Cho \(P=\sqrt{\frac{\left(x^2-3\right)^2+12x^2}{x^2}}+\sqrt{\left(x+2\right)^2-8x}\)
a) Rút gọn \(P\)
b) Tìm \(x\in Z\)để \(P\in Z\)
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Cô Hoàng Huyền
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28 tháng 8 2017
ĐKXĐ: \(x\ne0\)
\(y=\sqrt{\frac{x^4-6x^2+9+12x^2}{x^2}}+\sqrt{x^2+4x+4-8x}\)
\(y=\sqrt{\frac{x^4+6x^2+9}{x^2}}+\sqrt{x^2-4x+4}\)
\(y=\sqrt{\frac{\left(x^2+3\right)^2}{x^2}}+\sqrt{\left(x-2\right)^2}\)
\(y=\left|\frac{x^2+3}{x}\right|+\left|x-2\right|\)
Ta có bảng xét dấu:
Với \(x< 0,y=\frac{x^2+3}{-x}+2-x=\frac{2x^2-2x+3}{-x}\)
Với \(0< x\le2,y=\frac{x^2+3}{x}+2-x=\frac{2x+3}{x}\)
Với \(x>2,y=\frac{x^2+3}{x}+x-2=\frac{2x^2-2x+3}{x}\)
- Ta thấy ngay, với cả ba trường hợp thì \(y\in Z\Leftrightarrow x\in U\left(3\right)=\left\{-3;-1;1;3\right\}\)
BA
30 tháng 7 2019
\(ĐKXĐ:\)
\(\hept{\begin{cases}x-9\ne0\\\sqrt{x}-2\ne0\\\sqrt{x}+3\ne0;x\ge0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x\ne9\\x\ne4\\x\ge0\end{cases}}\)
Vậy...................................................
BA
30 tháng 7 2019
\(A=\left(\frac{x-3\sqrt{x}}{x-9}-1\right):\left(\frac{9-x}{x+\sqrt{x}-6}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}+3}\right)\)
\(=\left(\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-1\right):\left(\frac{9-x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}+3}\right)\)
\(=\frac{\sqrt{x}-\sqrt{x}-3}{\left(\sqrt{x}+3\right)}:\left(\frac{9-x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}+\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\right)\)
\(=\frac{-3}{\sqrt{x}+3}:\left(\frac{9-x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}+\frac{x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}-\frac{x-4}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\right)\)
\(=\frac{-3}{\sqrt{x}+3}:\frac{9-x+x-9-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{-3}{\sqrt{x}+3}:\frac{-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{-3}{\sqrt{x}+3}.\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{4-x}\)
\(=\frac{3\left(2-\sqrt{x}\right)}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\)
\(=\frac{3}{\left(2+\sqrt{x}\right)}\)
5 tháng 8 2018
Bài 1:
\(\sqrt{24+8\sqrt{15}-\sqrt{9-4\sqrt{5}}}\)
\(=\sqrt{24+8\sqrt{15}-\left(\sqrt{5}-2\right)}\)
\(=\sqrt{26+8\sqrt{15}-\sqrt{5}}\)
Bài 2:
\(A=\sqrt{\frac{\left(x^2-3\right)^2+12x^2}{x^2}}+\sqrt{\left(x+2\right)^2-8x}\)
\(A=\sqrt{\frac{x^4+6x^2+9}{x^2}}\)
\(A=\frac{\sqrt{x^4+6x^2+9}}{\sqrt{x^2}}\)
\(A=\frac{\sqrt{\left(x^2+3\right)^2}}{x}\)
\(A=\frac{x^2+3}{x}\)
\(A=\frac{x^2+3}{x}+x-2\)
\(A=\frac{2x^2+3}{x}-2\)
9 tháng 8 2018
wrecking ball sai rồi \(\frac{\sqrt{\left(x^2+3\right)^2}}{x}=\frac{trituyetdoix^2+3}{x}\) bằng
12 tháng 9 2020
Đk: x \(\ge\)0; x \(\ne\)1; x \(\ne\)9
1) \(B=\left(\frac{2x+3}{\sqrt{x^3}-1}-\frac{1}{\sqrt{x}-1}\right):\left(1-\frac{x+4}{x+\sqrt{x}+1}\right)\)
\(B=\frac{2x+3-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}:\frac{x+\sqrt{x}+1-x-4}{x+\sqrt{x}+1}\)
\(B=\frac{-x-\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\frac{x+\sqrt{x}+1}{\sqrt{x}-3}\)
\(B=\frac{-\left(x+2\sqrt{x}-\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}\)
\(B=\frac{-\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}=\frac{\sqrt{x}+2}{3-\sqrt{x}}\)
2. \(B=\frac{\sqrt{x}+2}{3-\sqrt{x}}=\frac{-\left(3-\sqrt{x}\right)+5}{3-\sqrt{x}}=-1+\frac{5}{3-\sqrt{x}}\)
Để B \(\in\)Z <=> 5 \(⋮\)\(3-\sqrt{x}\)
<=> \(3-\sqrt{x}\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)
Do \(3-\sqrt{x}\le\)3 => 3 - \(\sqrt{x}\)\(\in\){1; -1; -5}
Lập bảng:
| \(3-\sqrt{x}\) | 1 | -1 | -5 |
| x | 4 | 16 | 64 |
Vậy ...
