bài 1: tìm x thuộc Z để A thuộc Z
a, A = \(\frac{x-1}{x+1}\)
b, A = \(\frac{x+1}{x-2}\)
c, A = \(\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
d, A = \(\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
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23 tháng 10 2016
a) \(B=\frac{\sqrt{x}-3}{x+\sqrt{x}+1}\left(ĐK:x\ge0\right)\)
\(=\frac{\sqrt{81}-3}{81+\sqrt{81}+1}=\frac{9-3}{81+9+1}=\frac{6}{91}\)
b) \(A=\frac{2x+1}{\sqrt{x^3}-1}-\frac{1}{\sqrt{x}-1}\left(ĐK:x\ge0;x\ne1\right)\)
\(=\frac{2x+1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\frac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\frac{\sqrt{x}}{x+\sqrt{x}+1}\)
c) \(P=\frac{A}{B}\)
\(=\frac{\sqrt{x}}{x+\sqrt{x}+1}:\frac{\sqrt{x}-3}{x+\sqrt{x}+1}\left(ĐK:x\ge0;x\ne9\right)\)
\(=\frac{\sqrt{x}}{\sqrt{x}-3}\)
\(=\frac{\left(\sqrt{x}-3\right)+3}{\sqrt{x}-3}=1+\frac{3}{\sqrt{x}-3}\)
Vậy để P nguyên thì: \(\sqrt{x}-3\inƯ\left(3\right)\)
\(\Leftrightarrow\sqrt{x}-3\in\left\{-1;1;-3;3\right\}\)
+) \(\sqrt{x}-3=-1\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\left(tm\right)\)
+) \(\sqrt{x}-3=1\Leftrightarrow\sqrt{x}=4\Leftrightarrow x=16\left(tm\right)\)
+) \(\sqrt{x}-3=-3\Leftrightarrow\sqrt{x}=0\Leftrightarrow x=0\left(tm\right)\)
+) \(\sqrt{x}-3=3\Leftrightarrow\sqrt{x}=6\Leftrightarrow x=36\left(tm\right)\)
Vậy...........
19 tháng 7 2018
\(1,\frac{\sqrt{x}+1}{\sqrt{x}-3}=\frac{\sqrt{x}-3+4}{\sqrt{x}-3}=1+\frac{4}{\sqrt{x}-3}\)
Để \(\frac{\sqrt{x}+1}{\sqrt{x}-3}\in Z\Rightarrow\frac{4}{\sqrt{x}-3}\in Z\)
\(\Rightarrow\sqrt{x}-3\in\left(1;4;-1;-4\right)\)
\(\Rightarrow\sqrt{x}\in\left(4;7;2;-1\right)\)
\(\Rightarrow\sqrt{x}=4\Leftrightarrow x=2\)
19 tháng 7 2018
\(4,A=x+\sqrt{x}+1\)
\(A=\left(\sqrt{x}\right)^2+2.\frac{1}{2}.\sqrt{x}+\left(\frac{1}{2}\right)^2+\frac{3}{4}\)
\(A=\left(\sqrt{x}+\frac{1}{2}\right)^2+\frac{3}{4}\)
\(\Rightarrow A\ge\frac{3}{4}.\left(\sqrt{x}+\frac{1}{2}\right)^2\ge0\)
Dấu "=" xảy ra khi :
\(\sqrt{x}+\frac{1}{2}=0\Leftrightarrow\sqrt{x}=-\frac{1}{2}\)
Vậy Min A = 3/4 khi căn x = -1/2
9 tháng 8 2021
a, Với \(x\ge0;x\ne1\)
\(B=\frac{1}{\sqrt{x}-1}=2\Rightarrow2\sqrt{x}-2=1\Leftrightarrow2\sqrt{x}-3=0\Leftrightarrow x=\frac{9}{4}\)
b, Ta có : \(A.B=\frac{x+3}{\sqrt{x}+1}.\frac{1}{\sqrt{x}-1}=\frac{x+3}{x-1}=\frac{x-1+4}{x-1}=1+\frac{4}{x-1}\)
