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\(a.A=\dfrac{2}{x^2-y^2}.\sqrt{\dfrac{3x^2+6xy+3y^2}{4}}=\dfrac{2}{\left(x-y\right)\left(x+y\right)}.\dfrac{\left(x+y\right)\sqrt{3}}{2}=\dfrac{\sqrt{3}}{x-y}\) ( x # y )
\(b.\dfrac{1}{2x-1}.\sqrt{5a^4\left(1-4x+4a^2\right)}=\dfrac{1}{2a-1}.\left(2a-1\right)a^2\sqrt{5}=a^2\sqrt{5}\) ( a # \(\dfrac{1}{2}\) )
Bài 1:
a: \(A=\left|2a-1\right|-2a\)
TH1: a>=1/2
A=2a-1-2a=-1
TH2: a<1/2
A=1-2a-2a=1-4a
b: \(B=x-2y-\left|x-2y\right|\)
TH1: x>=2y
A=x-2y-x+2y=0
TH2: x<2y
A=x-2y+x-2y=2x-4y
c: \(=x^2+\left|x^2-4\right|\)
TH1: x>=2 hoặc x<=-2
\(A=x^2+x^2-4=2x^2-4\)
TH2: -2<x<2
\(A=x^2+4-x^2=4\)
d: \(D=2x-1-\dfrac{\left|x-5\right|}{x-5}\)
TH1: x>5
\(D=2x-1-1=2x-2\)
TH2: x<5
D=2x-1+1=2x
a)\(\left(\sqrt{12}+\sqrt{75}+\sqrt{27}\right)\div\sqrt{15}=\left(2\sqrt{3}+5\sqrt{3}+3\sqrt{3}\right)\div\sqrt{3}\sqrt{5}=10\sqrt{3}\div\sqrt{3}\sqrt{5}=\sqrt{2}\sqrt{5}\div\sqrt{5}=\sqrt{2}\)b)\(\sqrt{252}-\sqrt{700}+\sqrt{1008}-\sqrt{448}=\sqrt{4}\sqrt{9}\sqrt{7}-\sqrt{100}\sqrt{7}+\sqrt{16}\sqrt{9}\sqrt{7}-\sqrt{64}\sqrt{7}=2\cdot3\cdot\sqrt{7}-10\cdot\sqrt{7}+4\cdot3\cdot\sqrt{7}-8\sqrt{7}=6\sqrt{7}-10\sqrt{7}+12\sqrt{7}-8\sqrt{7}=0\)
c)\(\sqrt{27^2-23^2}+\sqrt{37^2-35^2}=\sqrt{\left(27-23\right)\left(27+23\right)}+\sqrt{\left(37-35\right)\left(37+35\right)}=\sqrt{4\cdot50}\cdot\sqrt{2\cdot72}=\sqrt{4\cdot50\cdot2\cdot72}=\sqrt{2^2\cdot2\cdot25\cdot2\cdot36\cdot2}=\sqrt{16}\cdot\sqrt{25}\cdot\sqrt{36}=4\cdot5\cdot6=120\)
d)\(\left(\sqrt{\dfrac{1}{7}}+\sqrt{\dfrac{16}{7}}+\sqrt{\dfrac{9}{7}}\right)\div\sqrt{7}=\left(\dfrac{1}{\sqrt{7}}+\dfrac{4}{\sqrt{7}}+\dfrac{3}{\sqrt{7}}\right)\cdot\dfrac{1}{\sqrt{7}}=\dfrac{7}{\sqrt{7}}\cdot\dfrac{1}{\sqrt{7}}=1\)
\(A=\dfrac{2}{x^2-y^2}\cdot\sqrt{\dfrac{3x^2+6xy+3y^2}{4}}=\dfrac{2}{x^2-y^2}\cdot\sqrt{\dfrac{3\left(x^2++2xy+y^2\right)}{4}}=\dfrac{2}{x^2-y^2}\cdot\sqrt{\dfrac{3\left(x-y\right)^2}{4}}=\dfrac{2}{\left(x-y\right)\left(x+y\right)}\cdot\dfrac{\sqrt{3}\left(x-y\right)}{2}=\dfrac{\sqrt{3}}{x+y}\)
\(B=\dfrac{1}{2a-1}\cdot\sqrt{5a^4\left(1-4a+4a^2\right)}=\dfrac{1}{2a-1}\cdot\sqrt{5a^4\left(2a-1\right)^2}=\dfrac{1}{2a-1}\cdot\sqrt{5}a^2\left(2a-1\right)=\sqrt{5}\cdot a^2\)
Giải:
\(\dfrac{1}{2a-1}.\sqrt{5a^4.\left(1-4a+4a^2\right)}\)
\(=\dfrac{1}{2a-1}.\sqrt{5a^4}.\sqrt{1-4a+4a^2}\)
\(=\dfrac{1}{2a-1}.a^2\sqrt{5}.\sqrt{\left(1-2a\right)^2}\)
\(=\dfrac{1}{2a-1}.a^2\sqrt{5}.\left|1-2a\right|\)
\(=\dfrac{\left|2a-1\right|.a^2\sqrt{5}}{2a-1}\left(1\right)\)
Chắc đề thiếu điều kiện, mình cho thêm để ra kết quả đẹp
ĐK: \(a\ge1\Leftrightarrow2a\ge2\Leftrightarrow2a-1\ge1>0\)
\(\left(1\right)=\dfrac{\left(2a-1\right).a^2\sqrt{5}}{2a-1}\)
\(=a^2\sqrt{5}\)
Vậy ...
