a) Ta có : Vì \(x\ge0\)và \(y\ge0\)nên \(x+y\ge0\)\(\Leftrightarrow\left|x+y\right|=x+y\)
\(\frac{2}{x^2-y^2}\sqrt{\frac{3\left(x+y\right)^2}{2}}\)
\(=\frac{2}{x^2-y^2}\sqrt{\frac{3}{2}.\left(x+y\right)^2}\)
\(=\frac{2}{x^2-y^2}.\sqrt{\frac{3}{2}}.\left|x+y\right|\)
\(=\frac{2}{\left(x-y\right)\left(x+y\right)}.\sqrt{\frac{3}{2}}.\left(x+y\right)\)
\(=\frac{2}{x-y}.\sqrt{\frac{3}{2}}\)
\(=\frac{1}{x-y}.2.\sqrt{\frac{3}{2}}\)
\(=\frac{1}{x-y}.\sqrt{\frac{2^2.3}{2}}\)
\(=\frac{1}{x-y}.\sqrt{6}=\frac{\sqrt{6}}{x-y}\)
a, \(\frac{2}{x^2-y^2}\sqrt{\frac{3\left(x+y\right)^2}{2}}=\frac{2}{x^2-y^2}\frac{\sqrt{3}\left|x+y\right|}{\sqrt{2}}=\frac{2\sqrt{3}\left(x+y\right)}{\left(x-y\right)\left(x+y\right)\sqrt{2}}\)
do \(x\ge0;y\ge0\)
\(=\frac{2\sqrt{3}}{\sqrt{2}\left(x-y\right)}=\frac{2\sqrt{6}}{2\left(x-y\right)}=\frac{\sqrt{6}}{x-y}\)
b) Ta có :
\(\frac{2}{2a-1}\sqrt{5a^2\left(1-4a+4a^2\right)}\)
\(=\frac{2}{2a-1}\sqrt{5a^2\left(1-2.2a+2^2.a^2\right)}\)
\(=\frac{2}{2a-1}\sqrt{5a^2\left[1^2-2.1.2a+\left(2a\right)^2\right]}\)
\(=\frac{2}{2a-1}\sqrt{5a^2\left(1-2a\right)^2}\)
\(=\frac{2}{2a-1}.\sqrt{5}.\sqrt{a^2}.\sqrt{\left(1-2a\right)^2}\)
\(=\frac{2}{2a-1}\sqrt{5}.\left|a\right|.\left|1-2a\right|\)
Vì a > 0,5 <=> a > 0 <=> | a | = a
Vì a > 0,5 <=> 2a > 2 . 0,5 <=> 2a > 1 hay 1 < 2a
<=> 1 - 2a < 0 <=> | 1 - 2a | = - ( 1 - 2a )
= -1 + 2a = 2a - 1
Thay vào trên ta được :
\(=\frac{2}{2a-1}\sqrt{5}.\left|a\right|.\left|1-2a\right|=\frac{2}{2a-1}\sqrt{5}.a.\left(2a-1\right)=2a\sqrt{5}\)
Vậy : \(\frac{2}{2a-1}\sqrt{5a^2\left(1-4a+4a^2\right)}=2a\sqrt{5}\)
b, \(\frac{2}{2a-1}\sqrt{5a^2\left(1-4a+4a^2\right)}\)
\(=\frac{2}{2a-1}\sqrt{5a^2\left(1-2a\right)^2}=\frac{2}{2a-1}\sqrt{5a^2\left(2a-1\right)^2}\)
\(=\frac{2\sqrt{5}a\left(2a-1\right)}{2a-1}=2\sqrt{5}a\)
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a) \(\dfrac{2}{x^2-y^2}\sqrt{\dfrac{3\left(x+y\right)^2}{2}}=\dfrac{2}{\left(x+y\right)\left(x-y\right)}.\dfrac{\sqrt{3}.\left(x+y\right)}{\sqrt{2}}=\dfrac{\sqrt{6}}{x-y}\)
b) \(\dfrac{2}{2a-1}\sqrt{5a^2\left(1-4a+4a^2\right)}=\dfrac{2}{2a-1}.\sqrt{5}a.\left(2a-1\right)=2\sqrt{5}a\)
a)\(\dfrac{\sqrt{6}}{x-y}\)
b)\(2a\sqrt{5}\)
a, \(\dfrac{2}{x^2-y^2}\sqrt{\dfrac{3(x+y)^2}{2}với}x\ge0,y\ge0\)\(vàx\ne y\)
