cho biểu thức A=\(\left(\dfrac{4\sqrt{y}}{2+\sqrt{y}}+\dfrac{8y}{4-y}\right):\left(\dfrac{\sqrt{y}-1}{y-2\sqrt{y}}-\dfrac{2}{\sqrt{y}}\right)\)
( Với điều kiện y>0, y khác 4, y khác 9)
a, Rút gọn A
b, tìm y để A=-2
A=\(\left(\dfrac{4\sqrt{y}}{2+\sqrt{y}}+\dfrac{8y}{4-y}\right):\left(\dfrac{\sqrt{y}-1}{y-2\sqrt{y}}-\dfrac{2}{\sqrt{y}}\right)\)
1. Rút gọn biểu thức A
2. Tìm y để A= - 2
\(A=\left(\dfrac{4\sqrt{y}}{2+\sqrt{y}}+\dfrac{8y}{4-y}\right):\left(\dfrac{\sqrt{y}-1}{y-2\sqrt{y}}-\dfrac{2}{\sqrt{y}}\right)\\ =\left(\dfrac{4\sqrt{y}.\left(2-\sqrt{y}\right)+8y}{\left(2+\sqrt{y}\right)\left(2-\sqrt{y}\right)}\right):\left(\dfrac{\sqrt{y}-1-2\left(\sqrt{y}-2\right)}{\sqrt{y}\left(\sqrt{y}-2\right)}\right)\\ =\left(\dfrac{4\sqrt{y}\left(2+\sqrt{y}\right)}{\left(2+\sqrt{y}\right)\left(2-\sqrt{y}\right)}\right):\left(\dfrac{3-\sqrt{y}}{\sqrt{y}\left(\sqrt{y}-2\right)}\right)\\ =.\dfrac{4\sqrt{y}.\left(-\sqrt{y}\right)\left(2-\sqrt{y}\right)}{\left(2-\sqrt{y}\right)\left(3-\sqrt{y}\right)}\\ =\dfrac{-4y}{3-\sqrt{y}}\)
Ta có:
\(A=\dfrac{-4y}{3-\sqrt{y}}=-2\Rightarrow-4y=-6+2\sqrt{y}\Rightarrow-4y+4\sqrt{y}-6\sqrt{y}+6=0\\ \Rightarrow-4\sqrt{y}\left(\sqrt{y}-1\right)-6\left(\sqrt{y}-1\right)=0\\ \Rightarrow\left(\sqrt{y}-1\right)\left(-4\sqrt{y}-6\right)=0\Rightarrow\sqrt{y}-1=0\Rightarrow y=1\)
Rút gọn biểu thức A
A=(\(\dfrac{4\sqrt{y}}{2+\sqrt{y}}+\dfrac{8y}{4-y}\)) : (\(\dfrac{\sqrt{y}-1}{y-2\sqrt{y}}-\dfrac{2}{\sqrt{y}}\))
với y>0; y≠4; y≠9
\(A=\left(\dfrac{4\sqrt{y}}{2+\sqrt{y}}+\dfrac{8y}{4-y}\right):\left(\dfrac{\sqrt{y}-1}{y-2\sqrt{y}}-\dfrac{2}{\sqrt{y}}\right)\)
\(A=\dfrac{4\sqrt{y}\left(2-\sqrt{y}\right)+8y}{\left(2+\sqrt{y}\right)\left(2-\sqrt{y}\right)}:\dfrac{\sqrt{y}-1-2\left(\sqrt{y}-2\right)}{\sqrt{y}\left(\sqrt{y}-2\right)}\)
\(A=\dfrac{8\sqrt{y}+4y}{\left(2+\sqrt{y}\right)\left(2-\sqrt{y}\right)}.\dfrac{\sqrt{y}\left(\sqrt{y}-2\right)}{-\sqrt{y}+3}\)
\(A=\dfrac{4\sqrt{y}}{2-\sqrt{y}}.\dfrac{\sqrt{y}\left(2-\sqrt{y}\right)}{\sqrt{y}-3}\)
\(A=\dfrac{4y}{\sqrt{y}-3}\)
Chúc bạn học tốt ^^
Rút gọn biểu thức 1) \(\dfrac{\sqrt{14}-\sqrt{21}}{\sqrt{7}}\) .
