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a) \(ĐKXĐ:\hept{\begin{cases}x\ne0;x\ne2\\x\ne-1\end{cases}}\)
\(Q=1+\left(\frac{x+1}{x^3+1}-\frac{1}{x-x^2-1}-\frac{2}{x+1}\right):\frac{x^3-2x^2}{x^3-x^2+x}\)
\(\Leftrightarrow Q=1+\left(\frac{x+1}{x^3+1}+\frac{1}{x^2-x+1}-\frac{2}{x+1}\right):\frac{x^2\left(x-2\right)}{x\left(x^2-x+1\right)}\)
\(\Leftrightarrow Q=1+\frac{\left(x+1\right)+\left(x+1\right)-2\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}:\frac{x\left(x-2\right)}{x^2-x+1}\)
\(\Leftrightarrow Q=1+\frac{x+1+x+1-2x^2+2x-2}{\left(x+1\right)\left(x^2-x+1\right)}.\frac{x^2-x+1}{x\left(x-2\right)}\)
\(\Leftrightarrow Q=1+\frac{-2x^2+4x}{x\left(x+1\right)\left(x-2\right)}\)
\(\Leftrightarrow Q=1+\frac{-2x\left(x-2\right)}{x\left(x+1\right)\left(x-2\right)}\)
\(\Leftrightarrow Q=1+\frac{-2}{x+1}\)
\(\Leftrightarrow Q=\frac{x-1}{x+1}\)
b) \(\left|x-\frac{3}{4}\right|=\frac{5}{4}\)
\(\Leftrightarrow\orbr{\begin{cases}x-\frac{3}{4}=\frac{5}{4}\\x-\frac{3}{4}=-\frac{5}{4}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\left(ktm\right)\\x=-\frac{1}{2}\left(tm\right)\end{cases}}\)
Thay \(x=-\frac{1}{2}\)vào Q, ta được :
\(Q=\frac{-\frac{1}{2}-1}{-\frac{1}{2}+1}\)
\(\Leftrightarrow Q=\frac{-\frac{3}{2}}{\frac{1}{2}}\)
\(\Leftrightarrow Q=-3\)
c) Để \(Q\inℤ\)
\(\Leftrightarrow x-1⋮x+1\)
\(\Leftrightarrow x+1-2⋮x+1\)
\(\Leftrightarrow2⋮x+1\)
\(\Leftrightarrow x+1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
\(\Leftrightarrow x\in\left\{-2;0;-3;1\right\}\)
Vậy để \(Q\inℤ\Leftrightarrow x\in\left\{-2;0;-3;1\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(ĐKXĐ:x\ne\pm2\)
\(D=\frac{3x}{x-2}+\frac{2}{x+2}-\frac{14x-4}{x^2-4}:\frac{x\left(x-1\right)}{x+2}\)
\(\Leftrightarrow D=\frac{3x^2+6x+2x-4-14x+4}{\left(x-2\right)\left(x+2\right)}\cdot\frac{x+2}{x\left(x-1\right)}\)
\(\Leftrightarrow D=\frac{3x^2-6x}{x\left(x-1\right)\left(x-2\right)}\)
\(\Leftrightarrow D=\frac{3x\left(x-2\right)}{x\left(x-1\right)\left(x-2\right)}\)
\(\Leftrightarrow D=\frac{3}{x-1}\)
b) Khi \(\left|x-1\right|-3=0\)
\(\Leftrightarrow\left|x-1\right|=3\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=3\\1-x=3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=4\left(tm\right)\\x=-2\left(ktm\right)\end{cases}}\)
Thay \(x=4\)vào D ta được :\(D=\frac{3}{4-1}=1\)
c) Để D có giá trị nguyên
\(\Leftrightarrow\frac{3}{x-1}\)có giá trị nguyên
\(\Leftrightarrow x-1\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
\(\Leftrightarrow x\in\left\{0;2;-2;4\right\}\)
Loại bỏ giá trị \(x=\pm2\)không làm cho biểu thức có nghĩa
