Chứng minh
(\(\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x-\sqrt{y}}}+\sqrt{xy}:\sqrt{x}+\sqrt{y}=1\)
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 $x\neq y; x,y\geq 0$
$\Rightarrow A< 1$
1. Tính:
\(\sqrt{\dfrac{x-1+\sqrt{2x-3}}{x+2-\sqrt{2x+3}}}\)
2. Chứng minh:
a) \(\dfrac{\left(3\sqrt{xy}-6y.2x\sqrt{y}+4y\sqrt{x}\right)\left(3\sqrt{y}+2\sqrt{xy}\right)}{y\left(\sqrt{x}-2\sqrt{y}\right)\left(y-4x\right)}=1\)
b) \(\left(\sqrt{x}-\sqrt{y}-\dfrac{\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\right)\left(\dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{y}}+\dfrac{y}{\sqrt{x}-\sqrt{y}}-\dfrac{2\sqrt{xy}}{xy}\right)=\sqrt{x}+\sqrt{y}\)
1.
\(\sqrt{\dfrac{x-1+\sqrt{2x-3}}{x+2-\sqrt{2x+3}}}\Leftrightarrow\)\(\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\\sqrt{\dfrac{\left(\sqrt{2x-3}+1\right)^2}{\left(\sqrt{2x+3}-1\right)^2}}\end{matrix}\right.\)\(\Leftrightarrow\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\\dfrac{\sqrt{2x-3}+1}{\sqrt{2x+3}-1}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\\dfrac{\left(\sqrt{2x-3}+1\right)\left(\sqrt{2x+3}+1\right)}{2\left(x+1\right)}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\\dfrac{\sqrt{4x^2-9}+\sqrt{2x-3}+\sqrt{2x+3}+1}{2\left(x+1\right)}\end{matrix}\right.\)
hết tối giải rồi
Chứng minh rằng biểu thức sau không phụ thuộc vào biến
A = \(\left(\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}}-\dfrac{\sqrt{x}-\sqrt{y}}{\sqrt{x}+\sqrt{y}}\right):\dfrac{\sqrt{xy}}{x-y}\)
B = \(\dfrac{x-y}{\sqrt{x}-\sqrt{y}}-\dfrac{\sqrt{x^3}-\sqrt{y^3}}{x+\sqrt{xy}+y}-2\sqrt{y}\)
A/
\(A=\frac{(\sqrt{x}+\sqrt{y})^2-(\sqrt{x}-\sqrt{y})^2}{(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})}.\frac{x-y}{\sqrt{xy}}\\ =\frac{x+y+2\sqrt{xy}-(x+y-2\sqrt{xy})}{x-y}.\frac{x-y}{\sqrt{xy}}\\ =\frac{4\sqrt{xy}}{x-y}.\frac{x-y}{\sqrt{xy}}=4\)
Vậy biểu thức A không phụ thuộc giá trị vào biến.
B/
\(B=\frac{(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})}{\sqrt{x}-\sqrt{y}}-\frac{(\sqrt{x}-\sqrt{y})(x+\sqrt{xy}+y)}{x+\sqrt{xy}+y}-2\sqrt{y}\\
=\sqrt{x}+\sqrt{y}-(\sqrt{x}-\sqrt{y})-2\sqrt{y}\\
=2\sqrt{y}-2\sqrt{y}=0\)
Vậy giá trị của biểu thức B không phụ thuộc vào giá trị của biến.
Chứng minh rằng biểu thức sau không phụ thuộc vào biến
A = \(\left(\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}}-\dfrac{\sqrt{x}-\sqrt{y}}{\sqrt{x}+\sqrt{y}}\right):\dfrac{\sqrt{xy}}{x-y}\)
B = \(\dfrac{x-y}{\sqrt{x}-\sqrt{y}}-\dfrac{\sqrt{x^3}-\sqrt{y^3}}{x+\sqrt{xy}+y}-2\sqrt{y}\)
\(A=\left(\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}}-\dfrac{\sqrt{x}-\sqrt{y}}{\sqrt{x}+\sqrt{y}}\right):\dfrac{\sqrt{xy}}{x-y}\left(dkxd:x,y\ge0,x\ne y\right)\)
\(=\dfrac{\left(\sqrt{x}+\sqrt{y}\right)^2-\left(\sqrt{x}-\sqrt{y}\right)^2}{\sqrt{x^2}-\sqrt{y^2}}.\dfrac{x-y}{\sqrt{xy}}\)
\(=\dfrac{x+2\sqrt{xy}+y-x+2\sqrt{xy}-y}{x-y}.\dfrac{x-y}{\sqrt{xy}}\)
\(=\dfrac{4\sqrt{xy}}{\sqrt{xy}}=4\)
\(B=\dfrac{x-y}{\sqrt{x}-\sqrt{y}}-\dfrac{\sqrt{x^3}-\sqrt{y^3}}{x+\sqrt{xy}+y}-2\sqrt{y}\left(dkxd:x,y\ge0,x\ne y\right)\)
\(=\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}-\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{x+\sqrt{xy}+y}-2\sqrt{y}\)
\(=\sqrt{x}+\sqrt{y}-\sqrt{x}+\sqrt{y}-2\sqrt{y}\\ =0\)
Vậy biểu thức A và B không phụ thuộc vào biến.
