Tìm x để \(A=\frac{x+1}{x^2+x+1}\) có giá trị nguyên !!
Làm hộ mik vs ><!
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a,\(A=\left(\frac{2x-x^2}{2\left(x^2+4\right)}-\frac{2x^2}{\left(x^2+4\right)\left(x-2\right)}\right)\left(\frac{2x+x^2\left(1-x\right)}{x^3}\right)\left(ĐKXĐ:x\ne2;x\ne0\right)\)
\(A=\frac{\left(2x-x^2\right)\left(x-2\right)-4x^2}{2\left(x^2+4\right)\left(x-2\right)}.\frac{-x^3+x^2+2x}{x^3}\)
\(=\frac{-x^3-4x}{2\left(x^2+4\right)\left(x-2\right)}.\frac{x^2-x-2}{-x^2}\)
\(=\frac{-x\left(x^2+4\right)}{2\left(x^2+4\right)\left(x-2\right)}.\frac{\left(x-2\right)\left(x+1\right)}{-x^2}=\frac{x+1}{2x}\)
b, \(A=x\Leftrightarrow\frac{x+1}{2x}=x\Rightarrow2x^2=x+1\Leftrightarrow2x^2-x-1=0\)
\(\Leftrightarrow\left(2x+1\right)\left(x-1\right)=0\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=1\end{cases}}\)(thỏa mãn điều kiện)
c, \(A\in Z\Leftrightarrow\frac{x+1}{2x}\in Z\Leftrightarrow x+1⋮\left(2x\right)\)
\(\Leftrightarrow2x+2⋮2x\Leftrightarrow2⋮2x\Leftrightarrow1⋮x\Leftrightarrow x=\pm1\) (thỏa mãn ĐKXĐ)
a, 2x-1 thuộc ước của 2,rồi giải ra
b,c tương tự
d\(\frac{x^2-64-123}{x+8}=\frac{\left(x+8\right)\left(x-8\right)-123}{x+8}=x-8-\frac{123}{X+8}\) .........rồi làm tương tự như câu a,,,,,,,,,,,,còn câu e cũng gần giống câu d
a) ĐKXĐ \(\hept{\begin{cases}x-1\ne0\\x+1\ne0\\x\ne0\end{cases}}\Rightarrow\hept{\begin{cases}x\ne1\\x\ne-1\\x\ne0\end{cases}}\)
b)\(\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}+\frac{x^2-4x-1}{x^2-1}\right)\frac{x+2003}{x}\)
\(=\frac{\left(x+1\right)^2-\left(x-1\right)^2+x^2-4x-1}{\left(x-1\right).\left(x+1\right)}.\frac{x+2003}{x}\)
\(\frac{\left(x+1-x+1\right)\left(x+1+x-1\right)+x^2-4x-1}{\left(x-1\right)\left(x+1\right)}.\frac{x+2003}{x}\)
\(\frac{4x+x^2-4x-1}{\left(x-1\right)\left(x+1\right)}.\frac{x+2003}{x}\)
\(=\frac{x^2-1}{\left(x-1\right)\left(x+1\right)}.\frac{x+2003}{x}=\frac{\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}.\frac{x+2003}{x}\)
\(=\frac{x+2003}{x}\)
c) Ta có \(K=\frac{x+2003}{x}\)
Để K nguyên thì x + 2003 ⋮ x
Ta có x ⋮ x => 2003 ⋮ x
=> x thuộc Ư(2003) = { 1; -1; 2003; -2003 }
Vậy khi x thuộc { 1; -1; 2003; -2003 } thì K nguyên
a: \(A=\dfrac{x-1+2x^2+2x+2-x^2-2x}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{1}{x-1}\)
d) \(A>0\Leftrightarrow\frac{-1}{x-2}>0\)
\(\Leftrightarrow x-2< 0\) ( vì \(-1< 0\))
\(\Leftrightarrow x< 2\)
\(A=\left(\frac{x}{x^2-4}+\frac{2}{2-x}+\frac{1}{x+2}\right):\left(x-2+\frac{10-x^2}{x+2}\right)\)
\(A=\)\(\left[\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{x-2}{\left(x-2\right)\left(x+2\right)}\right]\)
\(:\left[\frac{\left(x-2\right)\left(x+2\right)}{x+2}+\frac{10-x^2}{x+2}\right]\)
\(A=\frac{x-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}:\left[\frac{x^2-4+10-x^2}{x+2}\right]\)
\(A=\frac{-6}{\left(x-2\right)\left(x+2\right)}:\frac{6}{x+2}\)
\(A=\frac{-6}{\left(x-2\right)\left(x+2\right)}.\frac{x+2}{6}\)
\(A=\frac{-1}{x-2}\)