Rút gọn biểu thức: \(\dfrac{2}{x^2-y^2}\sqrt{\dfrac{9\left(x^2+2xy+y^2\right)}{4}}\)(x ≠ +- y).
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QN
1
10 tháng 9 2018
Có \(\frac{2}{x^2-y^2}\sqrt{\frac{9\left(x+2xy+y\right)}{4}}\)
=\(\frac{2}{\left(x+y\right)\left(x-y\right)}\sqrt{\frac{3^2.\left(x+y\right)^2}{2^2}}\)
=\(\frac{2}{\left(x+y\right)\left(x-y\right)}\frac{\sqrt{3^2}.\sqrt{\left(x+y\right)^2}}{\sqrt{2^2}}\)
=\(\frac{2}{\left(x+y\right)\left(x-y\right)}.\frac{3.\left(x+y\right)}{2}\)
=\(\frac{2.3.\left(x+y\right)}{\left(x+y\right)\left(x-y\right).2}\) =\(\frac{3}{x-y}\)
20 tháng 11 2023
1: \(C=\left(x-\dfrac{4xy}{x+y}+y\right):\left(\dfrac{x}{x+y}+\dfrac{y}{y-x}+\dfrac{2xy}{x^2-y^2}\right)\)
\(=\dfrac{\left(x+y\right)^2-4xy}{x+y}:\left(\dfrac{x}{x+y}-\dfrac{y}{x-y}+\dfrac{2xy}{\left(x-y\right)\left(x+y\right)}\right)\)
\(=\dfrac{x^2+2xy+y^2-4xy}{x+y}:\dfrac{x\left(x-y\right)-y\left(x+y\right)+2xy}{\left(x+y\right)\left(x-y\right)}\)
\(=\dfrac{x^2-2xy+y^2}{x+y}:\dfrac{x^2-xy-xy-y^2+2xy}{\left(x+y\right)\left(x-y\right)}\)
\(=\dfrac{\left(x-y\right)^2}{x+y}\cdot\dfrac{x^2-y^2}{x^2-y^2}=\dfrac{\left(x-y\right)^2}{x+y}\)
2: \(\left(x^2-y^2\right)\cdot C=-8\)
=>\(\left(x-y\right)\left(x+y\right)\cdot\dfrac{\left(x-y\right)^2}{x+y}=-8\)
=>\(\left(x-y\right)^3=-8\)
=>x-y=-2
=>x=y-2
\(M=x^2\left(x+1\right)-y^2\left(y-1\right)-3xy\left(x-y+1\right)+xy\)
\(=\left(y-2\right)^2\left(y-2+1\right)-y^2\left(y-1\right)-3xy\left(-2+1\right)+xy\)
\(=\left(y-1\right)\left[\left(y-2\right)^2-y^2\right]+3xy+xy\)
\(=\left(y-1\right)\left(-4y+4\right)+4xy\)
\(=-4\left(y-1\right)^2+4y\left(y-2\right)\)
\(=-4y^2+8y-4+4y^2-8y\)
=-4
H9
HT.Phong (9A5)
CTVHS
15 tháng 7 2023
\(P=\dfrac{4\sqrt{xy}}{x-y}:\left(\dfrac{1}{y-x}+\dfrac{1}{x+2\sqrt{x}\sqrt{y}+y}\right)-2x\) (với \(x\ne y,x,y\ge0\))
\(P=\dfrac{4\sqrt{xy}}{x-y}:\left(\dfrac{1}{\left(\sqrt{y}-\sqrt{x}\right)\left(\sqrt{y}+\sqrt{x}\right)}+\dfrac{1}{\left(\sqrt{x}+\sqrt{y}\right)^2}\right)-2x\)
\(P=\dfrac{4\sqrt{xy}}{x-y}:\left(\dfrac{\sqrt{y}+\sqrt{x}}{\left(\sqrt{y}+\sqrt{x}\right)^2\left(\sqrt{y}-\sqrt{x}\right)}+\dfrac{\sqrt{y}-\sqrt{x}}{\left(\sqrt{x}+\sqrt{y}\right)^2\left(\sqrt{y}-\sqrt{x}\right)}\right)-2x\)
\(P=\dfrac{4\sqrt{xy}}{x-y}:\left(\dfrac{\sqrt{y}+\sqrt{x}+\sqrt{y}-\sqrt{x}}{\left(\sqrt{y}-\sqrt{x}\right)\left(\sqrt{x}+\sqrt{y}\right)^2}\right)-2x\)
\(P=\dfrac{4\sqrt{xy}}{x-y}:\left(\dfrac{2\sqrt{y}}{\left(y-x\right)\left(\sqrt{x}+\sqrt{y}\right)}\right)-2x\)
\(P=\dfrac{4\sqrt{xy}}{x-y}\cdot\dfrac{\left(y-x\right)\left(\sqrt{x}+\sqrt{y}\right)}{2\sqrt{y}}-2x\)
\(P=\dfrac{4\sqrt{xy}\cdot\left(y-x\right)\left(\sqrt{x}+\sqrt{y}\right)}{\left(x-y\right)\cdot2\sqrt{y}}-2x\)
\(P=\dfrac{4\sqrt{xy}\cdot\left(y-x\right)\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)\cdot2\sqrt{y}}-2x\)
\(P=\dfrac{2\sqrt{x}\left(y-x\right)}{\sqrt{x}-\sqrt{y}}-2x\)
