Toán Lớp 9: chứng minh các đẳng thức sau
a, 9-4√5=(√5-2)2
b, 4-2√3=(√3-1)2
c, 6-2√5= (√5-1)2
d, √7+4√3+√7-4√3=4
e, √9-4√5- √9+4√5=3
g, (2+√5×2-√5)=-1
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Toán Lớp 9: chứng minh các đẳng thức sau
a, 9-4√5=(√5-2)2
b, 4-2√3=(√3-1)2
c, 6-2√5= (√5-1)2
d, √7+4√3+√7-4√3=4
e, √9-4√5- √9+4√5=3
g, (2+√5×2-√5)=-1
Comments ( 1 )
a)9 – 4\sqrt 5 = 5 – 2.2.\sqrt 5 + 4\\
= {\left( {\sqrt 5 – 2} \right)^2}\\
b)4 – 2\sqrt 3 = 3 – 2.\sqrt 3 .1 + 1\\
= {\left( {\sqrt 3 – 1} \right)^2}\\
c)6 – 2\sqrt 5 = 5 – 2\sqrt 5 + 1\\
= {\left( {\sqrt 5 – 1} \right)^2}\\
d)\sqrt {7 + 4\sqrt 3 } + \sqrt {7 – 4\sqrt 3 } \\
= \sqrt {4 + 2.2\sqrt 3 + 3} + \sqrt {4 – 2.2.\sqrt 3 + 3} \\
= \sqrt {{{\left( {2 + \sqrt 3 } \right)}^2}} + \sqrt {{{\left( {2 – \sqrt 3 } \right)}^2}} \\
= 2 + \sqrt 3 + 2 – \sqrt 3 \\
= 4\\
e)\sqrt {9 – 4\sqrt 5 } – \sqrt {9 + 4\sqrt 5 } \\
= \sqrt {{{\left( {\sqrt 5 – 2} \right)}^2}} – \sqrt {{{\left( {\sqrt 5 + 2} \right)}^2}} \\
= \sqrt 5 – 2 – \left( {\sqrt 5 + 2} \right)\\
= – 4\\
g)\left( {2 + \sqrt 5 } \right).\left( {2 – \sqrt 5 } \right)\\
= {2^2} – {\left( {\sqrt 5 } \right)^2}\\
= 4 – 5\\
= – 1
\end{array}$