28 tháng 6 2018
ĐKCĐ: \(x\ge0;x\ne9,x\ne4\)
\(A=\left(\frac{x-3\sqrt{x}}{x-9}-1\right):\left(\frac{9-x}{x+\sqrt{x}-6}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}+3}\right)\\ \)
\(=\left(\frac{\sqrt{x}.\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right).\left(\sqrt{x}+3\right)}-1\right):\left(\frac{\left(3-\sqrt{x}\right).\left(3+\sqrt{x}\right)}{\left(\sqrt{x}-2\right).\left(\sqrt{x+3}\right)}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)
\(=\left(\frac{\sqrt{x}}{\sqrt{x}+3}-1\right):\left(\frac{3-\sqrt{x}}{\sqrt{x}-2}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)
\(=-\frac{3}{\sqrt{x}+3}:\left(-\frac{\sqrt{x}-2}{\sqrt{x}+3}\right)=-\frac{3}{\sqrt{x}+3}:\frac{-\left(\sqrt{x}-2\right)}{\sqrt{x}+3}=\frac{3}{\sqrt{x}-2}\)
b, \(A\inℤ\Leftrightarrow\frac{3}{\sqrt{x}-2}\inℤ\)
Nếu x không là số chính phương thì \(\sqrt{x}\)là số vô tỉ thì \(\sqrt{x}-2\)là số vô tỉ\(\Rightarrow A=\frac{3}{\sqrt{x}-2}\)là số vô tỉ
Nếu x là số chính phương thì \(\sqrt{x}\)là số nguyên thì \(\sqrt{x}-2\inℤ\Rightarrow\sqrt{x}-2\inƯ\left(3\right)\Rightarrow\sqrt{x}-2\in\left\{\pm1;\pm3\right\}\Rightarrow\sqrt{x}\in\left\{1;3;5\right\}\)\(\Rightarrow x\in\left\{1;9;25\right\}\)
Mà theo ĐKXĐ có x khác 9 => \(x\in\left\{1,25\right\}\)
\(P=\sqrt{\frac{\left(x^2-3\right)^2+12x^2}{x^2}}+\sqrt{\left(x+2\right)^2-8x}\) Đk \(x\ne0\)
\(=\frac{\sqrt{x^4-6x^2+9+12x^2}}{\sqrt{x^2}}+\sqrt{x^2+4x+4-8x}\)
\(=\frac{\sqrt{x^4+6x^2+9}}{\sqrt{x^2}}+\sqrt{x^2-4x+4}\)
\(=\frac{\sqrt{\left(x^2+3\right)^2}}{\sqrt{x^2}}+\sqrt{\left(x-2\right)^2}\)
\(=\frac{x^2+3}{x}+x-2\)
\(=\frac{x^2+3+x\left(x-2\right)}{x}=\frac{x^2+3+x^2-2x}{x}\)
\(=\frac{2x^2-2x+3}{x}\)
b, \(P=\frac{2x^2-2x+3}{x}=2x-2+\frac{3}{x}\)
Để \(P\in z\)thì \(x\inƯ\left(3\right)=\left(-3;-1;1;3\right)\)