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a) \(\sqrt{6-\sqrt{11}}\cdot\sqrt{6+\sqrt{11}}\)
\(=\sqrt{\left(6-\sqrt{11}\right)\left(6+\sqrt{11}\right)}\)
\(=\sqrt{6^2-\left(\sqrt{11}\right)^2}\)
\(=\sqrt{36-11}\)
\(=\sqrt{25}\)
\(=\sqrt{5^2}\)
\(=5\)
b) \(\sqrt{8+\sqrt{15}}\cdot\sqrt{8-\sqrt{15}}\)
\(=\sqrt{\left(8+\sqrt{15}\right)\left(8-\sqrt{15}\right)}\)
\(=\sqrt{8^2-\left(\sqrt{15}\right)^2}\)
\(=\sqrt{64-15}\)
\(=\sqrt{49}\)
\(=\sqrt{7^2}\)
\(=7\)
a: \(=\sqrt{6^2-11}=\sqrt{25}=5\)
b: \(=\sqrt{8^2-15}=\sqrt{49}=7\)
Lời giải:
a.
\(=\sqrt{5+2.2\sqrt{5}+2^2}-\sqrt{5-2.2\sqrt{5}+2^2}\)
$=\sqrt{(\sqrt{5}+2)^2}-\sqrt{(\sqrt{5}-2)^2}$
$=|\sqrt{5}+2|-|\sqrt{5}-2|=(\sqrt{5}+2)-(\sqrt{5}-2)=4$
b.
$=\sqrt{3-2.3\sqrt{3}+3^2}+\sqrt{3+2.3.\sqrt{3}+3^2}$
$=\sqrt{(\sqrt{3}-3)^2}+\sqrt{(\sqrt{3}+3)^2}$
$=|\sqrt{3}-3|+|\sqrt{3}+3|$
$=(3-\sqrt{3})+(\sqrt{3}+3)=6$
c.
$=\sqrt{2+2.3\sqrt{2}+3^2}-\sqrt{2-2.3\sqrt{2}+3^2}$
$=\sqrt{(\sqrt{2}+3)^2}-\sqrt{(\sqrt{2}-3)^2}$
$=|\sqrt{2}+3|-|\sqrt{2}-3|$
$=(\sqrt{2}+3)-(3-\sqrt{2})=2\sqrt{2}$
a: \(\left(3+\sqrt{2}\right)^2=3^2+2\cdot3\cdot\sqrt{2}+\left(\sqrt{2}\right)^2\)
\(=9+6\sqrt{2}+2=11+6\sqrt{2}\)
b: \(\sqrt{11+6\sqrt{2}}+\sqrt{11-6\sqrt{2}}\)
\(=\sqrt{\left(3+\sqrt{2}\right)^2}+\sqrt{\left(3-\sqrt{2}\right)^2}\)
\(=3+\sqrt{2}+3-\sqrt{2}=6\)
c: \(\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}\)
\(=\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}\)
\(=\sqrt{7}-1-\sqrt{7}-1=-2\)
d: \(\sqrt{49-12\sqrt{5}}-\sqrt{49+12\sqrt{5}}\)
\(=\sqrt{45-2\cdot3\sqrt{5}\cdot2+4}-\sqrt{45+2\cdot3\sqrt{5}\cdot2+4}\)
\(=\sqrt{\left(3\sqrt{5}-2\right)^2}-\sqrt{\left(3\sqrt{5}+2\right)^2}\)
\(=3\sqrt{5}-2-3\sqrt{5}-2=-4\)
a) \(\left(3+\sqrt{2}\right)^2=9+6\sqrt{2}+2=11+6\sqrt{2}\)
b) \(\sqrt{11+6\sqrt{2}}+\sqrt{11-6\sqrt{2}}\)
\(=\sqrt{\left(3+\sqrt{2}\right)^2}+\sqrt{\left(3-\sqrt{2}\right)^2}\)
\(=3+\sqrt{2}+3-\sqrt{2}=6\)
c) \(\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}\)
\(=\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}\)
\(=\sqrt{7}-1-\sqrt{7}-1=-2\)
d) \(\sqrt{49-12\sqrt{5}}-\sqrt{49+12\sqrt{5}}\)
\(=\sqrt{\left(3\sqrt{5}-2\right)^2}-\sqrt{\left(3\sqrt{5}+2\right)^2}\)
\(=3\sqrt{5}-2-3\sqrt{5}-2=-4\)
b: =căn 10-3+4-căn 10=1
