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\(a,A=\dfrac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}:\dfrac{x-2-x+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\\ A=\dfrac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}\\ A=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
a, ĐKXĐ : \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
Ta có : \(P=\left(\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}\right).\dfrac{\left(x-1\right)^2}{2}\)
\(=\left(\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right).\dfrac{\left(x-1\right)^2}{2}\)
\(=\dfrac{x-2\sqrt{x}+\sqrt{x}-2-x-2\sqrt{x}+\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\dfrac{\left(x-1\right)^2}{2}\)
\(=\dfrac{-2\sqrt{x}}{\left(x-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\left(x-1\right)^2}{2}\)
\(=\dfrac{-\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)}=-\sqrt{x}\left(\sqrt{x}-1\right)\)
b, Ta có : \(P=-x+\sqrt{x}=-x+\dfrac{2.\sqrt{x}.1}{2}-\dfrac{1}{4}+\dfrac{1}{4}\)
\(=-\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\le\dfrac{1}{4}\)
Vậy \(Max=\dfrac{1}{4}\Leftrightarrow x=\dfrac{1}{4}\)
Lời giải:
ĐKXĐ: $x\geq 0; x\neq 1$
a.
\(A=\frac{(\sqrt{x}-2)(x-1)}{2}-\frac{(\sqrt{x}+2)(1-x)^2}{2(x+2\sqrt{x}+1)}=\frac{(\sqrt{x}-2)(x-1)}{2}-\frac{(\sqrt{x}+2)(\sqrt{x}-1)^2(\sqrt{x}+1)^2}{2(\sqrt{x}+1)^2}\)
\(=\frac{(\sqrt{x}-2)(x-1)}{2}-\frac{(\sqrt{x}+2)(\sqrt{x}-1)^2}{2}=\frac{2\sqrt{x}-2x}{2}=\sqrt{x}-x\)
b.
$\sqrt{x}-x=\frac{1}{4}-(x-\sqrt{x}+\frac{1}{4})$
$=\frac{1}{4}-(\sqrt{x}-\frac{1}{2})^2$
$\leq \frac{1}{4}$
Vậy GTLN của biểu thức là $\frac{1}{4}$. Giá trị này đạt tại $\sqrt{x}-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{4}$ (thỏa đkxđ)
a: \(=\dfrac{x-2\sqrt{x}+1-x-2\sqrt{x}-1}{x-1}\cdot\left(\dfrac{1-x}{2\sqrt{x}}\right)^2\)
\(=\dfrac{-4\sqrt{x}}{x-1}\cdot\dfrac{\left(x-1\right)^2}{4x}=\dfrac{-\left(x-1\right)}{\sqrt{x}}\)
b: \(\dfrac{P}{\sqrt{x}}>2\)
=>\(-\dfrac{\left(x-1\right)}{\sqrt{x}}\cdot\dfrac{1}{\sqrt{x}}>2\)
=>\(\dfrac{-x+1}{x}-2>0\)
=>\(\dfrac{-x+1-2x}{x}>0\)
\(\Leftrightarrow\dfrac{3x-1}{x}< 0\)
=>0<x<1/3
a) đk: \(x\ne0;4\); \(x>0\)
P = \(\left[\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{1}{\sqrt{x}-2}\right]\times\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)\)
= \(\dfrac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\times\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)\)
= \(\dfrac{1}{\sqrt{x}\left(\sqrt{x}-2\right)}.\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
b) Để P < \(\dfrac{1}{2}\)
<=> \(\dfrac{\sqrt{x}-1}{\sqrt{x}}< \dfrac{1}{2}\)
<=> \(1-\dfrac{1}{\sqrt{x}}< \dfrac{1}{2}\)
<=> \(\dfrac{1}{\sqrt{x}}>\dfrac{1}{2}\)
<=> \(\sqrt{x}< 2\)
<=> x < 4
<=> 0 < x < 4
Nếu có thêm điều kiện \(y>1\) thì kết quả là \(\dfrac{1}{x-1}\)
a) Ta có:
\(P=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3x+3}{x-9}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(=\left(\frac{2\sqrt{x}\left(\sqrt{x-3}\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x-3}\right)}+\frac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\frac{3x+3}{x-9}\right):\frac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)
\(=\left(\frac{2x-6}{x-9}+\frac{x+3\sqrt{x}}{x-9}-\frac{3x+3}{x-9}\right):\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(=\frac{2x-6+x+3\sqrt{x}-3x-3}{x-9}.\frac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\frac{3\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}+3}\)
\(=\frac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}+3}\)
\(=\frac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)^2}\)
b) \(P< \frac{-1}{2}\Rightarrow\frac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)^2}< \frac{-1}{2}\)
.....Chưa nghĩ ra....
