P/s gọi a = x cho dễ viết nhé
a, Với \(x\ge0;x\ne1;x\ne4\)
\(P=\left(\frac{1}{\sqrt{x}-1}-\frac{1}{\sqrt{x}}\right):\left(\frac{\sqrt{x}+1}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}-1}\right)\)
\(=\left(\frac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\frac{x-1-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\frac{3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}=\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{3\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{\sqrt{x}-2}{3\sqrt{x}}\)
chỗ này mình nghĩ ko phải trục căn thức đâu ha :D
b, Ta có P > 1/6 hay \(\frac{\sqrt{x}-2}{3\sqrt{x}}>\frac{1}{6}\Leftrightarrow\frac{\sqrt[]{x}-2}{3\sqrt{x}}-\frac{1}{6}>0\)
\(\Leftrightarrow\frac{6\sqrt{x}-12-3\sqrt{x}}{18\sqrt{x}}>0\Leftrightarrow\frac{3\sqrt{x}-12}{18\sqrt{x}}>0\)
\(\Leftrightarrow3\sqrt{x}-12>0\)( vì \(18\sqrt{x}>0\))
\(\Leftrightarrow3\sqrt{x}>12\Leftrightarrow\sqrt{x}>4\Leftrightarrow x>16\)
Vậy \(x>16\)
cho mình hỏi đề có sai ko ? \(P< \frac{1}{6}\)mình nghĩ sẽ hợp lí hơn
んuﾘ ｲ hãy thuận theo ý thầy :)) và nhớ chú ý đến ĐKXĐ
\(P=\left(\frac{1}{\sqrt{a}-1}-\frac{1}{\sqrt{a}}\right)\div\left(\frac{\sqrt{a}+1}{\sqrt{a}-2}-\frac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
ĐKXĐ : \(\hept{\begin{cases}x>0\\x\ne1\\x\ne4\end{cases}}\)
\(=\left(\frac{\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}-\frac{\sqrt{a}-1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right)\div\left(\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}-\frac{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\right)\)
\(=\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\div\left(\frac{a-1}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}-\frac{a-4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\right)\)
\(=\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\div\frac{3}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\times\frac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}=\frac{\sqrt{a}-2}{3\sqrt{a}}\)
Để P > 1/6 thì \(\frac{\sqrt{a}-2}{3\sqrt{a}}>\frac{1}{6}\)
<=> \(\frac{\sqrt{a}-2}{3\sqrt{a}}-\frac{1}{6}>0\)
<=> \(\frac{2\sqrt{a}-4}{6\sqrt{a}}-\frac{\sqrt{a}}{6\sqrt{a}}>0\)
<=> \(\frac{\sqrt{a}-4}{6\sqrt{a}}>0\)
Dễ thấy \(6\sqrt{a}>0\forall x>0\)
=> \(\sqrt{a}-4>0\)<=> \(\sqrt{a}>4\)<=> \(a>16\)
Vậy với a > 16 thì P > 1/6
a) P=√a−23√a .
b) Để P>16 thì √a−23√a >16 .
Vì √a>0 thỏa mãn điều kiện xác định nên để √a−46√a >0 thì √a−4>0.
Câu hỏi thuộc chủ đề: R
P= \(\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
a, \(\dfrac{\sqrt{a}-2}{\sqrt{a}}\)
b, a ≥ 6
a) .
b) Để thì .
