Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có: \(x^4+y^4+\frac{x^4y^4}{\left(x^2+y^2\right)^2}\)
\(=\left(x^4+2x^2y^2+y^4\right)-2x^2y^2+\frac{x^4y^4}{\left(x^2+y^2\right)}\)
\(=\left(x^2+y^2\right)^2-2x^2y^2+\left(\frac{x^2y^2}{x^2+y^2}\right)^2\)
\(=\left(x^2+y^2-\frac{x^2y^2}{x^2+y^2}\right)^2\)
Thay vào ta tính được:
\(P=\sqrt{\frac{x^2y^2}{x^2+y^2}+\frac{x^2y^2}{\left(x+y\right)^2}+\sqrt{\left(x^2+y^2-\frac{x^2y^2}{x^2+y^2}\right)^2}}\)
Mà \(x^2+y^2-\frac{x^2y^2}{x^2+y^2}=\frac{\left(x^2+y^2\right)^2-x^2y^2}{x^2+y^2}=\frac{x^4+x^2y^2+y^4}{x^2+y^2}>0\left(\forall x,y\right)\)
Khi đó:
\(P=\sqrt{\frac{x^2y^2}{x^2+y^2}+\frac{x^2y^2}{\left(x+y\right)^2}+x^2+y^2-\frac{x^2y^2}{x^2+y^2}}\)
\(P=\sqrt{x^2+y^2+\frac{x^2y^2}{\left(x+y\right)^2}}\)
\(P=\sqrt{\left(x^2+2xy+y^2\right)-2xy+\frac{x^2y^2}{\left(x+y\right)^2}}\)
\(P=\sqrt{\left(x+y\right)^2-2xy+\left(\frac{xy}{x+y}\right)^2}\)
\(P=\sqrt{\left(x+y-\frac{xy}{x+y}\right)^2}\)
\(P=\left|x+y-\frac{xy}{x+y}\right|=\left|\frac{x^2+xy+y^2}{x+y}\right|=\frac{x^2+xy+y^2}{x+y}\)
Vậy \(P=\frac{x^2+xy+y^2}{x+y}\)
\(a,\frac{\sqrt{108x^3}}{\sqrt{12x}}=\frac{\sqrt{36.3.x^3}}{\sqrt{3.4.x}}=\frac{6\sqrt{3}.\sqrt{x}^3}{2\sqrt{3}.\sqrt{x}}=3\sqrt{x}^2=3x\)
\(b,\frac{\sqrt{13x^4y^6}}{\sqrt{208x^6y^6}}=\frac{\sqrt{13}.\sqrt{x^4}.\sqrt{y^6}}{\sqrt{16.13}.\sqrt{x^6}.\sqrt{y^6}}=\frac{\sqrt{13}.x^2y^3}{4\sqrt{13}x^3y^3}=\frac{1}{4x}\)
\(c,\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}+\sqrt{y}\right)^2\)
\(=\frac{\sqrt{x}^3+\sqrt{y}^3}{\sqrt{x}+\sqrt{y}}-\left(x+2\sqrt{xy}+y\right)\)
\(=\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)}{\sqrt{x}+\sqrt{y}}-x-2\sqrt{xy}-y\)
\(=x-\sqrt{xy}+y-x-2\sqrt{xy}-y=-3\sqrt{xy}\)
\(d,\sqrt{\frac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}=\frac{\sqrt{\left(\sqrt{x}-1\right)^2}}{\sqrt{\left(\sqrt{x}+1\right)^2}}=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)
Đk chỗ này là \(\sqrt{x}-1\ge0\Rightarrow\sqrt{x}\ge\sqrt{1}\Rightarrow x\ge1\)nhé
\(e,\frac{x-1}{\sqrt{y}-1}.\sqrt{\frac{\left(y-2\sqrt{y}+1\right)^2}{\left(x-1\right)^4}}=\frac{x-1}{\sqrt{y}-1}.\frac{y-2\sqrt{y}+1}{\left(x-1\right)^2}\)
\(=\frac{\left(x-1\right)\left(\sqrt{y}-1\right)^2}{\left(\sqrt{y}-1\right)\left(x-1\right)^2}=\frac{\sqrt{y}-1}{x-1}\)
1, với x > 0 ; x khác 1 ; 4
a, \(P=\left(\dfrac{x+\sqrt{x}-x-2}{\sqrt{x}+1}\right):\left(\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)+\sqrt{x}-4}{x-1}\right)\)
