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\(M=\dfrac{7\sqrt{a}-2}{2\sqrt{a}+1}\left(đk:a\ge0\right)=\dfrac{3\left(2\sqrt[]{a}+1\right)+\sqrt{a}-5}{2\sqrt{a}+1}=3+\dfrac{\sqrt{a}-5}{2\sqrt{a}+1}\)
Để \(M\in Z,M>0\) thì \(\sqrt{a}-5\ge0\Leftrightarrow a\ge25\) và:
\(\left\{{}\begin{matrix}\sqrt{a}-5⋮2\sqrt{a}+1\\2\sqrt{a}+1⋮2\sqrt{a}+1\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}2\sqrt{a}-10⋮2\sqrt{a}+1\\2\sqrt{a}+1⋮2\sqrt{a}+1\end{matrix}\right.\)
\(\Rightarrow\left(2\sqrt{a}+1\right)-\left(2\sqrt{a}-10\right)⋮2\sqrt{a}+1\)
\(\Rightarrow11⋮2\sqrt{a}+1\Rightarrow2\sqrt{a}+1\inƯ\left(11\right)=\left\{-11;-1;1;11\right\}\)
Do \(\sqrt{a}\ge0\forall a\)
\(\Rightarrow\sqrt{a}\in\left\{0;5\right\}\)
\(\Rightarrow a\in\left\{0\left(loại\right);25\left(nhận\right)\right\}\)
Bài 1
a) \(P=\frac{3a+\sqrt{9a}-3}{a+\sqrt{a}-2}-\frac{\sqrt{a}+1}{\sqrt{a}+2}+\frac{\sqrt{a}-2}{1-\sqrt{a}}\) (ĐK : x\(\ge0\) ; x\(\ne\) 1)
\(=\frac{3a+\sqrt{9a}-3}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}-\frac{\sqrt{a}+1}{\sqrt{a}+2}-\frac{\sqrt{a}-2}{\sqrt{a}-1}\)
\(=\frac{3a+\sqrt{9a}-3-\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)-\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}\)
\(=\frac{3a+\sqrt{9a}-3-a+1-a+4}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}\)
\(=\frac{a+3\sqrt{a}+2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}\)
\(=\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}+2\right)}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}\)
\(=\frac{\sqrt{a}+1}{\sqrt{a}-1}\)
b) \(P=\frac{\sqrt{a}+1}{\sqrt{a}-1}=\frac{\sqrt{a}-1+2}{\sqrt{a}-1}=1+\frac{2}{\sqrt{a}-1}\)
Vậy để P là số nguyên thì: \(\sqrt{a}-1\inƯ\left(2\right)\)
Mà Ư(2)={-1;1;2;-1}
=> \(\sqrt{a}-1\in\left\{1;-1;2;-2\right\}\)
Ta có bảng sau:
| \(\sqrt{a}-1\) | 1 | -1 | 2 | -2 |
| a | 4 | 0 | 9 | \(\sqrt{a}=-1\) (ktm) |
vậy a={0;4;9} thì P nguyên
Bài 2
\(P=\frac{\sqrt{a+4\sqrt{a-4}}+\sqrt{a-4\sqrt{a-4}}}{\sqrt{1-\frac{8}{a}+\frac{16}{a^2}}}\)(ĐK:a\(\ge\)8)
\(=\frac{\sqrt{\left(a-4\right)+4\sqrt{a-4}+4}+\sqrt{\left(a-4\right)-4\sqrt{a-4}+4}}{\sqrt{\left(1-\frac{4}{a}\right)^2}}\)
\(=\frac{\sqrt{\left(\sqrt{a-4}+2\right)^2}+\sqrt{\left(\sqrt{a-4}-2\right)^2}}{1-\frac{4}{a}}\)
\(=\sqrt{a-4}+2+\sqrt{a-4}-2:\frac{a-4}{a}\)
\(=2\sqrt{a-4}\cdot\frac{a}{a-4}\)
\(=\frac{2a}{\sqrt{a-4}}\)
