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câu 1
a)\(ĐKXĐ:x^3-8\ne0=>x\ne2\)
b)\(\frac{3x^2+6x+12}{x^3-8}=\frac{3\left(x^2-2x+4\right)}{\left(x-2\right)\left(x^2-2x+4\right)}=\frac{3}{x-2}\left(#\right)\)
Thay \(x=\frac{4001}{2000}\)zô \(\left(#\right)\)ta được
\(\frac{3}{\frac{4001}{2000}-2}=\frac{3}{\frac{4001}{2000}-\frac{4000}{2000}}=\frac{3}{\frac{1}{2000}}=6000\)
Trước tiên ta đi rút gọn biểu thức trên :
Đặt \(A=\left(\frac{x^2}{x^3-4x}+\frac{6}{6-3x}+\frac{1}{x+2}\right):\left(x-2+\frac{10-x^2}{x+2}\right)\)
ĐKXĐ : \(x\ne\pm2,x\ne0\)
Ta có : \(A=\left(\frac{x^2}{x^3-4x}+\frac{6}{6-3x}+\frac{1}{x+2}\right):\left(x-2+\frac{10-x^2}{x+2}\right)\)
\(=\left(\frac{x^2}{x\left(x^2-4\right)}+\frac{6}{3\left(2-x\right)}+\frac{1}{x+2}\right):\left(\frac{\left(x-2\right)\left(x+2\right)+10-x^2}{x+2}\right)\)
\(=\left(\frac{x\cdot3-6\cdot\left(x+2\right)+3\cdot\left(x-2\right)}{3\left(x-2\right)\left(x+2\right)}\right):\left(\frac{x^2-4+10-x^2}{x+2}\right)\)
\(=\frac{-18}{3\left(x-2\right)\left(x+2\right)}:\left(-\frac{6}{x+2}\right)\)
\(=\frac{-6}{\left(x-2\right)\left(x+2\right)}\cdot\frac{x+2}{\left(-6\right)}=\frac{1}{x-2}\)
Để \(A\) nhận giá trị nguyên
\(\Leftrightarrow\frac{1}{x-2}\inℤ\) \(\Leftrightarrow1⋮x-2\) \(\Leftrightarrow x-2\inƯ\left(1\right)\)
\(\Leftrightarrow x-2\in\left\{-1,1\right\}\)
\(\Leftrightarrow x\in\left\{1,3\right\}\) ( Thỏa mãn ĐKXĐ )
Vậy : \(x\in\left\{1,3\right\}\) thì A nhận giá trị nguyên.
\(=\left(\frac{x}{\left(x+2\right)\left(x-2\right)}-\frac{2}{x-2}+\frac{1}{x+2}\right):\left(\frac{\left(x-2\right)\left(x+2\right)+10-x}{x+2}\right)\)
\(=\left(\frac{x-2\left(x+2\right)+\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}\right):\left(\frac{x^2-4+10-x}{x+2}\right)\)
Đổi 10-x lại thành\(10-x^2\) nha, mk thiếu! sorry!
\(=\left(\frac{x-2x-4+x-2}{\left(x+2\right)\left(x-2\right)}\right):\left(\frac{x^2-4+10-x^2}{x+2}\right)\)
\(=\frac{-6}{\left(x+2\right)\left(x-2\right)}.\frac{x+2}{6}\)
\(=\frac{-6\left(x+2\right)}{6\left(x-2\right)\left(x+2\right)}=-\frac{1}{x-2}\)
a) Đk: x > 0 và x khác +-1
Ta có: A = \(\left(\frac{x+1}{x}-\frac{1}{1-x}-\frac{x^2-2}{x^2-x}\right):\frac{x^2+x}{x^2-2x+1}\)
A = \(\left[\frac{\left(x-1\right)\left(x+1\right)+x-x^2+2}{x\left(x-1\right)}\right]:\frac{x\left(x+1\right)}{\left(x-1\right)^2}\)
A = \(\frac{x^2-1+x-x^2+2}{x\left(x-1\right)}\cdot\frac{\left(x-1\right)^2}{x\left(x+1\right)}\)
A = \(\frac{x+1}{x}\cdot\frac{x-1}{x\left(x+1\right)}=\frac{x-1}{x^2}\)
b) Ta có: A = \(\frac{x-1}{x^2}=\frac{1}{x}-\frac{1}{x^2}=-\left(\frac{1}{x^2}-\frac{1}{x}+\frac{1}{4}\right)+\frac{1}{4}=-\left(\frac{1}{x}-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\forall x\)
Dấu "=" xảy ra <=> 1/x - 1/2 = 0 <=> x = 2 (tm)
Vậy MaxA = 1/4 <=> x = 2
Ta có \(A=[\frac{2}{\left(x+1\right)^3}\left(\frac{1}{x}+1\right)+\frac{1}{x^2+2x+1}\left(\frac{1}{x^2}+1\right)]:\frac{x-1}{x^3}\)
\(\Leftrightarrow A=\left[\frac{2}{\left(x+1\right)^3}.\frac{x+1}{x}+\frac{1}{\left(x+1\right)^2}.\frac{x^2+1}{x^2}\right].\frac{x^3}{x-1}\)
\(\Leftrightarrow A=\left[\frac{2x+x^2+1}{x^2\left(x+1\right)^2}\right].\frac{x^3}{x+1}=\frac{x}{x+1}\)
Để \(A=\frac{x}{x+1}< 1\Leftrightarrow\frac{1}{x+1}>0\Leftrightarrow x>-1\)
Để \(A=1-\frac{1}{x+1}\text{ nguyên thì }\frac{1}{x+1}\text{ nguyên hay }x\in\left\{-2,0\right\} \)