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a) Để B có nghĩa thì \(\left\{{}\begin{matrix}y\ge0\\y\ne1\end{matrix}\right.\)
B=\(\left(\dfrac{1}{\sqrt{y}+1}-\dfrac{3\sqrt{y}}{\sqrt{y}-1}+3\right).\dfrac{\sqrt{y}+1}{\sqrt{y}+2}=\left[\dfrac{\sqrt{y}-1}{\left(\sqrt{y}+1\right)\left(\sqrt{y}-1\right)}-\dfrac{3\sqrt{y}\left(\sqrt{y}+1\right)}{\left(\sqrt{y}+1\right)\left(\sqrt{y}-1\right)}+\dfrac{3\left(\sqrt{y}+1\right)\left(\sqrt{y}-1\right)}{\left(\sqrt{y}+1\right)\left(\sqrt{y}-1\right)}\right].\dfrac{\sqrt{y}+1}{\sqrt{y}+2}=\left[\dfrac{\sqrt{y}-1}{\left(\sqrt{y}+1\right)\left(\sqrt{y}-1\right)}-\dfrac{3y+3\sqrt{y}}{\left(\sqrt{y}+1\right)\left(\sqrt{y}-1\right)}+\dfrac{3y-3}{\left(\sqrt{y}+1\right)\left(\sqrt{y}-1\right)}\right].\dfrac{\sqrt{y}+1}{\sqrt{y}+2}=\dfrac{\sqrt{y}-1-3y-3\sqrt{y}+3y-3}{\left(\sqrt{y}+1\right)\left(\sqrt{y}-1\right)}.\dfrac{\sqrt{y}+1}{\sqrt{y}+2}=\dfrac{\left(-2\sqrt{y}-4\right)\left(\sqrt{y}+1\right)}{\left(\sqrt{y}+1\right)\left(\sqrt{y}-1\right)\left(\sqrt{y}+2\right)}=\dfrac{-2\left(\sqrt{y}+2\right)\left(\sqrt{y}+1\right)}{\left(\sqrt{y}+1\right)\left(\sqrt{y}-1\right)\left(\sqrt{y}+2\right)}=\dfrac{-2}{\sqrt{y}-1}=\dfrac{2}{1-\sqrt{y}}\)
b) Ta có y=\(3+2\sqrt{2}\Rightarrow P=\dfrac{2}{1-\sqrt{3+2\sqrt{2}}}=\dfrac{2}{1-\sqrt{2+2\sqrt{2}+1}}=\dfrac{2}{1-\sqrt{\left(\sqrt{2}+1\right)^2}}=\dfrac{2}{1-\sqrt{2}-1}=\dfrac{2}{-\sqrt{2}}=-\sqrt{2}\)
Vậy khi x=\(3+2\sqrt{2}\) thì \(P=-\sqrt{2}\)
\(\dfrac{\left(\sqrt{X}+\sqrt{Y}\right)\left(1+\sqrt{XY}\right)+\left(\sqrt{X}-\sqrt{Y}\right)\left(1-\sqrt{XY}\right)}{1-XY}\cdot\dfrac{1-XY}{1-XY+\sqrt{X}+\sqrt{Y}+2\sqrt{XY}}=\dfrac{\sqrt{X}+X\sqrt{Y}+\sqrt{Y}+Y\sqrt{X}+\sqrt{X}-X\sqrt{Y}-\sqrt{Y}+Y\sqrt{X}}{1-XY}\cdot\dfrac{1-XY}{XY+X+Y+1}=\dfrac{2\sqrt{X}\left(1+Y\right)}{\left(1+Y\right)\left(X+1\right)}=\dfrac{2\sqrt{X}}{X+1}\)
b: Thay \(x=\dfrac{2}{2+\sqrt{3}}=2\left(2-\sqrt{3}\right)=4-2\sqrt{3}\) vào P, ta được:
\(P=\dfrac{2\left(\sqrt{3}-1\right)}{4-2\sqrt{3}+1}=\dfrac{2\sqrt{3}-2}{5-2\sqrt{3}}=\dfrac{6\sqrt{3}+2}{13}\)
a) Ta có:
\(\dfrac{1}{\sqrt{n}+\sqrt{n+1}}=\dfrac{\sqrt{n}-\sqrt{n+1}}{n-n-1}=-\sqrt{n}+\sqrt{n+1}\)
\(\Rightarrow A=...=-1+\sqrt{2}-\sqrt{2}+\sqrt{3}-...-\sqrt{48}+\sqrt{49}=-1+7=6\)
Bài 1:
a. ta có \(\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\)
= \(\dfrac{\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)}{\sqrt{x}+\sqrt{y}}-x+2\sqrt{xy}-y\)
= \(x-\sqrt{xy}+y-x+2\sqrt{xy}-y\)
=\(\sqrt{xy}\)
b.ĐK: x ≠ 1
Ta có: A= \(\sqrt{\dfrac{x+2\sqrt{x}+1}{x-2\sqrt{x}+1}}\)=\(\sqrt{\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)^2}}\)=\(\dfrac{\sqrt{x}+1}{\left|\sqrt{x}-1\right|}\)