Ta có \(\left(\sqrt{a}+2\right)\left(1-\sqrt{a}\right)=a+\sqrt{a}-2\)
\(=\frac{3\text{a}+3\sqrt{a}-3}{a+\sqrt{a}-2}-\frac{\sqrt{a}+1}{\sqrt{a}+2}-\frac{\sqrt{a}-2}{\sqrt{a}-1}\)
\(=\frac{3\text{a}+3\sqrt{a}-3-a+1+a-4}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}\)
\(=\frac{3\text{a}+3\sqrt{a}-6}{a+\sqrt{a}-2}\)
\(=\frac{3\left(a+\sqrt{a}-2\right)}{a+\sqrt{a}-2}\)
\(=3\)
b/ Ta có 3 là số nguyên nên biểu thức P luôn nguyên với mọi x
TICK CHO MÌNH NHA
\(B=\frac{\sqrt{a}\left(\sqrt{a}+3\right)}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}-\frac{3\left(\sqrt{a}-3\right)}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}-\frac{a-2}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}\)
\(=\frac{a+3\sqrt{a}-3\sqrt{a}+9-a+2}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}=\frac{11}{a-9}\)
Để B nguyên thì \(\frac{11}{a-9}\inℤ\Leftrightarrow a-9\inƯ\left(11\right)=\left\{\pm1;\pm11\right\}\)
đến đây bạn tự lập bảng xét ước nhé :c chú ý ĐK giùm mình không lại sai :>
a) P = \(\left(\frac{3\sqrt{a}}{a+\sqrt{a}+b}-\frac{3a}{a\sqrt{a}-b\sqrt{b}}+\frac{1}{\sqrt{a}-\sqrt{b}}\right):\frac{\left(a-1\right).\left(\sqrt{a}-\sqrt{b}\right)}{\left(2.a+2.\sqrt{ab}+2.b\right)}\)
= \(\left(\frac{3\sqrt{a}.\left(\sqrt{a}-\sqrt{b}\right)-3.a+a+\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right).\left(a+\sqrt{ab}+b\right)}\right).\frac{2.\left(a+\sqrt{ab}+b\right)}{\left(a-1\right).\left(\sqrt{a}-\sqrt{b}\right)}\)
= \(\frac{a-2.\sqrt{ab}+b}{\sqrt{a}-\sqrt{b}}.\frac{2}{\left(a-1\right).\left(\sqrt{a}-\sqrt{b}\right)}\)
= \(\frac{2}{a-1}\)
b) P nguyên <=> \(\frac{2}{a-1}\)nguyên => 2 \(⋮\)a - 1
=> ( a- 1 ) = { \(\pm\)1 ; \(\pm\) 2} => a = { -1 ; 0 ; 2 ;3 }
a. ĐK: \(x\ge0,x\ne49\)
\(M=\frac{3\left(\sqrt{x}+7\right)-\left(\sqrt{x}-7\right)}{\left(\sqrt{x}-7\right)\left(\sqrt{x}+7\right)}:\frac{2\sqrt{x}+6}{x-49}\)
\(=\frac{2\sqrt{x}+28}{x-49}.\frac{x-49}{2\sqrt{x}+6}=\frac{2\sqrt{x}+28}{2\sqrt{x}+6}\)
b. M nguyên \(\Leftrightarrow\frac{2\sqrt{x}+28}{2\sqrt{x}+6}\in Z\Rightarrow\frac{2\sqrt{x}+6+22}{2\sqrt{x}+6}\in Z\Rightarrow1+\frac{22}{2\sqrt{x}+6}\in Z\Rightarrow\frac{22}{2\sqrt{x}+6}\in Z\Rightarrow\left(2\sqrt{x}+6\right)\inƯ\left(22\right)\)
Đến đây đã rất dễ dàng rồi nhé ^^
đề không cho tìm x NGUYÊN để m nguyên mà chỉ tìm các điểm x để m nguyên thôi