\(=\dfrac{\sqrt{6}}{(x-y)}\)
b, \(=2a\sqrt{5}\)
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a căn 6 trên (x-y)
b 2a căn 5
b)\(\dfrac{2}{2a-1}\sqrt{5a^2\left(1-4a+4a^2\right)}=\dfrac{2\left|a\right|}{2a-1}\sqrt{5\left(1-2.2a+\left(2a\right)^2\right)}=\dfrac{2a}{2a-1}\sqrt{5\left(1-2a\right)^2}=\dfrac{2a\left|1-2a\right|}{2a-1}\sqrt{5}=\dfrac{2a\left(2a-1\right)}{2a-1}\sqrt{5}=2a\sqrt{5}\)
a) \(\dfrac{2}{x^2-y^2}\sqrt{\dfrac{3\left(x+y\right)^2}{2}}=\dfrac{\left|x+y\right|}{x^2-y^2}\sqrt{\dfrac{3.2^2}{2}}=\dfrac{\left(x+y\right)}{\left(x-y\right)\left(x+y\right)}\sqrt{6}=\dfrac{1}{x-y}\sqrt{6}\)
a)\(\dfrac{2}{x^2-y^2}\sqrt{\dfrac{3\left(x+y\right)^2}{2}}=\dfrac{\left|x+y\right|}{x^2-y^2}\sqrt{\dfrac{3.2^2}{2}}=\dfrac{x+y}{\left(x-y\right)\left(x+y\right)}\sqrt{6}=\dfrac{1}{x-y}\sqrt{6}\) b)\(\dfrac{2}{2a-1}\sqrt{5a^2\left(1-4a+4a^2\right)}=\dfrac{2\left|a\right|}{2a-1}\sqrt{5.\left(1-2a\right)^2}=\dfrac{2a\left|1-2a\right|}{2a-1}\sqrt{5}=2a\sqrt{5}\)
a)\(\dfrac{2}{x^2-y^2}\sqrt{\dfrac{3\left(x+y\right)^2}{2}}\)=\(\dfrac{2}{x^2-y^2}\sqrt{\dfrac{3}{2}\left(x+y\right)^2}\) =\(\dfrac{2}{x^2-y^2}\sqrt{\left(x+y\right)^2}.\sqrt{\dfrac{3}{2}}\)=\(\dfrac{2}{x^2-y^2}.\left|x+y\right|\sqrt{\dfrac{3}{2}}\)= \(\dfrac{\left|x+y\right|}{x^2-y^2}.\sqrt{\dfrac{3.2^2}{2}}=\dfrac{x+y}{\left(x-y\right)\left(x+y\right)}.\sqrt{6}\) (do x\(\ge\)0 y\(\ge\)0 \(\Rightarrow x+y\ge\)nên\(\left|x+y\right|=x+y\)) b)\(\dfrac{2}{2a-1}\sqrt{5a^2\left(1-4a+4a^2\right)}\)=\(\dfrac{2}{2a-1}\sqrt{5a^2\left(1-2.1.2a+\left(2a\right)^2\right)}\)
a)\(\dfrac{\sqrt{6}}{x-y}\)
b)\(2a\sqrt{5}\)
a= 2/(x-y) . (x+y) .\(\sqrt{ }\)3(x+y)^2.2/2.2 = 2/(x+y).(x-y) . \(\sqrt{ }\)6(x+y)^2/\(\sqrt{ }\)4 = 2/(x+y). (x-y) . \(\sqrt{ }\)6(x+y)/2 = \(\sqrt{ }\)6/x-y b = 2/2a-1 . \(\sqrt{ }\)5a^2 . (1-2a)^2 = 2/2a-1 . \(\sqrt{ }\)5 . \(\sqrt{ }\)a^2 . \(\sqrt{ }\)(1-2a)^2 = 2/2a-1 . \(\sqrt{ }\)5 . a . (2a-1) = 2\(\sqrt{ }\)5-a
a)\(\dfrac{2}{x^2-y^2}\cdot\sqrt{\dfrac{3\cdot\left(x+y\right)^2}{2}}\) (voi \(x\ge0;x\ne y\) ) =\(\dfrac{2}{x^2-y^2}\cdot\dfrac{\sqrt{3}\cdot\sqrt{\left(x+y\right)^2}}{\sqrt{2}}\) =\(\dfrac{2}{\left(x-y\right)\cdot\left(x+y\right)}\cdot\dfrac{\sqrt{3}\cdot\left|x+y\right|}{\sqrt{2}}\) =\(\dfrac{2\sqrt{3}}{\sqrt{2}\cdot\left(x-y\right)}\) =\(\sqrt{\dfrac{12}{2}}\cdot\dfrac{1}{x-y}\)