\(=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}:\dfrac{x-4}{x-1}=\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\)
b, Ta có P > 0 => \(\sqrt{x}-1>0\Leftrightarrow x>1\)
Kết hợp đk vậy x > 1 ; x khác 4
a. Ta có:\(\frac{x}{y}\sqrt{\frac{y^2}{x^4}=}\) \(\frac{x}{y}.\frac{\left|y\right|}{x^2}=\frac{x.y}{x^2y}\)\(=\frac{1}{x}\)(Vì \(x\ne0;y>0\))
b \(3x^2\sqrt{\frac{8}{x^2}}=3x^2\frac{2\sqrt{2}}{\left|x\right|}=\frac{6x^2\sqrt{2}}{-x}=-6x\sqrt{2}\)( Vì \(x< 0\))
a) \(\frac{\left(\sqrt{x}\right)^3-\left(\sqrt{y}\right)^3}{\left(\sqrt{x}-\sqrt{y}\right)}-\left(\sqrt{x}-\sqrt{y}\right)^2=\frac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{\sqrt{x}-\sqrt{y}}-x+2\sqrt{xy}-y\)
\(=3\sqrt{xy}\)
b) \(\frac{x-y}{\sqrt{y}-1}.\sqrt{\frac{\left(\sqrt{y}-1\right)^4}{\left(x-1\right)^4}}=\frac{x-y}{\sqrt{y}-1}.\frac{\left(\sqrt{y}-1\right)^2}{\left(x-1\right)^2}=\frac{\left(x-y\right)\left(\sqrt{y}-1\right)}{\left(x-1\right)^2}\)
a) \(=\frac{\left(\sqrt{x}\right)^3-\left(\sqrt{y}\right)^3}{\sqrt{x}-\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2=x+\sqrt{xy}+y-x+2\sqrt{xy}-y=3\sqrt{xy}\)
\(A=\sqrt{\frac{x^2y^2}{x^2+y^2}+\frac{x^2y^2}{\left(x+y\right)^2}+\sqrt{\frac{\left(x^4+x^2y^2\right)^2+\left(y^4+x^2y^2\right)^2+x^4y^4}{\left(x^2+y^2\right)^2}}}\)
\(=\sqrt{\frac{x^2y^2}{x^2+y^2}+\frac{x^2y^2}{\left(x+y\right)^2}+\sqrt{\frac{\left(x^4+x^2y^2\right)^2+2x^4y^4+2x^2y^6+y^8}{\left(x^2+y^2\right)^2}}}\)
\(=\sqrt{\frac{x^2y^2}{x^2+y^2}+\frac{x^2y^2}{\left(x+y\right)^2}+\sqrt{\frac{\left(x^4+x^2y^2\right)^2+2\left(x^4+x^2y^2\right)y^4+y^8}{\left(x^2+y^2\right)^2}}}\)
\(=\sqrt{\frac{x^2y^2}{x^2+y^2}+\frac{x^2y^2}{\left(x+y\right)^2}+\sqrt{\frac{\left(x^4+x^2y^2+y^4\right)^2}{\left(x^2+y^2\right)^2}}}\)
\(=\sqrt{\frac{x^2y^2}{x^2+y^2}+\frac{x^2y^2}{\left(x+y\right)^2}+\frac{x^4+x^2y^2+y^4}{x^2+y^2}}\)
\(=\sqrt{\frac{x^2y^2}{\left(x+y\right)^2}+\frac{x^4+2x^2y^2+y^4}{x^2+y^2}}=\sqrt{\frac{x^2y^2}{\left(x+y\right)^2}+\frac{\left(x^2+y^2\right)^2}{x^2+y^2}}\)
\(=\sqrt{\frac{x^2y^2}{\left(x+y\right)^2}+x^2+y^2}=\sqrt{\frac{\left(x^2+xy\right)^2+\left(y^2+xy\right)^2+x^2y^2}{\left(x+y\right)^2}}\)
\(=\sqrt{\frac{\left(x^2+xy\right)^2+2x^2y^2+2xy^3+y^4}{\left(x+y\right)^2}}=\sqrt{\frac{\left(x^2+xy\right)^2+2\left(x^2+xy\right)y^2+y^4}{\left(x+y\right)^2}}\)
\(=\sqrt{\frac{\left(x^2+xy+y^2\right)^2}{\left(x+y\right)^2}}=\frac{x^2+xy+y^2}{x+y}\)
Mik cảm ơn bạn nhìu nhé