Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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)
\(\(b)\frac{\sqrt{a}+a\sqrt{b}-\sqrt{b}-b\sqrt{a}}{ab-1}\left(a,b\ge0;a,b\ne1\right)\)\)
\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)+\left(a\sqrt{b}-b\sqrt{a}\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab+1}\right)}\)\)
\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)+\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}\)\)
\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{ab}+1\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}\)\)
\(\(=\frac{\sqrt{a}-\sqrt{b}}{\left(\sqrt{ab}-1\right)}\left(a,b\ge0.a,b\ne1\right)\)\)
_Minh ngụy_
\(\(c)\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\)\)( tự ghi điều kiện )
\(\(=\frac{x\sqrt{x}+y\sqrt{y}-\left(\sqrt{x}-\sqrt{y}\right)^2.\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)\)
\(\(=\frac{x\sqrt{x}+y\sqrt{y}-\left(x\sqrt{x}+x\sqrt{y}-2x\sqrt{y}-2y\sqrt{x}+y\sqrt{x}+y\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)\)
\(\(=\frac{x\sqrt{y}+y\sqrt{x}}{\sqrt{x}+\sqrt{y}}\)\)( phá ngoặc và tính )
\(\(=\frac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}=\sqrt{xy}\)\)
_Minh ngụy_
\(\frac{\left(\sqrt{x}+\sqrt{y}\right)}{x\sqrt{x}+y\sqrt{y}}\left(\frac{x-y}{\sqrt{x}-\sqrt{y}}-\frac{x\sqrt{x}+y\sqrt{y}}{x-y}\right)\)
\(=\frac{\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}^3+\sqrt{y}^3}\left(\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}-\frac{\sqrt{x}^3+\sqrt{y}^3}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}\right)\)
\(=\frac{1}{x-\sqrt{xy}+y}\left(\sqrt{x}+\sqrt{y}-\frac{x-\sqrt{xy}+y}{\sqrt{x}-\sqrt{y}}\right)\)
\(=\frac{1}{x-\sqrt{xy}+y}\left(\frac{x-y}{\sqrt{x}-\sqrt{y}}-\frac{x-\sqrt{xy}+y}{\sqrt{x}-\sqrt{y}}\right)\)
\(=\frac{1}{x-\sqrt{xy}+y}\left(\frac{x-y-x+\sqrt{xy}-y}{\sqrt{x}-\sqrt{y}}\right)\)
\(=\frac{1}{x-\sqrt{xy}+y}\left(\frac{\sqrt{xy}-2y}{\sqrt{x}-\sqrt{y}}\right)\)
tự làm tiếp nh đến đây dễ rồi
Năm 1930 có sự kiện gì và năm 1945 có sự kiện gì toán lóp 4
\(P=\left[\frac{x-y}{\sqrt{x}-\sqrt{y}}+\frac{x\sqrt{y}-y\sqrt{x}}{y-x}\right]:\frac{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)
\(=\left[\frac{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}+\frac{\sqrt{x}\sqrt{y}\left(\sqrt{x}-\sqrt{y}\right)}{\left(\sqrt{y}-\sqrt{x}\right)\left(\sqrt{y}+\sqrt{x}\right)}\right]:\frac{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)
\(=\left[\sqrt{x}+\sqrt{y}-\frac{\sqrt{x}\sqrt{y}\left(\sqrt{y}-\sqrt{x}\right)}{\left(\sqrt{y}-\sqrt{x}\right)\left(\sqrt{y}+\sqrt{x}\right)}\right]:\frac{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)
\(=\left[\sqrt{x}+\sqrt{y}-\frac{\sqrt{x}\sqrt{y}}{\left(\sqrt{y}+\sqrt{x}\right)}\right]:\frac{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)
\(=\frac{\left(\sqrt{x}+\sqrt{y}\right)^2-\sqrt{x}\sqrt{y}}{\left(\sqrt{y}+\sqrt{x}\right)}:\frac{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)
\(=\frac{\left(\sqrt{x}+\sqrt{y}\right)^2-\sqrt{x}\sqrt{y}}{\left(\sqrt{y}+\sqrt{x}\right)}.\frac{\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}\)
\(=\frac{\left(\sqrt{x}+\sqrt{y}\right)^2-\sqrt{xy}}{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}\)
\(=\frac{x+2\sqrt{xy}+y-\sqrt{xy}}{x-2\sqrt{xy}+y+\sqrt{xy}}\)
\(=\frac{x+\sqrt{xy}+y}{x-\sqrt{xy}+y}\)
Lời giải:
ĐK: $x\neq y; x,y\geq 0$
a)
\(B=\left[\frac{(x-y)(\sqrt{x}+\sqrt{y})}{(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})}-\frac{x\sqrt{x}-y\sqrt{y}}{(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})}\right]:\frac{x-\sqrt{xy}+y}{\sqrt{x}+\sqrt{y}}\)
\(=\frac{x\sqrt{x}+x\sqrt{y}-y\sqrt{x}-y\sqrt{y}-x\sqrt{x}+y\sqrt{y}}{(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})}.\frac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}\)
\(=\frac{x\sqrt{y}-y\sqrt{x}}{\sqrt{x}-\sqrt{y}}.\frac{1}{x-\sqrt{xy}+y}=\frac{\sqrt{xy}(\sqrt{x}-\sqrt{y})}{\sqrt{x}-\sqrt{y}}.\frac{1}{x-\sqrt{xy}+y}=\frac{\sqrt{xy}}{x-\sqrt{xy}+y}\)
b) Ta thấy:
\(\sqrt{xy}\geq 0, \forall x,y\geq 0\)
\(x-\sqrt{xy}+y=(\sqrt{x}-\frac{\sqrt{y}}{2})^2+\frac{3}{4}y>0, \forall x,y\geq 0; x\neq y\)
\(\Rightarrow B=\frac{\sqrt{xy}}{x-\sqrt{xy}+y}\geq 0\) (đpcm)
c)
Áp dụng BĐT AM-GM: \(x+y\geq 2\sqrt{xy}\Rightarrow x-\sqrt{xy}+y\geq \sqrt{xy}\)
\(\Rightarrow B=\frac{\sqrt{xy}}{x-\sqrt{xy}+y}\leq 1\)
Dấu "=" xảy ra khi $x=y$. Mà $x\neq y$ nên $B< 1\Rightarrow \sqrt{B}< 1$
Do đó: \(B=\sqrt{B}.\sqrt{B}< \sqrt{B}\)
bạn ơi ở phần c áp dụng cái j vậy bạn?