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Ta có B = 7 − 5 2 3 + 19 − 6 2 = ( 1 − 2 ) 3 3 + ( 3 2 − 1 ) 2 = 1 − 2 + 3 2 − 1 = 2 2 .
A = 3 − 1 + 3 + 1 ( 3 + 1 ) ( 3 − 1 ) + 2 ( 2 − 3 ) 2 = 2 3 3 − 1 + 2 − 3 = 3 + 2 − 3 = 2
a) \(\sqrt{33-12\sqrt{6}}+\sqrt{15+6\sqrt{6}}=\sqrt{24-2.2\sqrt{6}.3+9}+\sqrt{6+2.\sqrt{6}.3+9}=\sqrt{\left(2\sqrt{6}-3\right)^2}+\sqrt{\left(\sqrt{6}+3\right)^2}=\left|2\sqrt{6}-3\right|+\left|\sqrt{6}+3\right|=2\sqrt{6}-3+\sqrt{6}+3=3\sqrt{6}\)
b) \(\dfrac{\sqrt{99}}{\sqrt{11}}+\dfrac{\sqrt{28}}{\sqrt{7}}-\sqrt{\sqrt{81}}=\sqrt{\dfrac{99}{11}}+\sqrt{\dfrac{28}{7}}-\sqrt{9}=\sqrt{9}+\sqrt{4}-\sqrt{9}=\sqrt{4}=2\)
a) \(\sqrt{33-12\sqrt{6}}\) + \(\sqrt{15+6\sqrt{6}}\)
= \(\sqrt{9-2.3.2\sqrt{6}+24}\)+\(\sqrt{9+2.3\sqrt{6}+6}\)
= \(\sqrt{\left(3-2\sqrt{6}\right)^2}\)+\(\sqrt{\left(3+\sqrt{6}\right)^2}\)
=\(\left|3-2\sqrt{6}\right|+\left|3+\sqrt{6}\right|\)
=\(2\sqrt{6}-3+3+\sqrt{6}\)
=\(\sqrt{6}\)
b)\(\dfrac{\sqrt{99}}{\sqrt{11}}\)+\(\dfrac{\sqrt{28}}{\sqrt{7}}\)\(-\sqrt{\sqrt{81}}\)
= \(\sqrt{\dfrac{99}{11}}+\sqrt{\dfrac{28}{7}}-3\)
=\(\sqrt{9}+\sqrt{4}-3\)
= 3+2-3
= 2
9: \(A=\dfrac{\sqrt{8+2\sqrt{15}}-\sqrt{14-6\sqrt{5}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{5}+\sqrt{3}-3+\sqrt{5}}{\sqrt{2}}=\dfrac{2\sqrt{10}+\sqrt{6}-3\sqrt{2}}{2}\)
10: \(A=\dfrac{\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{3}+1+\sqrt{3}-1}{\sqrt{2}}=\dfrac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\)
11: \(A=\dfrac{\sqrt{24-6\sqrt{7}}-\sqrt{24+6\sqrt{7}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{21}-\sqrt{3}-\sqrt{21}-\sqrt{3}}{\sqrt{2}}=-\dfrac{2\sqrt{3}}{\sqrt{2}}=-\sqrt{6}\)
12: \(B=\left(3+\sqrt{3}\right)\sqrt{12-6\sqrt{3}}\)
\(=\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)\)
=9-3=6
13: \(A=\sqrt{5}-2-\left(3-\sqrt{5}\right)\)
\(=\sqrt{5}-2-3+\sqrt{5}=2\sqrt{5}-5\)
a, A = 7 - 4 3 + 1 2 - 3 = 2 - 3 + 2 + 3 = 4
b, B = sin 2 19 0 + cos 2 19 0 + tan 19 0 - c o t 71 0
= sin 2 19 0 + cos 2 19 0 + tan 19 0 - tan 19 0 = 1
a: \(\sqrt{\left(4-\sqrt{15}\right)^2}+\sqrt{15}\)
\(=4-\sqrt{15}+\sqrt{15}=4\)
b: \(\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)
\(=2+\sqrt{3}-2+\sqrt{3}\)
\(=2\sqrt{3}\)
c: \(\sqrt{29+12\sqrt{5}}-\sqrt{29-12\sqrt{5}}\)
\(=\sqrt{\left(2\sqrt{5}+3\right)^2}-\sqrt{\left(2\sqrt{5}-3\right)^2}\)
\(=2\sqrt{5}+3-2\sqrt{5}+3=6\)
Phương trình -7 x 2 +4x=3 ⇔ 7x2 -4x+3 = 0 có hệ số a=7, b’=-2 , c=3
Ta có: ∆ ’ = b ' 2 – ac = - 2 2 -7.3 = 4- 21= -17 < 0
Vậy phương trình vô nghiệm