Vì thỏa mãn điều kiện xác định nên để thì
\(ĐKXĐ: \text{x ≥ 0 }\); \(x \ne 1\); \(x \ne 4\)
\(a, P=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(P=\left(\dfrac{\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{\sqrt{a}-1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right) : \left(\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}-\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\right)\)
\(P=\dfrac{\sqrt{a}-\left(\sqrt{a}-1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)} : \dfrac{a-1-\left(a-4\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(P=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)} : \dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(P=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)} : \dfrac{3}{\left(\sqrt{a}=2\right)\left(\sqrt{a}-1\right)} \)
\(P=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot \dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}\)
\(P= \dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
b) Để \(P>\dfrac{1}{6}\)
\(\Leftrightarrow\dfrac{\sqrt{a}-2}{3\sqrt{a}}>\dfrac{1}{6}\)
\(\Leftrightarrow\dfrac{\sqrt{a}-2}{3\sqrt{a}}-\dfrac{1}{6}>0\)
\(\Leftrightarrow\dfrac{2\left(\sqrt{a}-2\right)-\sqrt{a}}{6\sqrt{a}}>0\)
\(\Leftrightarrow\dfrac{2\sqrt{a}-4-\sqrt{a}}{6\sqrt{a}}>0\)
\(\Leftrightarrow\dfrac{\sqrt{a}-4}{6\sqrt{a}}>0 \left(1\right)\)
Ta có: \(a\ge0\Rightarrow\sqrt{a}\ge0\Rightarrow6\sqrt{a}\ge0\left(2\right)\)
Từ (1) và (2), suy ra:
\(\Rightarrow\sqrt{a}-4>0 \)
\(\Leftrightarrow\sqrt{a}>4\)
\(\Leftrightarrow a>16\)
Kết hợp với điều kiện xác định
Vậy để \(P>\dfrac{1}{6}\) thì \(a>16\)
P=\(\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
a=2
a, P=\(\dfrac{1}{\sqrt{a}\cdot\left(\sqrt{a}-1\right)}\)/\(\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\cdot\left(\sqrt{a}-1\right)}\)=\(\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
b,để P >\(\dfrac{1}{6}\) thì \(\dfrac{\sqrt{a}-2}{3\sqrt{a}}>\dfrac{1}{6}\)
⇒6*\(\sqrt{a}-2>3\sqrt{a}\)
⇔\(6\sqrt{a}-12-3\sqrt{a}>0\)
⇔\(3\sqrt{a}>12\)
⇔\(\sqrt{a}>4\)
⇔a>16
a,P=\(\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
b, để P>\(\dfrac{1}{6}\) khi a>16
Điều kiện xác định : \(\left\{{}\begin{matrix}a>0\\a\ne1\\a\ne2\end{matrix}\right.\)
a, Rút gọn bt:
P = \(\left(\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right).\left(\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{3}\right)\)
P = \(\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
Vậy P = \(\dfrac{\sqrt{a}-2}{3\sqrt{a}}\) với \(\left\{{}\begin{matrix}a>0\\a\ne1\\a\ne2\end{matrix}\right.\)
b, Ta có P > \(\dfrac{1}{6}\)
\(\Rightarrow\) \(\dfrac{\sqrt{a}-2}{3\sqrt{a}}\) > \(\dfrac{1}{6}\)
\(\Leftrightarrow\)\(\dfrac{\sqrt{a}-2}{3\sqrt{a}}-\dfrac{1}{6}>0\)
\(\Leftrightarrow\dfrac{2\sqrt{a}-4-\sqrt{a}}{6\sqrt{a}}>0\)
