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ĐKXĐ:
\(\left\{{}\begin{matrix}2-\sqrt{x}\\2+\sqrt{x}\\x-4\end{matrix}\right.\ne0\Leftrightarrow x\ne4\)
P=\(\dfrac{\left(2+\sqrt{x}\right)\left(2+\sqrt{x}\right)-\left(2-\sqrt{x}\right)\left(2-\sqrt{x}\right)+4x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\)
P=\(\dfrac{\left(4+4\sqrt{x}+x\right)-\left(4-4\sqrt{x}+x\right)+4x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\)
P=\(\dfrac{4+4\sqrt{x}+x-4+4\sqrt{x}-x+4x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\)
P=\(\dfrac{8\sqrt{x}+4x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\)
P=\(\dfrac{4\sqrt{x}\left(2+\sqrt{x}\right)}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\\ P=\dfrac{4\sqrt{x}}{2-\sqrt{x}}\)
b) Th P>0
<=> \(\dfrac{4\sqrt{x}}{2-\sqrt{x}}\)>0<=>\(4\sqrt{x}\)>0 <=> x>0(x\(\ne\)4)
TH P < 0
<=>\(\dfrac{4\sqrt{x}}{2-\sqrt{x}}\)<0 <=>\(4\sqrt{x}\)<0<=> \(\sqrt{x}< 0\)(vô lý)
c) |P|=1
=>P=1 hoặc P=-1
TH P=1
=>\(\dfrac{4\sqrt{x}}{2-\sqrt{x}}\)=1 <=> \(4\sqrt{x}\)=\(2-\sqrt{x}\) <=> x=\(\dfrac{4}{25}\)
TH P= -1
=>\(\dfrac{4\sqrt{x}}{2-\sqrt{x}}\)=-1<=> \(4\sqrt{x}\)=\(\sqrt{x}-2\)<=> \(\sqrt{x}=-\dfrac{2}{3}\)(vô lý)
a: ĐKXĐ: x>0; x<>1
\(P=\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}}\right):\dfrac{1}{x+\sqrt{x}}\)
\(=\dfrac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{1}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
b: Để P<0 thì \(\sqrt{x}-1< 0\)
=>0<x<1
c: Để P là số nguyên thì \(\sqrt{x}-1+2⋮\sqrt{x}-1\)
\(\Leftrightarrow\sqrt{x}-1\in\left\{1;-1;2\right\}\)
hay \(x\in\left\{4;0;9\right\}\)
a, \(A=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\right)\left(1-\dfrac{1}{\sqrt{x}}\right)\left(đkxđ:x>0,x\ne1\right)\)
\(A=\left(\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\left(\dfrac{\sqrt{x}}{\sqrt{x}}-\dfrac{1}{\sqrt{x}}\right)\)
\(A=\left(\dfrac{x+2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\left(\dfrac{\sqrt{x}-1}{\sqrt{x}}\right)\)
\(A=\left(\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
\(A=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
b, A=1
\(\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}}=1\)
\(\Rightarrow\sqrt{x}+1=\sqrt{x}\)
\(\Leftrightarrow\sqrt{x}-\sqrt{x}+1=0\)
\(\Leftrightarrow1=0\) ( vô lý)
vậy không có giá trị của x để A=1
a) \(ĐKXĐ:\hept{\begin{cases}x>0\\x\ne9\\x\ne4\end{cases}}\)
\(P=\left(\frac{2+\sqrt{x}}{2-\sqrt{x}}-\frac{2-\sqrt{x}}{2+\sqrt{x}}-\frac{4x}{x-4}\right):\frac{\sqrt{x}-3}{2\sqrt{x}-x}\)
\(\Leftrightarrow P=\frac{\left(2+\sqrt{x}\right)^2-\left(2-\sqrt{x}\right)^2+4x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}:\frac{\sqrt{x}-3}{\sqrt{x}\left(2-\sqrt{x}\right)}\)
\(\Leftrightarrow P=\frac{4+4\sqrt{x}+x-4+4\sqrt{x}-x+4x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\cdot\frac{\sqrt{x}\left(2-\sqrt{x}\right)}{\sqrt{x}-3}\)
\(\Leftrightarrow P=\frac{8\sqrt{x}+4x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\cdot\frac{\sqrt{x}\left(2-\sqrt{x}\right)}{\sqrt{x}-3}\)
\(\Leftrightarrow P=\frac{4x\left(2+\sqrt{x}\right)}{\left(2+\sqrt{x}\right)\left(\sqrt{x}-3\right)}\)
\(\Leftrightarrow P=\frac{4x}{\sqrt{x}-3}\)
b) Để P < 0
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}-3< 0\Leftrightarrow4x>0\\\sqrt{x}-3>0\Leftrightarrow4x< 0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}< 3\Leftrightarrow x>0\\\sqrt{x}>3\Leftrightarrow x< 0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x< 9\Leftrightarrow x>0\left(ktm\right)\\x>9\Leftrightarrow x< 0\left(ktm\right)\end{cases}}\)
Vậy để \(P< 0\Leftrightarrow x\in\varnothing\)
Để P > 0
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}-3>0\Leftrightarrow4x>0\\\sqrt{x}-3< 0\Leftrightarrow4x< 0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}>3\Leftrightarrow x>0\left(tm\right)\\\sqrt{x}< 3\Leftrightarrow x< 0\left(ktm\right)\end{cases}}\)
\(\Leftrightarrow x>9\Leftrightarrow x>0\left(tm\right)\)
Vậy để \(P>0\Leftrightarrow x>9\)
c) Để \(\left|P\right|=1\)
\(\Leftrightarrow\orbr{\begin{cases}P=1\left(tm\right)\\P=-1\left(ktm\right)\end{cases}}\)
\(\Leftrightarrow\frac{4x}{\sqrt{x}-3}=1\)
\(\Leftrightarrow4x=\sqrt{x}-3\)
\(\Leftrightarrow4x-\sqrt{x}+3=0\)
\(\Leftrightarrow\left(2\sqrt{x}-\frac{1}{4}\right)^2+\frac{47}{48}=0\left(ktm\right)\)
Vậy để \(\left|P\right|=1\Leftrightarrow x\in\varnothing\)
a) Bạn dư sức làm.