\(P=\dfrac{2\sqrt{x}\left(y-x\right)-2x\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}\)
\(P=\dfrac{2y\sqrt{x}-2x\sqrt{x}-2x\sqrt{x}+2x\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)
\(P=\dfrac{2y\sqrt{x}-4x\sqrt{x}+2x\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)
8 tháng 7 2021
a) \(\sqrt{\dfrac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}\sqrt{\dfrac{\left(\sqrt{x+1}\right)^2}{\left(\sqrt{x}+1\right)^2}}\)
=\(\dfrac{\sqrt{x}-1}{\sqrt{x}+1};x\ge0\)
8 tháng 7 2021
b) Ta có: \(\dfrac{x-1}{\sqrt{y}-1}\cdot\sqrt{\dfrac{\left(y-2\sqrt{y}+1\right)^2}{\left(x-1\right)^4}}\)
\(=\dfrac{x-1}{\sqrt{y}-1}\cdot\dfrac{\sqrt{y}-1}{\left(x-1\right)^2}\)
\(=\dfrac{1}{x-1}\)
H9
HT.Phong (9A5)
CTVHS
6 tháng 12 2023
a) \(B=\left(\dfrac{x-y}{\sqrt{x}-\sqrt{y}}+\dfrac{x\sqrt{x}-y\sqrt{y}}{y-x}\right):\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\left(x,y\ge0;x\ne y\right)\)
\(B=\left[\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}-\dfrac{\left(\sqrt{x}\right)^3-\left(\sqrt{y}\right)^3}{x-y}\right]:\dfrac{x-2\sqrt{xy}+y+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)
\(B=\left[\left(\sqrt{x}+\sqrt{y}\right)-\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\right]:\dfrac{x+\sqrt{xy}+y}{\sqrt{x}+\sqrt{y}}\)
\(B=\left[\left(\sqrt{x}+\sqrt{y}\right)-\dfrac{x+\sqrt{xy}+y}{\sqrt{x}+\sqrt{y}}\right]:\dfrac{x+\sqrt{xy}+y}{\sqrt{x}+\sqrt{y}}\)
\(B=\dfrac{\left(\sqrt{x}+\sqrt{y}\right)^2-x-\sqrt{xy}-y}{\sqrt{x}+\sqrt{y}}\cdot\dfrac{\sqrt{x}+\sqrt{y}}{x+\sqrt{xy}+y}\)
\(B=\dfrac{x+2\sqrt{xy}+y-x-\sqrt{xy}-y}{\sqrt{x}+\sqrt{y}}\cdot\dfrac{\sqrt{x}+\sqrt{y}}{x+\sqrt{xy}+y}\)
\(B=\dfrac{\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\cdot\dfrac{\sqrt{x}+\sqrt{y}}{x+\sqrt{xy}+y}\)
\(B=\dfrac{\sqrt{xy}}{x+\sqrt{xy}+y}\)
b) Xét tử:
\(\sqrt{xy}\ge0\forall x,y\) (xác định) (1)
Xét mẫu:
\(x+\sqrt{xy}+y\)
\(=\left(\sqrt{x}\right)^2+2\cdot\dfrac{1}{2}\sqrt{y}\cdot\sqrt{x}+\left(\dfrac{1}{2}\sqrt{y}\right)^2+\dfrac{3}{4}y\)
\(=\left(\sqrt{x}+\dfrac{1}{2}\sqrt{y}\right)^2+\dfrac{3}{4}y\)
Mà: \(\left(\sqrt{x}+\dfrac{1}{2}\sqrt{y}\right)^2\ge0\forall x,y\) (xác định), còn: \(\dfrac{3}{4}y\ge0\) vì theo đkxđ thì \(y\ge0\) (2)
Từ (1) và (2) ⇒ B luôn không âm với mọi x,y (\(B\ge0\)) (đpcm)
27 tháng 5 2017
Nếu có thêm điều kiện \(y>1\) thì kết quả là \(\dfrac{1}{x-1}\)
12 tháng 11 2021
5: \(=\dfrac{1}{x-y}\cdot x^3\cdot\left(x-y\right)^2=x^3\left(x-y\right)\)
Ta có \(\dfrac{2}{x^2-y^2}\sqrt{\dfrac{9\left(x^2+2xy+y^2\right)}{4}}=\dfrac{2}{\left(x-y\right)\left(x+y\right)}\dfrac{\sqrt{9\left(x+y\right)^2}}{\sqrt{4}}=\dfrac{2}{\left(x-y\right)\left(x+y\right)}.\dfrac{3\left|x+y\right|}{2}=\dfrac{3\left|x+y\right|}{\left(x-y\right)\left(x+y\right)}\)Nếu x+y>0 thì
\(\dfrac{3\left|x+y\right|}{\left(x-y\right)\left(x+y\right)}=\dfrac{3}{x-y}\)
Nếu x+y<0 thì
\(\dfrac{3\left|x+y\right|}{\left(x-y\right)\left(x+y\right)}=\dfrac{-3}{x-y}\)