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\(P=\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{\sqrt{x}-2}{\sqrt{x}-1}\)
ĐKXĐ : \(\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)
\(=\frac{\sqrt{x}+\sqrt{x}-2}{\sqrt{x}-1}\)
\(=\frac{2\sqrt{x}-2}{\sqrt{x}-1}\)
\(=\frac{2\left(\sqrt{x}-1\right)}{\sqrt{x}-1}=2\)
=> Với mọi \(\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)thì P = 2
Đề sai à --
a/ \(P=\left[\frac{1}{\sqrt{x}+1}-\frac{2\left(\sqrt{x}-1\right)}{\sqrt{x}\left(x-1\right)+x-1}\right]:\left[\frac{1}{\sqrt{x}-1}-\frac{2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right]\)
\(=\left[\frac{1}{\sqrt{x}+1}-\frac{2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right]:\left[\frac{\sqrt{x}+1-2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right]\)
\(=\left[\frac{1}{\sqrt{x}+1}-\frac{2}{\left(\sqrt{x}+1\right)^2}\right]:\left[\frac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right]\)
\(=\frac{\sqrt{x}+1-2}{\left(\sqrt{x}+1\right)^2}.\left(\sqrt{x}+1\right)=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)
b/ Ta có: \(P=\frac{\sqrt{x}-1}{\sqrt{x}+1}=1-\frac{2}{\sqrt{x}+1}\)
Để \(P\in Z\) thì \(\left(\sqrt{x}+1\right)\inƯ\left(2\right)=\left\{1;-1;2;-2\right\}\)
+ Với \(\sqrt{x}+1=1\Rightarrow\sqrt{x}=0\Rightarrow x=0\)
+ Với \(\sqrt{x}+1=-1\Rightarrow\sqrt{x}=-2\left(vn\right)\)
+ Với \(\sqrt{x}+1=2\Rightarrow\sqrt{x}=1\Rightarrow x=1\)(loại)
+ Với \(\sqrt{x}+1=-2\Rightarrow\sqrt{x}=-3\left(vn\right)\)
Vậy x = 0 thì P nguyên
a) \(P=\left(\frac{1}{\sqrt{x}+1}-\frac{2\sqrt{x}-2}{x\sqrt{x}-\sqrt{x}+x-1}\right):\left(\frac{1}{\sqrt{x}-1}-\frac{2}{x-1}\right)\)
\(=\frac{x-1-2\sqrt{x}+2}{\left(\sqrt{x}+1\right)\left(x-1\right)}:\frac{\sqrt{x}+1-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}-1\right)^2}{\left(x-1\right)\left(\sqrt{x}+1\right)}.\frac{x-1}{\sqrt{x}-1}=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)
b) \(P=\frac{\sqrt{x}-1}{\sqrt{x}+1}=\frac{\sqrt{x}+1-2}{\sqrt{x}+1}=1-\frac{2}{\sqrt{x}+1}\)
Để P nguyên thì \(\sqrt{x}+1\in\left\{1;2\right\}\Leftrightarrow x\in\left\{0\right\}\) (Vì x khác 1 - điều kiện)
c) \(\sqrt{x}+1\ge1\Leftrightarrow\frac{2}{\sqrt{x}+1}\le\frac{1}{2}\Leftrightarrow1-\frac{2}{\sqrt{x}+1}\ge\frac{1}{2}\)
\(\Rightarrow P\ge\frac{1}{2}\). Dấu đẳng thức xảy ra khi x = 0
Vậy Min P = 1/2 <=> x = 0
a)
ĐKXĐ: \(x-4\ge0\text{ (1)};\text{ }x+4\sqrt{x-4}\ge0\text{ (2); }\frac{16}{x^2}-\frac{8}{x}+1>0\text{ (3)}\)
\(\left(1\right)\Leftrightarrow x\ge4\)