a: \(=\sqrt{11-4\sqrt{6}+\sqrt{15}}\)
tuổi con HN là :
50 : ( 1 + 4 ) = 10 ( tuổi )
tuổi bố HN là :
50 - 10 = 40 ( tuổi )
hiệu của hai bố con ko thay đổi nên hiệu vẫn là 30 tuổi
ta có sơ đồ : bố : |----|----|----|
con : |----| hiệu 30 tuổi
tuổi con khi đó là :
30 : ( 3 - 1 ) = 15 ( tuổi )
số năm mà bố gấp 3 tuổi con là :
15 - 10 = 5 ( năm )
ĐS : 5 năm
mình nha
a) \(\sqrt{11+4\sqrt{7}}-\sqrt{11-4\sqrt{7}}\)
\(=\sqrt{7+4\sqrt{7}+4}-\sqrt{7-4\sqrt{7}+4}\)
\(=\sqrt{\left(\sqrt{7}+2\right)^2}-\sqrt{\left(\sqrt{7}-2\right)^2}\)
\(=\left|\sqrt{7}+2\right|-\left|\sqrt{7}-2\right|\)
\(=\sqrt{7}+2-\sqrt{7}+2=4\)
a) \(\sqrt{11+4\sqrt{7}}-\sqrt{11-4\sqrt{7}}=\sqrt{\left(2+\sqrt{7}\right)^2}-\sqrt{\left(\sqrt{7}-2\right)^2}=2+\sqrt{7}-\sqrt{7}+2=4\)
b) \(A=\sqrt{11-4\sqrt{6}}-\sqrt{11+4\sqrt{6}}\)
\(\Rightarrow A^2=11-4\sqrt{6}-2\sqrt{\left(11-4\sqrt{6}\right)\left(11+4\sqrt{6}\right)}+11+4\sqrt{6}\)
\(A^2=22-2\sqrt{121-96}\)
\(A^2=22-2\sqrt{25}=22-2.5=12\)
\(\Rightarrow A=-\sqrt{12}\)(Chú ý \(A< 0\))
\(A=\sqrt{12+\sqrt{12+\sqrt{12}}}+\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6}}}}< \sqrt{12+\sqrt{12+\sqrt{16}}}+\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{9}}}}\)\(=7\)
\(B=\sqrt{14}+\sqrt{11}>\sqrt{13,69}+\sqrt{10,89}=7\)
\(\Rightarrow A< B\)
Ta có:
\(12< 16\Rightarrow\sqrt{12}< \sqrt{16}=4\\ 6< 9\Rightarrow\sqrt{6}< \sqrt{9}=3\)
\(\Rightarrow A< \sqrt{12+\sqrt{12+4}}+\sqrt{6+\sqrt{6+\sqrt{6+3}}}=\sqrt{12+4}+\sqrt{6+3}=4+3=7\) (1)
Lại có :
\(B=\sqrt{14}+\sqrt{11}\Rightarrow B^2=25+2\sqrt{14.11}=25+2\sqrt{154}>25+2\sqrt{144}=25+2.12=49=7^2\)
Mà B > 0
\(\Rightarrow B>7\) (2)
Từ (1),(2) suy ra A<B
\(A=\sqrt{6-\sqrt{11}}-\sqrt{6+\sqrt{11}}=\dfrac{\sqrt{2}\left(\sqrt{6-\sqrt{11}}-\sqrt{6+\sqrt{11}}\right)}{\sqrt{2}}=\dfrac{\sqrt{12-2\sqrt{11}}-\sqrt{12+2\sqrt{11}}}{\sqrt{2}}=\dfrac{\sqrt{\left(\sqrt{11}-1\right)^2}-\sqrt{\left(\sqrt{11}+1\right)^2}}{\sqrt{2}}=\dfrac{\sqrt{11}-1-\sqrt{11}-1}{\sqrt{2}}=\dfrac{-2}{\sqrt{2}}=-\sqrt{2}\)
\(A=\sqrt{\left(\sqrt{\dfrac{11}{2}}-\sqrt{\dfrac{1}{2}}\right)^2}-\sqrt{\left(\dfrac{11}{2}+\sqrt{\dfrac{1}{2}}\right)^2}\\ A=\sqrt{\dfrac{11}{2}}-\sqrt{\dfrac{1}{2}}-\sqrt{\dfrac{11}{2}}-\sqrt{\dfrac{1}{2}}\\ A=-2\sqrt{\dfrac{1}{2}}=-\dfrac{2\sqrt{2}}{2}=-\sqrt{2}\)