*Nếu \(\sqrt{x}-1\ge0\Rightarrow\sqrt{x}\ge1\)
⇒ A = \(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
*Nếu \(\sqrt{x}-1< 0\Rightarrow\sqrt{x}< 1\)
⇒ A=\(\dfrac{\sqrt{x}+1}{-\sqrt{x}+1}\)
c.Ta có:
\(P=\left(\dfrac{\sqrt{x}-\sqrt{y}}{1+\sqrt{xy}}+\dfrac{\sqrt{x}+\sqrt{y}}{1-\sqrt{xy}}\right):\left(\dfrac{x+y+2xy}{1-xy}+1\right)\)
Điều kiện : \(xy\ge0\) hoặc \(xy\le0\) ; \(xy\ne1\); \(x\ge0\);\(y\ge0\)
\(P=\left(\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(1-\sqrt{xy}\right)+\left(\sqrt{x}+\sqrt{y}\right)\left(1+\sqrt{xy}\right)}{\left(1+\sqrt{xy}\right)\left(1-\sqrt{xy}\right)}\right):\left(\dfrac{x+2xy+y+1-xy}{1-xy}\right)\)
\(P=\left(\dfrac{\sqrt{x}-x\sqrt{y}-\sqrt{y}+y\sqrt{x}+\sqrt{x}+x\sqrt{y}+\sqrt{y}+y\sqrt{x}}{1-xy}\right):\left(\dfrac{x+xy+y+1}{1-xy}\right)\)
\(P=\left(\dfrac{2\sqrt{x}+2y\sqrt{x}}{1-xy}\right):\left(\dfrac{x\left(1+y\right)+\left(y+1\right)}{1-xy}\right)\)
\(P=\left(\dfrac{2\sqrt{x}\left(1+y\right)}{1-xy}\right):\left(\dfrac{\left(1+y\right)\left(x+1\right)}{1-xy}\right)\)
\(P=\dfrac{2\sqrt{x}\left(1+y\right)}{1-xy}.\dfrac{1-xy}{\left(1+y\right)\left(x+1\right)}\)
\(P=\dfrac{2\sqrt{x}}{x+1}\)
b) ta có :\(x=\dfrac{2}{2+\sqrt{3}}=\dfrac{2\left(2-\sqrt{3}\right)}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}=\dfrac{4-2\sqrt{3}}{4-3}=3-2\sqrt{3}+1=\left(\sqrt{3}-1\right)^2\)
thay \(x=\left(\sqrt{3}-1\right)^2\) vào biểu thức P
ta được : \(P=\dfrac{2\sqrt{\left(\sqrt{3}-1\right)^2}}{\left(\sqrt{3}-1\right)^2+1}\)
\(P=\dfrac{2\left|\sqrt{3}-1\right|}{4-2\sqrt{3}+1}=\dfrac{2\sqrt{3}-2}{5-2\sqrt{3}}\)
\(P=\dfrac{\left(2\sqrt{3}-2\right)\left(5+2\sqrt{3}\right)}{\left(5-2\sqrt{3}\right)\left(5+2\sqrt{3}\right)}=\dfrac{10\sqrt{3}+12-10-4\sqrt{3}}{25-12}\)
\(P=\dfrac{6\sqrt{3}+2}{13}\)
c) để P\(\le\)1 thì \(\dfrac{2\sqrt{x}}{x+1}\le1\)
\(\Leftrightarrow\dfrac{2\sqrt{x}}{x+1}-1\le0\)
\(\Leftrightarrow\dfrac{2\sqrt{x}-x-1}{x+1}\le0\)
\(\Leftrightarrow\dfrac{-\left(x-2\sqrt{x}+1\right)}{x+1}\le0\)
\(\Leftrightarrow\dfrac{-\left(x-1\right)^2}{x+1}\le0\)
Vì \(-\left(x-1\right)^2\le0\) nên x + 1 \(\ge\) 0
\(\Leftrightarrow\) x \(\ge\) -1
đúng thì cho xin 1 like nha
2
\(A=\sqrt{1-6x+9x^2}+\sqrt{9x^2-12x+4}\)
A= \(\sqrt{9x^2-6x+1}+\sqrt{9x^2-12x+4}\)
A= \(\sqrt{\left(3x-1\right)^2}+\sqrt{\left(3x-2\right)^2}=\left|3x-1\right|+\left|3x-2\right|\)
ta có |3x-1|+|3x-2|=|3x-1|+|2-3x| ≥ |3x-1+2-3x|=1
=> A ≥ 1
=> Min A =1 khi 1/3 ≤ x ≤ 2/3
đkxđ a>=0 a khác 1
\(C=\left(\frac{a}{\sqrt{a}\left(\sqrt{a}-1\right)}-\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\frac{\sqrt{a}+1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}+\frac{2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\)
