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 \(B\ge0\)
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.
1) Rút gọn biểu thứ
A=\(\left(\dfrac{x-y}{\sqrt{x}-\sqrt{y}}+\dfrac{\sqrt{x^3}-\sqrt{y^3}}{y-x}\right):\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)
a) Rút gọn A
b) Chứng minh A<1
Lời giải:
a) ĐK: $x\geq 0; y\geq 0; x\neq y$
\(A=\left[\frac{(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})}{\sqrt{x}-\sqrt{y}}-\frac{(\sqrt{x}-\sqrt{y})(x+\sqrt{xy}+y)}{(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})}\right]:\frac{x-\sqrt{xy}+y}{\sqrt{x}+\sqrt{y}}\)
\(=\left(\sqrt{x}+\sqrt{y}-\frac{x+\sqrt{xy}+y}{\sqrt{x}+\sqrt{y}}\right).\frac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}\)
\(=\frac{\sqrt{xy}}{\sqrt{x}+\sqrt{y}}.\frac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}=\frac{\sqrt{xy}}{x-\sqrt{xy}+y}\)
b) \(1-A=\frac{(\sqrt{x}-\sqrt{y})^2}{x-\sqrt{xy}+y}>0\) với mọi $x\neq y; x,y\geq 0$
$\Rightarrow A< 1$
\(P=\left(\sqrt{x}+\dfrac{y-\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\right)\):\(\left(\dfrac{x}{\sqrt{xy}+y}+\dfrac{y}{\sqrt{xy}-x}-\dfrac{x+y}{\sqrt{xy}}\right)\)
a) Với giá trị nào cùa x thì biểu thức có nghĩa
b) Rút gọn P
c) Tím P với x=3 và y=\(\dfrac{2}{2-\sqrt{3}}\)
Giúp với ạ
Cho biểu thức \(A=\left(\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\sqrt{xy}\right):\left(x-y\right)+\dfrac{2\sqrt{y}}{\sqrt{x}+\sqrt{y}};x\ge0,y\ge0,x\ne y\)
Chứng minh rằng giá trị của biểu thức A không phụ thuộc vào x, y
Ta có: \(A=\left(\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\sqrt{xy}\right):\left(x-y\right)+\dfrac{2\sqrt{y}}{\sqrt{x}+\sqrt{y}}\)
\(=\dfrac{\left(x-2\sqrt{xy}+y\right)}{x-y}+\dfrac{2\sqrt{y}}{\sqrt{x}+\sqrt{y}}\)
\(=\dfrac{\sqrt{x}-\sqrt{y}+2\sqrt{y}}{\sqrt{x}+\sqrt{y}}\)
=1
Cho biểu thức sau :
\(P=\left(\dfrac{x-y}{\sqrt{x}-\sqrt{y}}-\dfrac{x\sqrt{x}-y\sqrt{y}}{x-y}\right):\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)
a, Tìm điều kiện để biểu thức trên có nghĩa
b, Rút gọn P
a) tự làm.
b) \(P=\left(\dfrac{x-y}{\sqrt{x}-\sqrt{y}}-\dfrac{x\sqrt{x}-y\sqrt{y}}{x-y}\right)\div\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)
\(=\left(\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}-\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\right)\cdot\dfrac{\sqrt{x}+\sqrt{y}}{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}\)
\(=\left(\sqrt{x}+\sqrt{y}-\dfrac{x+\sqrt{xy}+y}{\sqrt{x}+\sqrt{y}}\right)\cdot\dfrac{\sqrt{x}+\sqrt{y}}{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)+\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)-\left(x+\sqrt{xy}+y\right)}{\sqrt{x}+\sqrt{y}}\cdot\dfrac{\sqrt{x}+\sqrt{y}}{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}\)
\(=\dfrac{x+\sqrt{xy}+\sqrt{xy}+y-x-\sqrt{xy}-y}{\sqrt{x}+\sqrt{y}}\cdot\dfrac{\sqrt{x}+\sqrt{y}}{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}\)
\(=\dfrac{\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\cdot\dfrac{\sqrt{x}+\sqrt{y}}{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}\)
\(=\sqrt{xy}\cdot\dfrac{1}{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}\)
\(=\dfrac{\sqrt{xy}}{x-2\sqrt{xy}+y+\sqrt{xy}}\)