\(\frac{\sqrt{3x^2+6xy+3y^2}}{x^2-y^2}\)
<=>\(\frac{\sqrt{3.\left(x+y\right)^2}}{\left(x-y\right).\left(x+y\right)}\)
<=>\(\frac{\sqrt{3}\left|x+y\right|}{\left(x-y\right).\left(x+y\right)}.\)
<=>\(\frac{\sqrt{3}}{x-y}\)
\(\text{a) }\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\\ =\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\\ =\dfrac{x\sqrt{x}+y\sqrt{y}-\left(\sqrt{x}-\sqrt{y}\right)\left(x-y\right)}{\sqrt{x}+\sqrt{y}}\\ =\dfrac{x\sqrt{x}+y\sqrt{y}-x\sqrt{x}+x\sqrt{y}+y\sqrt{x}-y\sqrt{y}}{\sqrt{x}+\sqrt{y}}\\ =\dfrac{x\sqrt{y}+y\sqrt{x}}{\sqrt{x}+\sqrt{y}}\\ =\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\\ =\sqrt{xy}\)
\(\text{b) }\sqrt{\dfrac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}=\sqrt{\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)^2}}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
\(\text{c) }\dfrac{x-1}{\sqrt{y}-1}\cdot\sqrt{\dfrac{\left(y-2\sqrt{y}+1\right)^2}{\left(x-1\right)^4}}\\ =\dfrac{x-1}{\sqrt{y}-1}\cdot\sqrt{\dfrac{\left(\sqrt{y}-1\right)^4}{\left(x-1\right)^4}}\\ =\dfrac{x-1}{\sqrt{y}-1}\cdot\dfrac{\left(\sqrt{y}-1\right)^2}{\left(x-1\right)^2}\\ =\dfrac{\sqrt{y}-1}{x-1}\)
a)\(\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\)
\(=\dfrac{\sqrt{x^3}+\sqrt{y^3}}{\sqrt{x}+\sqrt{y}}-\left(x-2\sqrt{x}\sqrt{y}+y\right)\)
\(=\dfrac{\left(\sqrt{x}+\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{\sqrt{x}+\sqrt{y}}-x+2\sqrt{xy}+y\)
\(=x+\sqrt{xy}+y-x+2\sqrt{xy}+y\)
\(=3\sqrt{xy}+2y\)
a) \(\dfrac{2}{x^2-y^2}\cdot\sqrt{\dfrac{3\left(x+y\right)^2}{2}}=\dfrac{2\cdot\left(x+y\right)\cdot\sqrt{3}}{\left(x+y\right)\cdot\left(x-y\right)\cdot\sqrt{2}}=\dfrac{2\sqrt{3}}{\left(x-y\right)\cdot\sqrt{2}}=\dfrac{2\sqrt{6}}{2\left(x-y\right)}=\dfrac{\sqrt{6}}{x-y}\)
b) \(\dfrac{2}{2a-1}\sqrt{5a^2\left(1-4a+4a^2\right)}=\dfrac{2}{2a-1}\cdot\sqrt{5a^2\left[\left(2a\right)^2-2\cdot2\cdot a+1^2\right]}=\dfrac{2}{2a-1}\cdot\sqrt{5a^2\left(2a-1\right)^2}=\dfrac{2}{2a-1}\cdot a\cdot\left(2a-1\right)\cdot\sqrt{5}=\dfrac{2a\left(2a-1\right)\sqrt{5}}{2a-1}=2a\sqrt{5}\)