\(C=\frac{a-1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\frac{\sqrt{a}+3}{a-1}\)
\(C=\frac{\left(a-1\right).\left(\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}+3\right)}\)
b)
\(a=4-2\sqrt{3}=3-2\sqrt{3}+1=\left(\sqrt{3}-1\right)^2\)
\(\sqrt{a}=\sqrt{3}-1\)
thay vào nha
c) \(C=\frac{\left(a-1\right).\left(\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}+3\right)}\)
để c<0 thì \(\frac{\left(a-1\right).\left(\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}+3\right)}< 0\)
mà \(\sqrt{a}\left(\sqrt{a}+3\right)>0\)
\(\left(a-1\right)\left(\sqrt{a}+1\right)< 0\)
mà \(\sqrt{a}+1>0\)
nên a-1<0
\(0\le a< 1\)
\(a,\dfrac{x+2\sqrt{x}-3}{\sqrt{x}-1}\)
\(\Leftrightarrow\dfrac{x+3\sqrt{x}-\sqrt{x}-3}{\sqrt{x}-1}\)
\(\Leftrightarrow\dfrac{\sqrt{x}.\left(\sqrt{x}+3\right)-\left(\sqrt{x}+3\right)}{\sqrt{x}-1}\)
\(\Leftrightarrow\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\)
\(\Rightarrow\sqrt{x}+3\)
\(b,\dfrac{4y+3\sqrt{y}-7}{4\sqrt{y}+7}\)
\(\Leftrightarrow\dfrac{4y+7\sqrt{y}-4\sqrt{y}-7}{4\sqrt{y}+7}\)
\(\Leftrightarrow\dfrac{\sqrt{y}.\left(4\sqrt{y}\right)-\left(4\sqrt{y}+7\right)}{4\sqrt{y}+7}\)
\(\Leftrightarrow\dfrac{\left(4\sqrt{y}+7\right).\left(\sqrt{y}-1\right)}{4\sqrt{y}+7}\)
\(\Rightarrow\sqrt{y}-1\)
\(c,\dfrac{x\sqrt{y}-y\sqrt{x}}{\sqrt{x}-\sqrt{y}}\)
\(\Leftrightarrow\dfrac{\sqrt{xy}.\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}\)
\(\Rightarrow\sqrt{xy}\)
\(d,\dfrac{x-3\sqrt{x}-4}{x-\sqrt{x}-12}\)
\(\Leftrightarrow\dfrac{x+\sqrt{x}-4\sqrt{x}-4}{x+3\sqrt{x}-4\sqrt{x}-12}\)
\(\Leftrightarrow\dfrac{\sqrt{x}.\left(\sqrt{x}+1\right)-4\left(\sqrt{x}+1\right)}{\sqrt{x}.\left(x+3\right)-4\left(\sqrt{x}+3\right)}\)
\(\Leftrightarrow\dfrac{\left(\sqrt{x}+1\right).\left(\sqrt{x}-4\right)}{\left(\sqrt{x}+3\right).\left(\sqrt{x}-4\right)}\)
\(\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)
\(\Rightarrow\dfrac{x-2\sqrt{x}-3}{x-9}\)
\(e,\dfrac{1+\sqrt{x}+\sqrt{y}+\sqrt{xy}}{1+\sqrt{4}}\)
\(\Leftrightarrow\dfrac{1+\sqrt{x}+\sqrt{y}+\sqrt{xy}}{1+2}\)
\(\Rightarrow\dfrac{1+\sqrt{x}+\sqrt{y}+\sqrt{xy}}{3}\)
a: ĐKXĐ: \(\left\{{}\begin{matrix}y\ge0\\y\ne1\end{matrix}\right.\)
Ta có: \(P=\left(\dfrac{1}{1-\sqrt{y}}+\dfrac{1}{1+\sqrt{y}}\right):\left(\dfrac{1}{1-\sqrt{y}}-\dfrac{1}{1+\sqrt{y}}\right)+\dfrac{1}{1-\sqrt{y}}\)
\(=\dfrac{1+\sqrt{y}+1-\sqrt{y}}{\left(1-\sqrt{y}\right)\left(1+\sqrt{y}\right)}:\dfrac{1+\sqrt{y}-1+\sqrt{y}}{\left(1-\sqrt{y}\right)\left(1+\sqrt{y}\right)}+\dfrac{1}{1-\sqrt{y}}\)
\(=\dfrac{2}{2\sqrt{y}}-\dfrac{1}{\sqrt{y}-1}\)
\(=\dfrac{\sqrt{y}-1-\sqrt{y}}{\sqrt{y}\left(\sqrt{y}-1\right)}\)
\(=\dfrac{-1}{\sqrt{y}\left(\sqrt{y}-1\right)}\)