Cho biểu thức
P=\(\frac{x+12}{x-4}\)+\(\frac{1}{\sqrt{x}+2}\)-\(\frac{4}{\sqrt{x}-2}\) với x\(\ge\)0, x\(\ne\)4 .a) Rút gọn P b) Tính P sau khi x=25 c) Tìm x \(\in\)Z để \(\frac{1}{P}\)là số nguyên d) Tìm x để P nhận giá trị nguyên
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a/ \(P=\frac{2}{\sqrt{x}-2}:\left(\frac{\sqrt{x}}{x-4}+\frac{1}{\sqrt{x}-2}\right)\left(ĐKXĐ:x\ge0,x\ne4\right)\)
\(\Leftrightarrow P=\frac{2}{\sqrt{x}-2}:\left(\frac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right)\)
\(\Leftrightarrow P=\frac{2}{\sqrt{x}-2}.\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{2\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}+2}{\sqrt{x}+1}\)
b/ \(P=\frac{\sqrt{x}+1+1}{\sqrt{x}+1}=1+\frac{1}{\sqrt{x}+1}\)
$P$ đạt giá trị lớn nhất \(\Leftrightarrow\left(\sqrt{x}+1\right)\) đạt GTNN
Vì \(\sqrt{x}\ge0\Leftrightarrow\sqrt{x}+1\ge1\) đạt giá trị nhỏ nhất là $1$ tại \(x=0\)
Vậy \(MaxP=2\Leftrightarrow x=0\)
KL: ...................
a/ \(P=\frac{2}{\sqrt{x}-2}:\left(\frac{\sqrt{x}}{x-4}+\frac{1}{\sqrt{x}-2}\right)\left(ĐKXĐ:x\ge0,x\ne4\right)\)
\(\Leftrightarrow P=\frac{2}{\sqrt{x}-2}:\left(\frac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right)\)
\(\Leftrightarrow P=\frac{2}{\sqrt{x}-2}.\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{2\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}+2}{\sqrt{x}+1}\)
b/ \(P=\frac{\sqrt{x}+1+1}{\sqrt{x}+1}=1+\frac{1}{\sqrt{x}+1}\)
$P$ đạt giá trị lớn nhất \(\Leftrightarrow\left(\sqrt{x}+1\right)\) đạt GTNN
Vì \(\sqrt{x}\ge0\Leftrightarrow\sqrt{x}+1\ge1\) đạt giá trị nhỏ nhất là $1$ tại \(x=0\)
Vậy \(MaxP=2\Leftrightarrow x=0\)
KL: ...................
a)\(M=\left(\frac{\sqrt{x}+3}{\sqrt{x}-2}+\frac{\sqrt{x}+2}{3-\sqrt{x}}+\frac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right):\left(1-\frac{\sqrt{x}}{\sqrt{x}+1}\right)\)
\(=\left(\frac{x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{x-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\frac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right):\left(\frac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}+1}\right)\)
\(=\frac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}.\left(\sqrt{x}+1\right)\)
\(=\frac{\sqrt{x}+1}{\sqrt{x}-2}\)
b)\(\frac{1}{M}=\frac{\sqrt{x}-2}{\sqrt{x}+1}=\frac{\sqrt{x}+1-3}{\sqrt{x}+1}=1-\frac{3}{\sqrt{x}+1}\)
Ta có: \(\sqrt{x}\ge0,\forall x\ge0\)
\(\Leftrightarrow\sqrt{x}+1\ge1\)
\(\Leftrightarrow\frac{1}{\sqrt{x}+1}\le1\)
\(\Leftrightarrow\frac{3}{\sqrt{x}+1}\le3\)
\(\Leftrightarrow-\frac{3}{\sqrt{x}+1}\ge-3\)
\(\Leftrightarrow1-\frac{3}{\sqrt{x}+1}\ge-2\)
Dấu "=" xảy ra khi x=0
Vậy \(Min_{\frac{1}{M}}=-2\) khi x=0
\(B=\left(\frac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right).\left(\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}+2}\right)\)
\(=\frac{\left(2\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)}=\frac{2\sqrt{x}+2}{\sqrt{x}+2}\)
\(C=\frac{\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)}\left(\frac{2\sqrt{x}+2}{\sqrt{x}+2}-2\right)=\frac{\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)}.\frac{-2}{\left(\sqrt{x}+2\right)}=\frac{-2}{\sqrt{x}+2}\)
Để C nguyên \(\Rightarrow\sqrt{x}+2=Ư\left(2\right)=2\)
\(\Rightarrow\sqrt{x}=0\Rightarrow x=0\)
a/ ĐKXĐ:...
\(B=\left(\frac{\sqrt{x}+\sqrt{x}+2}{x-4}\right).\frac{x-4}{\sqrt{x}+2}=\frac{2\sqrt{x}+2}{\sqrt{x}+2}\)
\(C=\frac{\sqrt{x}+2}{\sqrt{x}-2}\left(\frac{2\sqrt{x}+2}{\sqrt{x}+2}-2\right)\)
\(C=\frac{\sqrt{x}+2}{\sqrt{x}-2}.\frac{2\sqrt{x}+2-2\sqrt{x}-4}{\sqrt{x}+2}=\frac{-2}{\sqrt{x}-2}\)
Để C đạt GT nguyên
\(\Leftrightarrow\sqrt{x}-2\inƯ_{\left(-2\right)}=\left\{\pm2;\pm1\right\}\)
\(\left[{}\begin{matrix}x=0\\x=9\\x=1\\x=16\end{matrix}\right.\)
Sửa đề :
a) \(A=\left(\frac{x-\sqrt{x}}{x-\sqrt{x}-2}+\frac{4}{\sqrt{x}-2}\right):\left(\frac{\sqrt{x}+2}{\sqrt{x}+1}-\frac{x-\sqrt{x}-5}{x-\sqrt{x}-2}\right)\)
\(\Leftrightarrow A=\frac{x-\sqrt{x}+4\sqrt{x}+4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}:\frac{x-4-x+\sqrt{x}+5}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
\(\Leftrightarrow A=\frac{x+3\sqrt{x}+4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}:\frac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
\(\Leftrightarrow A=\frac{x+3\sqrt{x}+4}{\sqrt{x}+1}\)
b) \(A=4\)
\(\Leftrightarrow\frac{x+3\sqrt{x}+4}{\sqrt{x}+1}=4\)
\(\Leftrightarrow x+3\sqrt{x}+4=4\sqrt{x}+4\)
\(\Leftrightarrow x-\sqrt{x}=0\)
\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}=0\\\sqrt{x}=1\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
Vậy \(A=4\Leftrightarrow x\in\left\{0;1\right\}\)
cho hỏi là mẫu biểu thức A là\(\sqrt{x}-3\) hay\(\sqrt{x-3}\)