=\(\sqrt{6}\cdot\dfrac{1}{x-y}\)
b)\(\dfrac{2}{2a-1}\cdot\sqrt{5a^2\cdot\left(1-4a+4a^2\right)}\) ( Voi a>0,5 )
= \(\dfrac{2}{2a-1}\cdot\sqrt{5a^2\cdot\left(1-2a\right)^2}\)
=\(\dfrac{2}{2a-1}\cdot\sqrt{5a^2}\cdot\left|1-2a\right|\)
=\(\dfrac{2}{2a-1}\cdot\sqrt{5a^2}\cdot\left(2a-1\right)\)
= \(2\sqrt{5a^2}\)
= \(2a\sqrt{5}\)
a) 2x2−y23(x+y)22\dfrac{2}{x^2-y^2}\sqrt{\dfrac{3(x+y)^2}{2}}
=2x2−y232.(x+y)2=\dfrac{2}{x^2-y^2}\sqrt{\dfrac{3}{2}.(x+y)^2}
=2x2−y2(x+y)2.32=\dfrac{2}{x^2-y^2}\sqrt{(x+y)^2}.\sqrt{\dfrac{3}{2}}
=2x2−y2.∣x+y∣32=\dfrac{2}{x^2-y^2}.|x+y|\sqrt{\dfrac{3}{2}}
=∣x+y∣x2−y23.222=\dfrac{|x+y|}{x^{2}-y^{2}}\sqrt{\dfrac{3.2^2}{2}}
=x+y(x−y)(x+y).6=\dfrac{x+y}{(x-y)(x+y)}.\sqrt{6}
(Do x≥ 0,y≥ 0x \ge 0, y \ge 0 ⇒\Rightarrow x+y≥0x+y \ge 0 nên ∣x+y∣=x+y|x+y|=x+y)
=6x−y=\dfrac{\sqrt{6}}{x-y}
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b) 22a−15a2(1−4a+4a2)\dfrac{2}{2 a-1} \sqrt{5a^2(1-4 a+4 a^2)}
=22a−15a2(1−2.1.2a+(2a)2)=\dfrac{2}{2 a-1} \sqrt{5a^2(1-2.1.2a+(2a)^2)}
=22a−15a2(1−2a)2=\dfrac{2}{2 a-1} \sqrt{5a^2(1-2a)^2}
=22a−15.a2.(1−2a)2=\dfrac{2}{2 a-1} \sqrt{5}.\sqrt{a^2}.\sqrt{(1-2a)^2}
=22a−15.∣a∣.∣1−2a∣=\dfrac{2}{2 a-1} \sqrt{5}.|a|.|1-2a|
=22a−15.a.(2a−1)=\dfrac{2}{2 a-1} \sqrt{5}.a.(2a-1)
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a) \(\dfrac{2}{x^2-y^2}\)\(\sqrt{\dfrac{3\left(x+y\right)^2}{2}}\)
= \(\dfrac{2}{\left(x-y\right).\left(x+y\right)}\)\(\sqrt{\dfrac{3}{2}.\left(x+y\right)^2}\)
= \(\dfrac{2}{\left(x+y\right).\left(x-y\right)}\).|x+y| \(\sqrt{\dfrac{3}{2}}\)
= \(\dfrac{x+y}{\left(x-y\right).\left(x+y\right)}\) .\(\sqrt{\dfrac{3.2^2}{2}}\)
= \(\dfrac{\sqrt{6}}{x-y}\)
b) \(\dfrac{2}{2a-1}\) \(\sqrt{5a^2\left(1-4a+4a^2\right)}\)
= \(\dfrac{2}{2a-1}\)\(\sqrt{5a^2\left(1-2a\right)^2}\)
=\(\dfrac{2}{2a-1}\)\(\sqrt{5}\).\(\sqrt{a^2}\).\(\sqrt{\left(1-2a\right)^2}\)
= \(\dfrac{2}{2a-1}\)\(\sqrt{5}\). |a| . |1-2a|
= \(\dfrac{2}{2a-1}\)\(\sqrt{5}\) . |a| .|1-2a|
= \(\dfrac{2}{2a-1}\)\(\sqrt{5}\).a.( 2a-1) ( vì a > 0,5 )
= 2a\(\sqrt{5}\)
=
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