\(\left(2\right)\Leftrightarrow\left(\sqrt{x-4}+2\right)^2\ge0\text{ (đúng }\forall x\ge4\text{)}\)
\(\left(3\right)\Leftrightarrow\left(\frac{4}{x}-1\right)^2>0\Leftrightarrow\frac{4}{x}-1\ne0\Leftrightarrow x\ne4\)
Vậy ĐKXĐ là \(x>4\)
b)
\(A=\frac{\left|\sqrt{x-4}+2\right|+\left|\sqrt{x-4}-2\right|}{\left|\frac{4}{x}-1\right|}=\frac{\sqrt{x-4}+2+\left|\sqrt{x-4}-2\right|}{1-\frac{4}{x}}=\frac{x\left(\sqrt{x-4}+2+\left|\sqrt{x-4}-2\right|\right)}{x-4}\)
\(+\sqrt{x-4}\le2\Leftrightarrow0<\)\(x-4\le4\)
thì \(A=\frac{x\left(\sqrt{x-4}+2+2-\sqrt{x-4}\right)}{x-4}=\frac{4x}{x-4}=4+\frac{16}{x-4}\)
A nguyên khi \(\frac{16}{x-4}\)nguyên hay \(x-4\inƯ\left(16\right)\)
Mà \(0<\)\(x-4\le4\)
Nên \(x-4\in\left\{2;4\right\}\Rightarrow x\in\left\{6;8\right\}\)
\(+\text{Xét }\sqrt{x-4}>2\Leftrightarrow x-4>4\)
\(A=\frac{x\left(\sqrt{x-4}+2+\sqrt{x-4}-2\right)}{x-4}=\frac{2x\sqrt{x-4}}{x-4}=\frac{2x}{\sqrt{x-4}}\)
Nếu \(\sqrt{x-4}\)là số vô tỉ thì A là số vô tỉ.
Để A là hữu tỉ thì \(\sqrt{x-4}=t\text{ }\left(t\in Z;\text{ }t>4\right)\Rightarrow x=t^2+4\)
Khi đó, \(A=\frac{2\left(t^2+4\right)}{t}=2t+\frac{8}{t}\)
A nguyên khi \(\frac{8}{t}\) nguyên hay \(t=8\text{ (do }t>4\text{)}\)
\(t=\sqrt{x-4}=8\Leftrightarrow x=8^2+4=68\)
Vậy \(x\in\left\{6;8;68\right\}\)
c/
\(+0<\sqrt{x-4}\)\(<2\)
Thì \(A=4+\frac{16}{x-4}>4+\frac{16}{4}=8\)
\(+\sqrt{x-4}\ge2\)
\(A=\frac{2x}{\sqrt{x-4}}=2t+\frac{8}{t}\text{ (}t=\sqrt{x-4}\ge2\text{)}\)
Mà \(t+\frac{4}{t}\ge2\sqrt{t.\frac{4}{t}}=4\)
\(\Rightarrow A\ge2.4=8\)
Dấu "=" xảy ra khi \(t=\frac{4}{t}\Leftrightarrow t=2\Leftrightarrow\sqrt{x-4}=2\Leftrightarrow x=8\)
Vậy GTNN của A là 8 khi x = 8.
Tự tìm ĐKXĐ nhé
\(P=\frac{1}{\sqrt{x}+2}-\frac{5}{x-\sqrt{x}-6}-\frac{\sqrt{x}-2}{3-\sqrt{x}}\)
\(=\frac{1}{\sqrt{x}+2}-\frac{5}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}-2}{\sqrt{x}-3}\)
\(=\frac{\sqrt{x}-3}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}-\frac{5}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}+\frac{x-4}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{\sqrt{x}-3-5+x-4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{x+\sqrt{x}-12}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+4\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{\sqrt{x}+4}{\sqrt{x}+2}\)
c, \(P=\frac{\sqrt{x}+4}{\sqrt{x}+2}=\frac{\sqrt{x}+2+2}{\sqrt{x}+2}=1+\frac{2}{\sqrt{x}+2}\)
Để \(P\in Z\Rightarrow1+\frac{2}{\sqrt{x}+2}\in Z\)
\(\Rightarrow\sqrt{x}+2\inƯ\left(2\right)=\left\{1;2;-1;-2\right\}\)
\(\Rightarrow\sqrt{x}=\left\{-1;0\right\}\)
\(\Rightarrow x=\left\{0\right\}\)
Kết hợp với ĐKXĐ =>...