Vậy để D có giá trị nguyên \(\Leftrightarrow x\in\left\{0;4\right\}\)
Khi làm bài thì chỉnh lại giúp bạn cái đề:
\(D=\left(\frac{3X}{X-2}+\frac{2}{X+2}-\frac{14X-4}{X^2-4}\right):\frac{X\left(X-1\right)}{X+2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(ĐKXĐ:\hept{\begin{cases}x^3+1\ne0\\x^3-2x^2\ne0\\x+1\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne0\\x\ne-1\\x\ne2\end{cases}}\)(chỗ chữ và là do OLM thiếu ngoặc 4 cái nên mk để thế nha! trình bày thì kẻ thêm 1 ngoặc nưax)
\(Q=1+\left(\frac{x+1}{x^3+1}-\frac{1}{x-x^2-1}-\frac{2}{x+1}\right):\frac{x^3-2x^2}{x^3-x^2+x}\)
\(=1+\left[\frac{x+1}{\left(x+1\right)\left(x^2-x+1\right)}+\frac{1}{x^2-x+1}-\frac{2}{x+1}\right]:\frac{x^2\left(x-2\right)}{x\left(x^2-x+1\right)}\)
\(=1+\frac{\left(x+1\right)+\left(x+1\right)-2\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}.\frac{x^2-x+1}{x\left(x-2\right)}\)
\(=1+\frac{4x-2x^2}{x+1}.\frac{1}{x\left(x-2\right)}\)
\(=1-\frac{2x\left(x-2\right)}{x\left(x+1\right)\left(x-2\right)}=1-\frac{2}{x+1}=\frac{x-1}{x+1}\)
b, Với \(x\ne0;x\ne-1;x\ne2\)Ta có:
\(|x-\frac{3}{4}|=\frac{5}{4}\)
*TH1:
\(x-\frac{3}{4}=\frac{5}{4}\Rightarrow x=2\)(ko thảo mãn)
*TH2:
\(x-\frac{3}{4}=-\frac{5}{4}\Rightarrow x=-\frac{1}{2}\)
\(\Rightarrow Q=\frac{-\frac{1}{2}-1}{-\frac{1}{2}+1}=-3\)
c,
\(Q=\frac{x-1}{x+1}=1-\frac{2}{x+1}\)
Để Q nguyên thì x+1 phải thuộc ước của 2!! tự làm tiếp dễ rồi!!
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a, \(M=\left(\frac{1}{x-1}-\frac{x}{1-x^3}.\frac{x^2+x+1}{x+1}\right):\frac{1}{x^2-1}\)
\(=\left(\frac{1}{x-1}-\frac{x}{\left(1-x\right)\left(x^2+x+1\right)}.\frac{x^2+x+1}{x+1}\right):\frac{1}{x^2-1}\)
\(=\left(\frac{1}{x-1}+\frac{x}{\left(x-1\right)\left(x+1\right)}\right):\frac{1}{x^2-1}\)
\(=\left(\frac{x+1+x}{\left(x-1\right)\left(x+1\right)}\right).\left(x-1\right)\left(x+1\right)=2x+1\)
b, Thay x = 1/2 vào biểu thức trên ta được : \(2.\frac{1}{2}+1=1\)
c, Để M luôn dương hay \(2x+1\ge0\Leftrightarrow x\ge-\frac{1}{2}\)
Vậy với x \(\ge-\frac{1}{2}\)thì \(M\ge0\)
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Câu 1:
\(F=\frac{\frac{x^3-x}{x+1}+\frac{2x-2}{1+\frac{x}{2}}}{\frac{x^3-3x^2}{x-3}-\frac{2x^2+8}{x+2}}\left(ĐKXĐ:x\ne3;-2;-1\right)\)
\(F=\frac{\frac{x\left(x-1\right)\left(x+1\right)}{x+1}+\frac{2x-2}{1+\frac{x}{2}}}{\frac{x^2\left(x-3\right)}{x-3}-\frac{2x^2+8}{x+2}}\)
\(F=\frac{\frac{\left(x^2-x\right)\left(1+\frac{x}{2}\right)+2x-2}{1+\frac{x}{2}}}{\frac{x^2\left(x+2\right)-2x^2-8}{x+2}}\)
\(F=\frac{\frac{x^2+\frac{x^3}{2}-x-\frac{x^2}{2}+2x-2}{1+\frac{x}{2}}}{\frac{x^3-8}{x+2}}\)