2) \(\dfrac{\sqrt{a^2+5a+6}}{\sqrt{a+3}}\)
3) \(\sqrt{3\left(x^2-10x+25\right)}.\sqrt{27}\) với x < 5
4)
\(\dfrac{y}{x}\sqrt{\dfrac{x^2}{y^4}}\) với x > 0; y < 0
5) \(\dfrac{1}{x-y}.\sqrt{x^6\left(x-y\right)^4}\) với x \(\ne\) y
5: \(=\dfrac{1}{x-y}\cdot x^3\cdot\left(x-y\right)^2=x^3\left(x-y\right)\)
Rút gọn các biểu thức sau:
a) A = \(\left(\dfrac{\sqrt{x}}{x-4}+\dfrac{2}{2-\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right):\left(\sqrt{x}-2+\dfrac{10-x}{\sqrt{x}+2}\right)\)
b) B = \(\left(\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\sqrt{xy}\right):\left(x-y\right)+\dfrac{2\sqrt{y}}{\sqrt{x}+\sqrt{y}}\)
c) C = \(\left(1-\dfrac{\sqrt{x}}{1+\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{2+\sqrt{x}}{3-\sqrt{x}}+\dfrac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right)\)
d) D = \(\sqrt{\dfrac{a+x^2}{x}-2\sqrt{a}}-\sqrt{\dfrac{a+x^2}{x}+2\sqrt{a}}\) với a > 0, x > 0.
A= \(\left(\dfrac{4\sqrt{y}}{2+\sqrt{y}}+\dfrac{8y}{4-y}\right)\) / \(\left(\dfrac{\sqrt{y}-1}{y-2\sqrt{y}}-\dfrac{2}{\sqrt{y}}\right)\)
rút gọn A
\(A=\left(\dfrac{4\sqrt{y}}{2+\sqrt{y}}+\dfrac{8y}{4-y}\right):\left(\dfrac{\sqrt{y}-1}{y-2\sqrt{y}}-\dfrac{2}{\sqrt{y}}\right)\)
\(=\dfrac{4\sqrt{y}\left(2-\sqrt{y}\right)+8y}{\left(2+\sqrt{y}\right)\left(2-\sqrt{y}\right)}:\dfrac{\sqrt{y}-1-2\left(\sqrt{y}-2\right)}{\sqrt{y}\left(\sqrt{y}-2\right)}\)
\(=\dfrac{8\sqrt{y}-4y+8y}{\left(2+\sqrt{y}\right)\left(2-\sqrt{y}\right)}\cdot\left(\dfrac{-\sqrt{y}\left(2-\sqrt{y}\right)}{-\left(\sqrt{y}-3\right)}\right)=\dfrac{4y\left(\sqrt{y}+2\right)}{\left(\sqrt{y}+2\right)\left(\sqrt{y}-3\right)}=\dfrac{4y}{\sqrt{y}-3}\)
Rút gọn các biểu thức sau:
b) \(\dfrac{x-1}{\sqrt{y}-1}\sqrt{\dfrac{\left(y-2\sqrt{y}+1\right)^2}{\left(x-1\right)^4}}\) x \(\ne\) 1, y \(\ne\) 1, y > 0
a) \(\sqrt{\dfrac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}\sqrt{\dfrac{\left(\sqrt{x+1}\right)^2}{\left(\sqrt{x}+1\right)^2}}\)
=\(\dfrac{\sqrt{x}-1}{\sqrt{x}+1};x\ge0\)
b) Ta có: \(\dfrac{x-1}{\sqrt{y}-1}\cdot\sqrt{\dfrac{\left(y-2\sqrt{y}+1\right)^2}{\left(x-1\right)^4}}\)
\(=\dfrac{x-1}{\sqrt{y}-1}\cdot\dfrac{\sqrt{y}-1}{\left(x-1\right)^2}\)
\(=\dfrac{1}{x-1}\)
Rút gọn biểu thức P=\(\sqrt{\dfrac{1}{x^2+y^2}+\dfrac{1}{\left(x+y\right)^2}+\sqrt{\dfrac{1}{x^4}+\dfrac{1}{y^4}+\dfrac{1}{\left(x^2+y^2\right)^2}}}\)
Mn giúp e với
Rút gọn biểu thức sau A = \(\left(\sqrt{\dfrac{x}{y}}+\sqrt{\dfrac{y}{x}}\right):\left(\sqrt{\dfrac{x}{y}}-\sqrt{\dfrac{y}{x}}\right)\) với x = 1 + a; y = 1 - a
Lời giải:
\(A=\frac{x+y}{\sqrt{xy}}: \frac{x-y}{\sqrt{xy}}=\frac{x+y}{\sqrt{xy}}.\frac{\sqrt{xy}}{x-y}=\frac{x+y}{x-y}\)
\(=\frac{1+a+1-a}{1+a-(1-a)}=\frac{2}{2a}=\frac{1}{a}\)
Cho biểu thức B = \(\left(\dfrac{x-y}{\sqrt{x}-\sqrt{y}}+\dfrac{x\sqrt{x}-y\sqrt{y}}{y-x}\right):\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)
a) Rút gọn biểu thức B
b) Chứng minh \(B\ge0\)
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)