\(\Rightarrow x-1\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)
| x - 1 | 1 | -1 | 2 | -2 | 4 | -4 |
| x | 2 | 0 | 3 | -1 | 5 | -3 |
c, Ta có : \(A=\frac{x+3}{\sqrt{x}+1}\le3\Leftrightarrow\frac{x+3}{\sqrt{x}+1}-3\le0\)
\(\Leftrightarrow\frac{x-3\sqrt{x}}{\sqrt{x}+1}\le0\Rightarrow\sqrt{x}-3\le0\Leftrightarrow x\le9\)
Kết hợp với đk vậy 0 =< x =< 9
a, \(A=\frac{x-1}{x+1}=\frac{x+1-1-1}{x+1}=\frac{x+1-2}{x+1}=1-\frac{2}{x+1}\)
Để \(A\in z\) thì \(x+1\inƯ\left(2\right)=\left(-2;-1:1;2\right)\)
\(x+1=-2\Rightarrow x=-3\)
\(x+1=-1\Rightarrow x=-2\)
\(x+1=1\Rightarrow x=0\)
\(x+1=2\Rightarrow x=1\)
Vậy \(x=\left(-3;-2;0;1\right)\)thì \(A\in z\)
b, \(A=\frac{x+1}{x-2}=1+\frac{3}{x-2}\)
Để \(A\in z\)thì \(x-2\inƯ\left(3\right)=\left(-3;-1;1;3\right)\)
\(x-2=-3\Rightarrow x=-1\)
\(x-2=-1\Rightarrow x=1\)
\(x-2=1\Rightarrow x=3\)
\(x-2=3\Rightarrow x=5\)
Vậy \(x=\left(-1;1;3;5\right)\)thì \(A\in z\)
c, \(A=\frac{\sqrt{x}+1}{\sqrt{x}-3}\)\(ĐK:\)\(x\ge0;x\ne9\)
\(A=\frac{\sqrt{x}+1}{\sqrt{x}-3}=1+\frac{4}{\sqrt{x}-3}\)
Để \(A\in z\)thì \(\sqrt{x}-3\inƯ\left(4\right)=\left(-4;-2;-1;1;2;4\right)\)
\(\sqrt{x}-3=-4\Rightarrow\sqrt{x}=-1VN\)
\(\sqrt{x}-3=-2\Rightarrow\sqrt{x}=1\Rightarrow x=1\)
\(\sqrt{x}-3=-1\Rightarrow\sqrt{x}=2\Rightarrow x=4\)
\(\sqrt{x}-3=1\Rightarrow\sqrt{x}=4\Rightarrow x=16\)
\(\sqrt{x}-3=2\Rightarrow\sqrt{x}=5\Rightarrow x=25\)
\(\sqrt{x}-3=4\Rightarrow\sqrt{x}=7\Rightarrow x=49\)
Vậy \(x=\left(1;4;16;25;49\right)\)thì \(A\in z\)
d, \(A=\frac{\sqrt{x}+1}{\sqrt{x}-1}\) \(ĐK:\)\(x\ge0;x\ne1\)
\(A=\frac{\sqrt{x}+1}{\sqrt{x}-1}=1+\frac{2}{\sqrt{x}-1}\)
Để \(A\in z\) thì \(\sqrt{x}-1\inƯ\left(2\right)=\left(-2;-1;1;2\right)\)
\(\sqrt{x}-1=-2\Rightarrow\sqrt{x}=-1VN\)
\(\sqrt{x}-1=-1\Rightarrow\sqrt{x}=0\Rightarrow x=0\)
\(\sqrt{x}-1=1\Rightarrow\sqrt{x}=2\Rightarrow x=4\)
\(\sqrt{x}-1=2\Rightarrow\sqrt{x}=3\Rightarrow x=9\)
Vậy \(x=\left(0,4,9\right)\)thì \(A\in z\)
\(a,A=\frac{x-1}{x+1}\)
Để \(A\in Z\)
\(\Rightarrow\frac{x-1}{x+1}\in Z\)
\(\Rightarrow\frac{x+1-2}{x+1}\in Z\)
\(\Rightarrow1-\frac{2}{x+1}\in Z\)
\(\Rightarrow\frac{2}{x+1}\in Z\)
\(\Rightarrow x+1\in U_{\left(2\right)}\)
\(\Rightarrow x+1=\left\{-2,-1,1,2\right\}\)
\(\Rightarrow x=\left\{-3,-2,0,1\right\}\)