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\(A=\left|-3\right|+2+\left|5x\right|\)
\(A=3+2+\left(-5x\right)=-5x+5\)
a: \(A=\dfrac{x^5}{x^3}\cdot\dfrac{y^{-2}}{y}=x^2\cdot y^{-1}=\dfrac{x^2}{y}\)
b: \(B=\dfrac{x^2\cdot y^{-3}}{x^3\cdot y^{-12}}=\dfrac{x^2}{x^3}\cdot\dfrac{y^{-3}}{y^{-12}}=\dfrac{1}{x}\cdot y^{-3+12}=\dfrac{y^9}{x}\)
a) \(A=\dfrac{x^5y^{-2}}{x^3y}=\dfrac{x^5}{x^3}.\dfrac{1}{y^{2-1}}=x^{5-3}y^{-1}=x^2y^{-1}\).
b) \(B=\dfrac{x^2y^{-3}}{\left(x^{-1}y^4\right)^{-3}}=\dfrac{x^2y^{-3}}{x^3y^{-12}}=x^{2-3}y^{-3-\left(-12\right)}=\dfrac{1}{xy^9}\)
a: =căn 3+căn 5-căn 3=căn5
b: \(=\sqrt{x-2-2\sqrt{x-2}+1}=\sqrt{\left(\sqrt{x-2}-1\right)^2}\)
\(=\left|\sqrt{x-2}-1\right|\)
a) \(A=\dfrac{x\sqrt{y}+y\sqrt{x}}{x+2\sqrt{xy}+y}\)
\(A=\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}+\sqrt{y}\right)^2}\)
\(A=\dfrac{\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)
b) \(B=\dfrac{x\sqrt{y}-y\sqrt{x}}{x-2\sqrt{xy}+y}\)
\(B=\dfrac{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)^2}\)
\(B=\dfrac{\sqrt{xy}}{\sqrt{x}-\sqrt{y}}\)
c) \(C=\dfrac{3\sqrt{a}-2a-1}{4a-4\sqrt{a}+1}\)
\(C=\dfrac{-\left(2a-3\sqrt{a}+1\right)}{\left(2\sqrt{a}\right)^2-2\sqrt{a}\cdot2\cdot1+1^2}\)
\(C=\dfrac{-\left(\sqrt{a}-1\right)\left(2\sqrt{a}-1\right)}{\left(2\sqrt{a}-1\right)^2}\)
\(C=\dfrac{-\sqrt{a}+1}{2\sqrt{a}-1}\)
d) \(D=\dfrac{a+4\sqrt{a}+4}{\sqrt{a}+2}+\dfrac{4-a}{\sqrt{a}-2}\)
\(D=\dfrac{\left(\sqrt{a}+2\right)^2}{\sqrt{a}+2}+\dfrac{\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)}{\sqrt{a}-2}\)
\(D=\sqrt{a}+2-\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{\sqrt{a}-2}\)
\(D=\left(\sqrt{a}+2\right)-\left(\sqrt{a}+2\right)\)
\(D=0\)
\(a,A=\dfrac{1}{2-\sqrt{3}}+\dfrac{1}{2+\sqrt{3}}\)
\(=\dfrac{2+\sqrt{3}+2-\sqrt{3}}{2^2-\sqrt{3}^2}\)
\(=\dfrac{4}{1}=4\)
Vậy \(A=4\)
\(b,B=\left(\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{1}{\sqrt{x}-1}\right).\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}\)
\(=\left(\dfrac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\right).\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
Vậy \(B=\dfrac{\sqrt{x}-1}{\sqrt{x}}\) với \(x>0,x\ne1\)
a: \(=2+\sqrt{3}+2-\sqrt{3}=4\)
b: \(=\dfrac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
\(=\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)=3-1=2\)
b: \(=\dfrac{\sqrt{x}-\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{2}{\sqrt{x}+1}=\dfrac{-4}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}\)
a, \(=\left(\dfrac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}+1\right)\left(\sqrt{3}-1\right)=\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)=2\)
b, với x > 0
\(=\left(\dfrac{\sqrt{x}-\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}+2\right)}\right)\left(\dfrac{2}{\sqrt{x+1}}\right)\)
\(=-\dfrac{-4}{\sqrt{x}\left(\sqrt{x}+2\right)\sqrt{x+1}}=\dfrac{4}{\left(\sqrt{x}+2\right)\sqrt{x^2+x}}\)
`a)(2x-1)^2+(x+3)^2-5(x-7)(x+7)`
`=4x^2-4x+1+x^2+6x+9-5(x^2-49)`
`=5x^2-5x^2-4x+6x+1+9+245`
`=2x+255`
`b)(x-2)(x^2+2x+4)-(25+x^3)`
`=x^3-8-x^3-25=-33`
Lời giải:
a.
$(2x-1)^2+(x+3)^2-5(x-7)(x+7)$
$=4x^2-4x+1+(x^2+6x+9)-5(x^2-49)$
$=5x^2+2x+10-(5x^2-245)=2x+255$
b.
$(x-2)(x^2+2x+4)-(25+x^3)=(x^3-2^3)-(25+x^3)$
$=-8-25=-33$
a) \(\sqrt{4a^2}=2\left|a\right|=-2a\) ( do a<0)
b) \(\sqrt{4x^2-12x+9}=\sqrt{\left(2x-3\right)^2}=\left|2x-3\right|=3-2x\)(do \(x< \dfrac{3}{2}\Leftrightarrow2x-3< 0\))
