a: ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x< >1\end{matrix}\right.\)
b: \(P=\left(\dfrac{\sqrt{x}+2}{\sqrt{x}-1}-\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\right)\cdot\dfrac{x-1}{\sqrt{x}+2}\)
\(=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{x-1}{\sqrt{x}+2}\)
\(=\dfrac{x+3\sqrt{x}+2-x+3\sqrt{x}-2}{\sqrt{x}+2}=\dfrac{6\sqrt{x}}{\sqrt{x}+2}\)
c: \(P=\dfrac{6\sqrt{x}+12-12}{\sqrt{x}+2}=6-\dfrac{12}{\sqrt{x}+2}\)
\(\sqrt{x}+2>=2\)
=>\(\dfrac{12}{\sqrt{x}+2}< =\dfrac{12}{2}=6\)
=>\(-\dfrac{12}{\sqrt{x}+2}>=-\dfrac{12}{2}=-6\)
=>\(-\dfrac{12}{\sqrt{x}+2}+6>=0\)
Dấu = xảy ra khi x=0
a) ĐKXĐ:x\(\ge0\) ,x\(\ne+-1\) ,x \(\ne-2\)
b) P=\(\left(\dfrac{\sqrt{x}+2}{\sqrt{x}-1}-\dfrac{\sqrt{x}-2}{\sqrt{x+1}}\right)\).\(\dfrac{x-1}{\sqrt{x}-2}\)
P=\(\left(\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\) \(-\)\(\left(\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\).\(\dfrac{x-1}{\sqrt{x}-2}\)
P=\(\left(\dfrac{x+3\sqrt{x}+2-x+3\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)\(.\dfrac{x-1}{\sqrt{x}-2}\)
P=\(\dfrac{sáu\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\). \(\dfrac{x-1}{\sqrt{x}-2}\)
P=\(\dfrac{sáu\sqrt{x}}{x-1}.\dfrac{x-1}{\sqrt{x}-2}\)
P=\(\dfrac{sáu\sqrt{x}}{\sqrt{x}-2}\)
Mình ko viết đc số sáu nha.Bn thông cảm nha
Máy mình bị liệt số sáu