\(F=\frac{\frac{x^2}{2}+\frac{x^3}{2}+x-2}{1+\frac{x}{2}}.\frac{x+2}{x^3-8}\)
Câu 2:
\(G=\frac{\frac{x^4+1}{x^3-1}-x}{\frac{x}{x^2+x+1}-\frac{2}{x-1}}\left(ĐKXĐ:x\ne1\right)\)
\(G=\frac{\frac{x^4+1-x\left(x^3-1\right)}{x^3-1}}{\frac{x\left(x-1\right)-2\left(x^2+x+1\right)}{x^3-1}}\)
\(G=\frac{x+1}{x^3-1}:\frac{x^2-x-2x^2-2x-2}{x^3+1}\)
\(G=\frac{x+1}{-x^2-3x-2}\)
\(G=\frac{x+1}{-\left(x+2\right)\left(x+1\right)}\)
\(G=-\frac{1}{x+2}\)Tại x=2017 ta đc:\(G=-\frac{1}{2+2017}=-\frac{1}{2019}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(ĐKXĐ:\hept{\begin{cases}x\ne\pm2\\x\ne0\end{cases}}\)
a) \(P=\left(\frac{x^2}{x^3-4x}+\frac{6}{6-3x}+\frac{1}{x+2}\right):\left(x-2+\frac{10-x^2}{x+2}\right)\)
\(\Leftrightarrow P=\left(\frac{x^2}{x\left(x-2\right)\left(x+2\right)}-\frac{6}{3\left(x-2\right)}+\frac{1}{x+2}\right):\frac{x^2-4+10-x^2}{x-2}\)
\(\Leftrightarrow P=\frac{x^2-2x\left(x+2\right)+x\left(x-2\right)}{x\left(x-2\right)\left(x+2\right)}:\frac{6}{x-2}\)
\(\Leftrightarrow P=\frac{x^2-2x^2-4x+x^2-2x}{x\left(x-2\right)\left(x+2\right)}\cdot\frac{x-2}{6}\)
\(\Leftrightarrow P=\frac{-6x}{6x\left(x+2\right)}\)
\(\Leftrightarrow P=\frac{-1}{x+2}\)
b) Khi \(\left|x\right|=\frac{3}{4}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{4}\\x=-\frac{3}{4}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}P=-\frac{1}{\frac{3}{4}+2}=-\frac{4}{11}\\P=-\frac{1}{-\frac{3}{4}+2}=-\frac{4}{5}\end{cases}}\)
c) Để P = 7
\(\Leftrightarrow-\frac{1}{x+2}=7\)
\(\Leftrightarrow7\left(x+2\right)=-1\)
\(\Leftrightarrow7x+14=-1\)
\(\Leftrightarrow7x=-15\)
\(\Leftrightarrow x=-\frac{15}{7}\)
Vậy để \(P=7\Leftrightarrow x=-\frac{15}{7}\)
d) Để \(P\inℤ\)
\(\Leftrightarrow1⋮x+2\)
\(\Leftrightarrow x+2\inƯ\left(1\right)=\left\{\pm1\right\}\)
\(\Leftrightarrow x\in\left\{-3;-1\right\}\)
Vậy để \(P\inℤ\Leftrightarrow x\in\left\{-3;-1\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)
2x-4=2(x-2)
2x+4=2(x+2)
x
Để P xác định thì
[2(x-2) => [2(x+2)
[2(x+2) =>[ 2(x-2)
[ (x-2)(x+2) => [(x+2)(x-2)
Vay 2(x+2) , 2(x-2), (x+2)(x-2) thi P xác định
Ta có: \(x+\frac{1}{x}=3\Leftrightarrow\left(x+\frac{1}{x}\right)^2=3^2\Leftrightarrow x^2+2+\frac{1}{x^2}=9\Leftrightarrow x^2+\frac{1}{x^2}=7\)
\(\Leftrightarrow\left(x^2+\frac{1}{x^2}\right)^2=7^2\Leftrightarrow x^4+2+\frac{1}{x^4}=49\Leftrightarrow x^4+\frac{1}{x^4}=47\)
\(x+\frac{1}{x}=3\)
=> \(\left(x+\frac{1}{x}\right)^2=9\)
<=> \(x^2+\frac{1}{x^2}+2=9\)
<=> \(x^2+\frac{1}{x^2}=7\)
=> \(\left(x^2+\frac{1}{x^2}\right)^2=49\)
<=> \(x^4+\frac{1}{x^4}+2=49\)
<=> \(x^4+\frac{1}{x^4}=47\)