b) \(A=\dfrac{x^2+\sqrt{x}}{x-\sqrt{x}+1}+1-\dfrac{2x+\sqrt{x}}{\sqrt{x}}\)
\(=\dfrac{\sqrt{x}\cdot\left(x\sqrt{x}+1\right)}{x-\sqrt{x}+1}+1-\dfrac{2x+\sqrt{x}}{\sqrt{x}}\)
\(=\dfrac{\sqrt{x}\cdot\left(\sqrt{x}+1\right)\cdot\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}+1-\dfrac{2x+\sqrt{x}}{\sqrt{x}}\)
\(=\sqrt{x}\cdot\left(\sqrt{x}+1\right)+1-\dfrac{2x+\sqrt{x}}{\sqrt{x}}\)
\(=x+\sqrt{x}+1-\dfrac{2x+\sqrt{x}}{\sqrt{x}}\)
\(=\dfrac{x\sqrt{x}+x+\sqrt{x}-\left(2x+\sqrt{x}\right)}{\sqrt{x}}\)
\(=\dfrac{x\sqrt{x}+x+\sqrt{x}-2x-\sqrt{x}}{\sqrt{x}}\)
\(=\dfrac{x\sqrt{x}-x}{\sqrt{x}}\)
\(=\dfrac{\left(x\sqrt{x}-x\right)\sqrt{x}}{x}\)
\(=\dfrac{x\cdot\left(\sqrt{x}-1\right)\sqrt{x}}{x}\)
\(=\left(\sqrt{x}-1\right)\sqrt{x}\)
\(=x-\sqrt{x}\)
Bài 1:
\(a,E=\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{2\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\\ =\dfrac{x-2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\\ =\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}\left(\sqrt{x}-1\right)}\\ =\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
\(b,E>0\Leftrightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}}>0\)
Mà: \(\sqrt{x}>0\\ \Rightarrow\sqrt{x}-1>0\\ \Leftrightarrow\sqrt{x}>1\\ \Leftrightarrow x>1\)
Bài 2:
\(a,G=\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{1}{1-\sqrt{x}}-\dfrac{2\sqrt{x}}{x-1}\right)\left(\sqrt{x}+1\right)\\ =\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}+\dfrac{1}{\sqrt{x}-1}-\dfrac{2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\left(\sqrt{x}+1\right)\\ =\left(\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)+\sqrt{x}+1-2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\left(\sqrt{x}+1\right)\\ =\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}.\left(\sqrt{x}+1\right)\\ =\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}\\ =\sqrt{x}-1\)
\(M=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right)\div\dfrac{\sqrt{x}-1}{2}\)
(ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\))
\(=\left[\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\right]\times\dfrac{2}{\sqrt{x}-1}\)
\(=\dfrac{\left(x+2\right)+\sqrt{x}\left(\sqrt{x}-1\right)-\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\times\dfrac{2}{\sqrt{x}-1}\)
\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{1}{x+\sqrt{x}+1}\)
\(M=\dfrac{1}{\left(\sqrt{x}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\le1\)
Dấu "=" xảy ra khi x = 0
Cảm ơn nhé! Nhưng tớ làm ra câu a,b rồi :( cậu biết làm c,d không?
Để C<1 thì C-1<0
\(\Leftrightarrow\dfrac{-1-\sqrt{x}+2}{\sqrt{x}-2}< 0\)
\(\Leftrightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}-2}>0\)
=>x>4 hoặc x<1