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a: \(=\dfrac{\sqrt{x}+1-\sqrt{x}+1-2\sqrt{x}}{x-1}=\dfrac{-2\left(\sqrt{x}-1\right)}{x-1}=\dfrac{-2}{\sqrt{x}+1}\)
b: \(=\dfrac{\sqrt{x}-x\sqrt{y}-\sqrt{y}+y\sqrt{x}+\sqrt{x}+x\sqrt{y}+\sqrt{y}+y\sqrt{x}}{1-xy}:\left(\dfrac{x+y+2xy+1-xy}{1-xy}\right)\)
\(=\dfrac{2\sqrt{x}+2y\sqrt{x}}{1-xy}\cdot\dfrac{1-xy}{x+y+xy+1}\)
\(=\dfrac{2\sqrt{x}\left(y+1\right)}{\left(y+1\right)\left(x+1\right)}=\dfrac{2\sqrt{x}}{x+1}\)
c: \(=\dfrac{3x+3\sqrt{x}-9+x+2\sqrt{x}-3-x+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{3x+5\sqrt{x}-8}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\dfrac{3\sqrt{x}+8}{\sqrt{x}+2}\)
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
a: \(=\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}-\sqrt{ab}=\sqrt{ab}-\sqrt{ab}=0\)
b: \(=\dfrac{\left(\sqrt{x}-2\sqrt{y}\right)^2}{\sqrt{x}-2\sqrt{y}}+\dfrac{\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)
\(=\sqrt{x}-2\sqrt{y}+\sqrt{y}=\sqrt{x}-\sqrt{y}\)
c: \(=\sqrt{x}+2-\dfrac{x-4}{\sqrt{x}-2}\)
\(=\sqrt{x}+2-\sqrt{x}-2=0\)
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
a: \(=x-\sqrt{xy}+y-x+2\sqrt{xy}-y=\sqrt{xy}\)
b: \(=\dfrac{1+\sqrt{a}}{a-\sqrt{a}}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)
\(P=\frac{x}{\left(\sqrt{x}+\sqrt{y}\right)\left(1-\sqrt{y}\right)}-\frac{y}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}+1\right)}-\frac{xy}{\left(\sqrt{x}+1\right)\left(1-\sqrt{y}\right)}\)
\(=\sqrt{xy}+\sqrt{x}-\sqrt{y}\)
\(P=2\Rightarrow\sqrt{xy}+\sqrt{x}-\sqrt{y}=2\)
\(\Rightarrow\left[{}\begin{matrix}x=y=2\\x=4;y=0\end{matrix}\right.\) (t/m)
làm thế nào để ra được P = \(\sqrt{xy}\)+ \(\sqrt{x}\)- \(\sqrt{y}\) vậy bn ?
Nếu có thêm điều kiện \(y>1\) thì kết quả là \(\dfrac{1}{x-1}\)
Bài 2:
a: ĐKXĐ: a>0 và b>0
b: \(P=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}=\sqrt{a}-\sqrt{b}\)
c: Khi a=4 và b=1 thì P=2-1=1
a) \(P=\dfrac{x}{\left(\sqrt{x}+\sqrt{y}\right)\left(1-\sqrt{y}\right)}-\dfrac{y}{\left(\sqrt{x}+\sqrt{y}\right)\left(1+\sqrt{x}\right)}-\dfrac{xy}{\left(\sqrt{x}+1\right)\left(1-\sqrt{y}\right)}\)
ĐK: \(\left\{{}\begin{matrix}x\ge0;y\ge0\\\sqrt{x}+\sqrt{y}\ne0\\\sqrt{x}+1\ne0\\1-\sqrt{y}\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge0;y\ge0\\x^2+y^2>0\\y\ne1\end{matrix}\right.\)
\(P=\dfrac{x\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(1-\sqrt{y}\right)\left(\sqrt{x}+1\right)}-\dfrac{y\left(1-\sqrt{y}\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(1-\sqrt{y}\right)\left(\sqrt{x}+1\right)}-\dfrac{xy\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(1-\sqrt{y}\right)\left(\sqrt{x}+1\right)}\)
\(P=\dfrac{x\sqrt{x}+x-y+y\sqrt{y}-xy\sqrt{x}-xy\sqrt{y}}{\left(\sqrt{x}+\sqrt{y}\right)\left(1-\sqrt{y}\right)\left(\sqrt{x}+1\right)}\)
\(P=\dfrac{\left(x\sqrt{x}+y\sqrt{y}\right)+\left(x-y\right)-\left(xy\sqrt{x}+xy\sqrt{y}\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(1-\sqrt{y}\right)\left(\sqrt{x}+1\right)}\)
\(P=\dfrac{\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)+\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)-xy\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(1-\sqrt{y}\right)\left(\sqrt{x}+1\right)}\)
\(P=\dfrac{\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y+\sqrt{x}-\sqrt{y}-xy\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(1-\sqrt{y}\right)\left(\sqrt{x}+1\right)}\)
\(P=\dfrac{\left(x-xy\right)+\left(-\sqrt{y}+y\right)+\left(\sqrt{x}-\sqrt{xy}\right)}{\left(1-\sqrt{y}\right)\left(\sqrt{x}+1\right)}\)
\(P=\dfrac{x\left(1-y\right)-\sqrt{y}\left(1-\sqrt{y}\right)+\sqrt{x}\left(1-\sqrt{y}\right)}{\left(1-\sqrt{y}\right)\left(\sqrt{x}+1\right)}\)
\(P=\dfrac{x\left(1-\sqrt{y}\right)\left(1+\sqrt{y}\right)-\sqrt{y}\left(1-\sqrt{y}\right)+\sqrt{x}\left(1-\sqrt{y}\right)}{\left(1-\sqrt{y}\right)\left(\sqrt{x}+1\right)}\)
\(P=\dfrac{\left(1-\sqrt{y}\right)\left(x+x\sqrt{y}-\sqrt{y}+\sqrt{x}\right)}{\left(1-\sqrt{y}\right)\left(\sqrt{x}+1\right)}\)
\(P=\dfrac{x+x\sqrt{y}-\sqrt{y}+\sqrt{x}}{\sqrt{x}+1}\)
\(P=\dfrac{\left(x+\sqrt{x}\right)+\left(x\sqrt{y}-\sqrt{y}\right)}{\sqrt{x}+1}\)
\(P=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)+\sqrt{y}\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}+1}\)
\(P=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+\sqrt{xy}-\sqrt{y}\right)}{\sqrt{x}+1}\)
\(P=\sqrt{x}+\sqrt{xy}-\sqrt{y}\)
b) \(P=2\) khi:
\(\sqrt{x}+\sqrt{xy}-\sqrt{y}=2\)
\(\Leftrightarrow\sqrt{x}\left(\sqrt{y}+1\right)-\sqrt{y}-1=2-1\)
\(\Leftrightarrow\sqrt{x}\left(\sqrt{y}+1\right)-\left(\sqrt{y}+1\right)=1\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(\sqrt{y}+1\right)=1\)
Mà: x,y là nguyên \(\Rightarrow\sqrt{x}-1,\sqrt{y}+1\inƯ\left(1\right)=\left\{1;-1\right\}\)
Mặt khác: \(\sqrt{y}+1\ge1\) nên ta có:
\(\sqrt{y}+1=1\Leftrightarrow y=0\) (tm)
\(\Rightarrow\sqrt{x}-1=1\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\) (tm)
Vậy: \(\left(x;y\right)=\left(4;0\right)\)