\(\Leftrightarrow\dfrac{\sqrt{a}-4}{6\sqrt{a}}>0\)
Ta có : \(6\sqrt{a}>0\)
\(\Rightarrow\sqrt{a}>4\)
\(\Leftrightarrow a>16\)
Vậy a > 16 để P > \(\dfrac{1}{6}\)
a Rút gọn biểu thức p =
a/ đkxd a>0; a≠1;a≠2
P = \(\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}.\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-1\right).\left(\sqrt{a}-2\right)}\)=\(\dfrac{1}{\sqrt{a}.\left(\sqrt{a}-1\right)}.\dfrac{\left(\sqrt{a}-1\right).\left(\sqrt{a}-2\right)}{3}=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
Vậy P= \(\dfrac{\sqrt{a}-2}{3\sqrt{a}}\) với a>0;a≠1;a≠2
b/ ta có P>\(\dfrac{1}{6}\)⇔\(\dfrac{\sqrt{a}-2}{3\sqrt{a}}-\dfrac{1}{6}=\dfrac{2\sqrt{a}-4-\sqrt{a}}{6\sqrt{a}}=\dfrac{\sqrt{a}-4}{6\sqrt{a}}\)
Ta có a>0⇒\(\sqrt{a}\)>0⇒6\(\sqrt{a}\)>0
Để \(\dfrac{\sqrt{a}-4}{6\sqrt{a}}>0\) mà 6\(\sqrt{a}\)>0⇒\(\sqrt{a}\)-4>0⇒\(\sqrt{a}\)>4⇒a>16
Vậy a>16 thì P>\(\dfrac{1}{6}\)
a)=\(\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
b)=\(a>16\)
a) Với a>0;a≠1;a≠4,ta có:
P=\([\dfrac{\sqrt{a}}{(\sqrt{a}-1)\times\sqrt{a}}-\dfrac{\sqrt{a}-1}{(\sqrt{a}-1)\times\sqrt{a}}]\div[\dfrac{(\sqrt{a}+1)\times(\sqrt{a}-1)}{(\sqrt{a}-2)\times(\sqrt{a}-1)}-\dfrac{(\sqrt{a}+2)\times(\sqrt{a}-2)}{(\sqrt{a}-2)\times(\sqrt{a}-1)}]\)
=\([\dfrac{\sqrt{a}-(\sqrt{a}-1)}{(\sqrt{a}-1)\times\sqrt{a}}]\div[\dfrac{(a-1)-(a-4)}{(\sqrt{a}-2)\times(\sqrt{a}-1)}]\)
=\(\dfrac{1}{(\sqrt{a}-1)\times\sqrt{a}}\div\dfrac{3}{(\sqrt{a}-2)\times(\sqrt{a}-1)}\)
=\(\dfrac{1}{(\sqrt{a}-1)\times\sqrt{a}}\times\dfrac{(\sqrt{a}-2)\times(\sqrt{a}-1)}{3}\)
=\(\dfrac{(\sqrt{a}-2)\times(\sqrt{a}-1)}{(\sqrt{a}-1)\times\sqrt{a}\times3}\)
P=\(\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
Vậy với a>0; a≠1;a≠4 thì P=\(\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
b) Để P>\(\dfrac{1}{6}\)
⇔\(\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)>\(\dfrac{1}{6}\)
⇔\(\dfrac{(\sqrt{a}-2)\times6}{3\sqrt{a}}\)>\(1\)
⇔\(6\sqrt{a}-12\)>\(3\sqrt{a}\)
⇔3\(\sqrt{a}\)>12
⇔\(3\sqrt{a}\)>12
⇔\(\sqrt{a}\)>4⇔a>16(vì 4>0;thỏa mãn ∀a>0;a≠1;a≠4)
Vậy với a>16 thì P>\(\dfrac{1}{6}\)
a) P=√a−23√a .
b) Để P>16 thì √a−23√a >16 .
Vì √a>0 thỏa mãn điều kiện xác định nên để √a−46√a >0 thì
a) P=\(\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}\):\(\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\)\(\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\)\(.\)\(\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}\)\(=\)\(\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
b) P\(>\)\(\dfrac{1}{6}\)⇔\(\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)\(>\)\(\dfrac{1}{6}\)
⇔\(\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)\(-\)\(\dfrac{1}{6}\)\(=\)0⇔\(\dfrac{2\sqrt{a}-4-\sqrt{a}}{6\sqrt{a}}\)\(=\)0⇔\(\dfrac{\sqrt{a}-4}{6\sqrt{a}}\)\(=\)0⇔\(\sqrt{a}-4\)\(=\)0⇔\(\sqrt{a}\)\(=\)4⇔\(a=2\)
