=\left(\sqrt{6}\right)^2-2\sqrt{6}+1\)
\(=\left(\sqrt{6}-1\right)^2\)
2/ \(10+2\sqrt{21}=\left(\sqrt{7}\right)^2+2.\sqrt{7}.\sqrt{3}+\left(\sqrt{3}\right)^2\)
\(=\left(\sqrt{7}+\sqrt{3}\right)^2\)
4/ \(10+4\sqrt{6}=2^2+2.2.\sqrt{6}+\left(\sqrt{6}\right)^2\)
\(=\left(2+\sqrt{6}\right)^2\)
5/ \(11-2\sqrt{30}=\left(\sqrt{6}\right)^2-2.\sqrt{6}.\sqrt{5}+\left(\sqrt{5}\right)^2\)
= \(\left(\sqrt{6}-\sqrt{5}\right)^2\)
8/ \(11+4\sqrt{7}=2^2+2.2.\sqrt{7}+\left(\sqrt{7}\right)^2\)
= \(\left(2+\sqrt{7}\right)^2\)
10/ \(12+6\sqrt{3}=3^2+2.3.\sqrt{3}+\left(\sqrt{3}\right)^2\)
= \(\left(3+\sqrt{3}\right)^2\)
\(A=2^5.5^2-8^2-7=800-64-7=729=27^2\)
\(B=2^3.4^2+3^2.3^2-40=128+81-40=169=13^2\)
\(C=11.2^4+6^2.19+40=176+684+40=900=30^2\)
\(D=4^3+6^3+7^3+2=64+216+343+2=625=25^2\)
b)\(27-10\sqrt{2}=5^2-2.5\sqrt{2}+2=\left(5-\sqrt{2}\right)^2\)
c)\(18-8\sqrt{2}=4^2-2.4\sqrt{2}+2=\left(4-\sqrt{2}\right)^2\)
d)\(4-2\sqrt{3}=3-2\sqrt{3}+1=\left(\sqrt{3}-1\right)^2\)
e)\(6\sqrt{5}+14=9+2.3\sqrt{5}+5=\left(3+\sqrt{5}\right)^2\)
f)\(20\sqrt{5}+45=5^2+2.5.2\sqrt{5}+20=\left(5+2\sqrt{5}\right)^2\)
g)\(7-2\sqrt{6}=6-2\sqrt{6}+1=\left(\sqrt{6}-1\right)^2\)
a) 87 trừ đi 4 rồi cộng với 40
Tức là:
87 - 4 + 40
= 83 + 40
= 123
b) Tích của 7 và 6 rồi trừ đi 2
Tức là:
7 x 6 - 2
= 42 - 2
= 40
A = 2⁵.(-5)² - 8² - 7
= 32.25 - 64 - 7
= 729
= 27²
B = 2³.(-4)² + (-3)².3² - 40
= 8.16 + 9.9 - 40
= 169
= 13²
C = (1/4 - 1/2 - 1)³ . (2 - 2/5)³
= (-5/4)³ . (8/5)³
= (-5/4 . 8/5)³
= (-2)³
D = (-1/4)² : (1/2 - 1/3)
= 1/16 : 1/6
= 3/8
E = 4 . (1/4)² + 25 . [(3/4)³ : (5/4)³] : (3/2)³
= 1/4 + 25 . (3/4 . 5/4)³ : (3/2)³
= 1/4 + 25 . (15/16)³ : 27/8
= 1/4 + 25 . 3375/4096 : 27/8
= 1/4 + 84375/4096 : 27/8
= 1/4 + 3125/512
= 3253/512
F = 2³ + 3.(1/2)⁰ - 1 + [(-2)² : 1/2] - 8
= 8 + 3.1 - 1 + (4 : 1/2) - 8
= 8 + 3 - 1 + 8 - 8
= 10
a, 48.84
= (22)8.(23)4
= 216.212
= 228
b, 415.515
= (4.5)15
= 2015
c, 210.15 + 210.85
= 210.(15 + 85)
= 210.100
=210.(2.5)2
= 212.52
d, 33.92
= 33 . (32)2
= 33.34
= 37
e, 512.7 - 511.10
= 511.(5.7 - 10)
= 511.25
=511.52
=513
f, \(x^1\).\(x^2\).\(x^3\)....\(x^{100}\)
= \(x^{1+2+3+...+100}\)
= \(x^{\left(1+100\right).100:2}\)
= \(x^{5050}\)
a) \(A=7+2\sqrt{10}\)
\(2A=14+4\sqrt{10}\)
\(2A=10+4\sqrt{10}+4\)
\(2A=\left(\sqrt{10}+2\right)^2\)
\(A=\frac{\left(\sqrt{10}+2\right)^2}{2}\)
b) \(B=11-2\sqrt{28}=11-4\sqrt{7}\)
\(B=7-4\sqrt{7}+4\)
\(B=\left(\sqrt{7}-2\right)^2\)
c) \(C=4-2\sqrt{3}=3-2\sqrt{3}+1=\left(\sqrt{3}-1\right)^2\)
d) \(D=7+4\sqrt{3}=3+4\sqrt{3}+4=\left(\sqrt{3}+2\right)^2\)