Question

[DUTREND123456789]

mọi người giúp được bài nào thì giúp mik gấp lắm 5:15 mik hok rồi

Answer 1

[HT.Phong (9A5)]

Bài 9:

a) \(P=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\) (ĐK: \(x\ne1;x\ne4;x>0\))

\(P=\left[\dfrac{\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{\sqrt{a}-1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right]:\left[\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}-\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}\right]\)

\(P=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}\)

\(P=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{3}{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}\)

\(P=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{3}\)

\(P=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\) 

b) \(P>\dfrac{1}{6}\) khi 

\(\Rightarrow\dfrac{\sqrt{a}-2}{3\sqrt{a}}>\dfrac{1}{6}\)

\(\Leftrightarrow\dfrac{\sqrt{a}-2}{3\sqrt{a}}-\dfrac{1}{6}>0\)

\(\Leftrightarrow\dfrac{2\left(\sqrt{a}-2\right)-\sqrt{a}}{2\cdot3\sqrt{a}}>0\)

\(\Leftrightarrow\dfrac{2\sqrt{a}-4-\sqrt{a}}{6\sqrt{a}}>0\)

\(\Leftrightarrow\dfrac{\sqrt{a}-4}{6\sqrt{a}}>0\) 

Mà: \(6\sqrt{a}>0\)

\(\Leftrightarrow\sqrt{a}-4>0\)

\(\Leftrightarrow\sqrt{a}>4\) 

\(\Leftrightarrow a>16\)

Vậy: ... 

Answer 2

[Nguyễn Lê Phước Thịnh]

20:

a: ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x< >1\end{matrix}\right.\)

\(P=\dfrac{x+2+\sqrt{x}\left(\sqrt{x}-1\right)-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{2}{\sqrt{x}-1}\)

\(=\dfrac{1-\sqrt{x}+x-\sqrt{x}}{\left(\sqrt{x}-1\right)^2\left(x+\sqrt{x}+1\right)}\cdot2\)

\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\cdot\dfrac{2}{x+\sqrt{x}+1}=\dfrac{2}{x+\sqrt{x}+1}\)

b: \(x+\sqrt{x}+1=\left(\sqrt{x}+1\right)\cdot\sqrt{x}+1>=1>0\)

2>0

Do đó: \(P=\dfrac{2}{x+\sqrt{x}+1}>0\)

8:

a: ĐKXĐ: \(\left\{{}\begin{matrix}a>0\\a< >1\end{matrix}\right.\)

\(P=\left(\dfrac{a-1}{2\sqrt{a}}\right)^2\cdot\left(\dfrac{\left(\sqrt{a}-1\right)^2-\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\)

\(=\dfrac{\left(a-1\right)^2}{4a}\cdot\dfrac{a-2\sqrt{a}+1-a-2\sqrt{a}-1}{a-1}\)

\(=\dfrac{a-1}{4a}\cdot\left(-4\sqrt{a}\right)=-\dfrac{a-1}{\sqrt{a}}\)

b: P<0

=>\(-\left(a-1\right)< 0\)

=>a-1>0

=>a>1

c: P=-2

=>\(\dfrac{a-1}{\sqrt{a}}=2\)

=>\(a-1=2\sqrt{a}\)

=>\(a-2\sqrt{a}-1=0\)

=>\(\left[{}\begin{matrix}\sqrt{a}=1+\sqrt{2}\left(nhận\right)\\\sqrt{a}=1-\sqrt{2}\left(loại\right)\end{matrix}\right.\Leftrightarrow a=3+2\sqrt{2}\)