\(\dfrac{\sqrt{y}}{x-\sqrt{ }xy}+\dfrac{\sqrt{x}}{y-\sqrt{ }xy}=-\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{xy}}\)
Chứng minh đẳng thức với x ; y lớn hơn 0 ; x khác y
chứng minh
\(\left(\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}}-\dfrac{\sqrt{x}-\sqrt{y}}{\sqrt{x}+\sqrt{y}}\right):\dfrac{\sqrt{xy}}{x-y}=4\)
\(\left(\dfrac{\left(\sqrt{x}+\sqrt{y}\right)^2-\left(\sqrt{x}-\sqrt{y}\right)^2}{x-y}\right):\dfrac{\sqrt{xy}}{x-y}\\ =\left(\dfrac{x+2\sqrt{xy}+y-x+2\sqrt{xy}-y}{x-y}\right):\dfrac{\sqrt{xy}}{x-y}\\ =\dfrac{4\sqrt{xy}}{x-y}.\dfrac{x-y}{\sqrt{xy}}\\ =4\left(đpcm\right)\)
\(\left(\dfrac{x-y}{\sqrt{x}-\sqrt{y}}+\dfrac{\sqrt{x^2}-\sqrt{y^2}}{y-x}\right):\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)
a) Rút gọn A
b) Chứng minh A ≥0
a:
Sửa đề: \(A=\left(\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}+\dfrac{\sqrt{x^3}-\sqrt{y^3}}{y-x}\right)\cdot\dfrac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}\)
\(A=\left(\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}+\dfrac{x\sqrt{x}-y\sqrt{y}}{y-x}\right)\cdot\dfrac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}\)
\(A=\left(\sqrt{x}+\sqrt{y}-\dfrac{x+\sqrt{xy}+y}{\sqrt{x}+\sqrt{y}}\right)\cdot\dfrac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}\)
\(=\dfrac{x+2\sqrt{xy}+y-x-\sqrt{xy}-y}{\sqrt{x}+\sqrt{y}}\cdot\dfrac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}\)
\(=\dfrac{\sqrt{xy}}{x-\sqrt{xy}+y}\)
b: căn xy>0
\(x-\sqrt{xy}+y=x-2\cdot\sqrt{x}\cdot\dfrac{1}{2}\sqrt{y}+\dfrac{1}{4}y+\dfrac{3}{4}y\)
\(=\left(\sqrt{x}-\dfrac{1}{2}\sqrt{y}\right)^2+\dfrac{3}{4}y>0\)
=>A>0
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\(\ge0\)
a \(A=\dfrac{\left(x-y\right)\left(\sqrt{x}+\sqrt{y}\right)-x\sqrt{x}+y\sqrt{y}}{x-y}\cdot\dfrac{\left(\sqrt{x}+\sqrt{y}\right)}{x-\sqrt{xy}+y}\)
\(=\dfrac{x\sqrt{x}+x\sqrt{y}-y\sqrt{x}-y\sqrt{y}-x\sqrt{x}+y\sqrt{y}}{\sqrt{x}-\sqrt{y}}\cdot\dfrac{1}{x-\sqrt{xy}+y}\)
\(=\dfrac{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)}=\dfrac{\sqrt{xy}}{x-\sqrt{xy}+y}\)
b: căn xy>=0
x-căn xy+y
=x-2*căn x*1/2*căn y+1/4*y+3/4y
=(căn x-1/2*căn y)^2+3/4y>0
=>A>=0
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)