\(=\dfrac{\sqrt{xy}}{x-\sqrt{xy}+y}\)
\(\dfrac{\sqrt{x}+\sqrt{y}-1}{x+\sqrt{xy}}+\dfrac{\sqrt{x}-\sqrt{y}}{2\sqrt{xy}}\cdot\left(\dfrac{\sqrt{y}}{x-\sqrt{xy}}+\dfrac{\sqrt{y}}{x+\sqrt[]{xy}}\right)\)
Thu gọn biểu thức trên
Ta có: \(\dfrac{\sqrt{y}}{x-\sqrt{xy}}+\dfrac{\sqrt{y}}{x+\sqrt{xy}}\)
\(=\dfrac{\sqrt{y}\left(x+\sqrt{xy}\right)+\sqrt{y}\left(x-\sqrt{xy}\right)}{x^2-xy}\)
\(=\dfrac{\sqrt{y}\left(x+\sqrt{xy}+x-\sqrt{xy}\right)}{x\left(x-y\right)}=\dfrac{2x\sqrt{y}}{x\left(x-y\right)}\)
\(=\dfrac{2\sqrt{y}}{x-y}=\dfrac{2\sqrt{y}}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)
\(\Rightarrow A=\dfrac{\sqrt{x}+\sqrt{y}-1}{x+\sqrt{xy}}+\dfrac{\sqrt{x}-\sqrt{y}}{2\sqrt{xy}}.\dfrac{2\sqrt{y}}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)
\(=\dfrac{\sqrt{x}+\sqrt{y}-1}{\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)}+\dfrac{1}{\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)}\)
\(=\dfrac{\sqrt{x}+\sqrt{y}-1+1}{\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)}=\dfrac{1}{\sqrt{x}}=\dfrac{\sqrt{x}}{x}\)
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:
\(A=\frac{x+y}{\sqrt{xy}}: \frac{x-y}{\sqrt{xy}}=\frac{x+y}{\sqrt{xy}}.\frac{\sqrt{xy}}{x-y}=\frac{x+y}{x-y}\)
\(=\frac{1+a+1-a}{1+a-(1-a)}=\frac{2}{2a}=\frac{1}{a}\)
rút gọn biểu thức: P=\(\dfrac{4\sqrt{xy}}{x-y}\):\(\left(\dfrac{1}{y-x}+\dfrac{1}{x+2\sqrt{x}\sqrt{y}+y^2}\right)\)-2x
\(P=\dfrac{4\sqrt{xy}}{x-y}:\left(\dfrac{1}{y-x}+\dfrac{1}{x+2\sqrt{x}\sqrt{y}+y}\right)-2x\) (với \(x\ne y,x,y\ge0\))
\(P=\dfrac{4\sqrt{xy}}{x-y}:\left(\dfrac{1}{\left(\sqrt{y}-\sqrt{x}\right)\left(\sqrt{y}+\sqrt{x}\right)}+\dfrac{1}{\left(\sqrt{x}+\sqrt{y}\right)^2}\right)-2x\)
\(P=\dfrac{4\sqrt{xy}}{x-y}:\left(\dfrac{\sqrt{y}+\sqrt{x}}{\left(\sqrt{y}+\sqrt{x}\right)^2\left(\sqrt{y}-\sqrt{x}\right)}+\dfrac{\sqrt{y}-\sqrt{x}}{\left(\sqrt{x}+\sqrt{y}\right)^2\left(\sqrt{y}-\sqrt{x}\right)}\right)-2x\)
\(P=\dfrac{4\sqrt{xy}}{x-y}:\left(\dfrac{\sqrt{y}+\sqrt{x}+\sqrt{y}-\sqrt{x}}{\left(\sqrt{y}-\sqrt{x}\right)\left(\sqrt{x}+\sqrt{y}\right)^2}\right)-2x\)
\(P=\dfrac{4\sqrt{xy}}{x-y}:\left(\dfrac{2\sqrt{y}}{\left(y-x\right)\left(\sqrt{x}+\sqrt{y}\right)}\right)-2x\)
\(P=\dfrac{4\sqrt{xy}}{x-y}\cdot\dfrac{\left(y-x\right)\left(\sqrt{x}+\sqrt{y}\right)}{2\sqrt{y}}-2x\)
\(P=\dfrac{4\sqrt{xy}\cdot\left(y-x\right)\left(\sqrt{x}+\sqrt{y}\right)}{\left(x-y\right)\cdot2\sqrt{y}}-2x\)
\(P=\dfrac{4\sqrt{xy}\cdot\left(y-x\right)\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)\cdot2\sqrt{y}}-2x\)
\(P=\dfrac{2\sqrt{x}\left(y-x\right)}{\sqrt{x}-\sqrt{y}}-2x\)
\(P=\dfrac{2\sqrt{x}\left(y-x\right)-2x\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}\)
\(P=\dfrac{2y\sqrt{x}-2x\sqrt{x}-2x\sqrt{x}+2x\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)
\(P=\dfrac{2y\sqrt{x}-4x\sqrt{x}+2x\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)
Cho biểu thức:
\(A=\left[\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}\right)\times\dfrac{2}{\sqrt{x}+\sqrt{y}}+\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}\right]:\dfrac{\sqrt{x^3}+y\sqrt{x}+x\sqrt{y}+\sqrt{y^3}}{\sqrt{x^3y}+\sqrt{xy^3}}\)
a, Rút gọn A
b, Biết xy=6 Tìm x, y để A đạt GTNN