c) Ta có: \(\frac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)^2}\ge0\)
Dấu "=" xảy ra \(\Leftrightarrow\sqrt{x}-3=0\Rightarrow x=9\)
Vậy Min P = 0 khi x =9.
k - kb với tớ nhia mn!
Bài 1:
a: \(A=\left(\dfrac{x-1}{2\sqrt{x}}\right)^2\cdot\dfrac{x-2\sqrt{x}+1-x-2\sqrt{x}-1}{x-1}\)
\(=\dfrac{\left(x-1\right)^2}{4x}\cdot\dfrac{-4\sqrt{x}}{x-1}=\dfrac{-\left(x-1\right)}{\sqrt{x}}\)
b: Để B<0 thì -x+1<0
=>-x<-1
hay x>1
c: Để B=2 thì \(-\left(x-1\right)=2\sqrt{x}\)
\(\Leftrightarrow-x+1-2\sqrt{x}=0\)
\(\Leftrightarrow x+\sqrt{x}-1=0\)
\(\Leftrightarrow\sqrt{x}=\dfrac{\sqrt{5}-1}{2}\)
hay \(x=\dfrac{6-2\sqrt{5}}{4}\)
\(P=\left(\dfrac{\sqrt{x}-2}{x-1}-\dfrac{\sqrt{x}+2}{x+2\sqrt{x}}\right).\dfrac{\left(1-x\right)^2}{2}\) (ĐK:\(x>0;x\ne1\))
\(=\left[\dfrac{\sqrt{x}-2}{x-1}-\dfrac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}+2\right)}\right].\dfrac{\left(x-1\right)^2}{2}\)
\(=\left[\dfrac{\left(\sqrt{x}-2\right)\sqrt{x}}{\left(x-1\right)\sqrt{x}}-\dfrac{x-1}{\sqrt{x}\left(x-1\right)}\right].\dfrac{\left(x-1\right)^2}{2}\)
\(=\dfrac{x-2\sqrt{x}-x+1}{\sqrt{x}\left(x-1\right)}.\dfrac{\left(x-1\right)^2}{2}=\dfrac{-2\sqrt{x}+1}{\sqrt{x}\left(x-1\right)}.\dfrac{\left(x-1\right)^2}{2}\)
\(=\dfrac{\left(-2\sqrt{x}+1\right)\left(x-1\right)}{2\sqrt{x}}\)
Sai đề ko em?
a) Ta có: \(P=\left(\dfrac{\sqrt{x}-2}{x-1}-\dfrac{\sqrt{x}+2}{x+2\sqrt{x}}\right)\cdot\dfrac{\left(1-x\right)^2}{2}\)
\(=\left(\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{1}{\sqrt{x}}\right)\cdot\dfrac{\left(x-1\right)^2}{2}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)-\left(x-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)^2\cdot\left(\sqrt{x}+1\right)^2}{2}\)
\(=\dfrac{x-2\sqrt{x}-x+1}{\sqrt{x}}\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{2}\)
\(=\dfrac{\left(-2\sqrt{x}+1\right)\left(x-1\right)}{2\sqrt{x}}\)