a,b) Đk để biểu thức A xác định là x > 4
\(A=\frac{x\left(\sqrt{\left(\sqrt{x-4}+2\right)^2}+\sqrt{\left(\sqrt{x-4}-2\right)^2}\right)}{\sqrt{\left(x-4\right)^2}}\)
\(A=\frac{x\left(|\sqrt{x-4}+2|+|\sqrt{x-4}-2|\right)}{|x-4|}\)
\(A=\frac{x\left(\sqrt{x-4}+2+|\sqrt{x-4}-2|\right)}{x-4}\)
+) Nếu 4 < x < 8 thì \(\sqrt{x-4}-2< 0\)nên \(A=\frac{x\left(\sqrt{x-4}+2+2-\sqrt{x-4}\right)}{x-4}=\frac{4x}{x-4}=4+\frac{16}{x-4}\)
Do 4 < x < 8 nên 0 < x - 4 < 4 => A > 88
+) Nếu \(x\ge8\)thì \(\sqrt{x-4}-2\ge0\)nên :
\(A=\frac{x\left(\sqrt{x-4}+2+\sqrt{x-4}-2\right)}{x-4}=\frac{2x\sqrt{x-4}}{x-4}=\frac{2x}{\sqrt{x-4}}=2\sqrt{x-4}+\frac{8}{\sqrt{x-4}}\ge2\sqrt{16}=8\)
( Theo bđt Cô si )
- Dấu " = " xảy ra khi và chỉ khi \(2\sqrt{x-4}=\frac{8}{\sqrt{x-4}}\Leftrightarrow x-4=4\Leftrightarrow x=8\)
Vậy Min của A = 8 khi x = 8
c) Xét 4 < x < 8 thì \(A=4+\frac{16}{x-4}\), ta thấy \(A\in Z\)khi và chỉ khi \(\frac{16}{x-4}\in Z\Leftrightarrow x-4\)là ước nguyên dương của 16
- Hay \(x-4\in\left\{1;2;4;16\right\}\Leftrightarrow x=\left\{5;6;8;12;20\right\}\)đối chiếu điều kiện => x = 5 hoặc x = 6
+) Xét \(x\ge8\)ta có : \(A=\frac{2x}{\sqrt{x-4}}\)
Đặt \(\sqrt{x-4}=m\Rightarrow\hept{\begin{cases}x=m^2+4\\m\ge2\end{cases}}\)khi đó ta có : \(A=\frac{2\left(m^2+4\right)}{m}=2m+\frac{8}{m}\)
\(\Rightarrow m\in\left\{2;4;8\right\}\Leftrightarrow x\in\left\{8;20;68\right\}\)
Vậy để A nhận giá trị nguyên thì \(x\in\left\{5;6;8;20;68\right\}\)
ĐK : x >= 0
\(B=\frac{\sqrt{x}-2}{\sqrt{x}+1}=\frac{\sqrt{x}+1-3}{\sqrt{x}+1}=1-\frac{3}{\sqrt{x}+1}\)
Để B nguyên thì \(\frac{3}{\sqrt{x}+1}\)nguyên hay \(\sqrt{x}+1\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\Leftrightarrow x\in\left\{0;4\right\}\)
Ta có : \(\sqrt{x}+1\ge1\forall x\ge0\Rightarrow\frac{3}{\sqrt{x}+1}\le3\Leftrightarrow1-\frac{3}{\sqrt{x}+1}\ge-2\)
Dấu "=" xảy ra <=> x = 0 . Vậy MinB = -2