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\(\text{a) ĐKXĐ: }a\ne1\)
\(\text{b) }M=\frac{a^2+1+a}{a^2+1}:\left[\frac{1}{a-1}-\frac{2a}{a^2\left(a-1\right)+\left(a-1\right)}\right]\)
\(M=\frac{a^2+a+1}{a^2+1}:\left[\frac{1}{a-1}-\frac{2a}{\left(a-1\right)\left(a^2+1\right)}\right]\)
\(M=\frac{a^2+a+1}{a^2+1}:\frac{a^2+1-2a}{\left(a-1\right)\left(a^2+1\right)}\)
\(M=\frac{a^2+a+1}{a^2+1}.\frac{\left(a-1\right)\left(a^2+1\right)}{\left(a-1\right)^2}\)
\(M=\frac{a^2+a+1}{a-1}\)
a: ĐKXĐ: x^3-3x-2<>0
=>x^3-x-2x-2<>0
=>x(x-1)(x+1)-2(x+1)<>0
=>(x+1)(x-2)(x+1)<>0
=>x<>2 và x<>-1
b: \(A=\dfrac{\left(x-1\right)^2\cdot\left(x+1\right)^2}{\left(x-2\right)\left(x+1\right)^2}=\dfrac{\left(x-1\right)^2}{x-2}\)
c:
A<1
=>A-1<0
\(A-1=\dfrac{x^2-2x+1-x+2}{x-2}=\dfrac{x^2-3x+3}{x-2}\)
=>x-2<0
=>x<2
a) Phân thức A được xác định khi: \(x^2-1\ne0\Rightarrow\left(x-1\right)\left(x+1\right)\ne0\Rightarrow\left\{{}\begin{matrix}x+1\ne0\\x-1\ne0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x\ne1\\x\ne-1\end{matrix}\right.\)
Vây ĐKXĐ của A là \(\left\{{}\begin{matrix}x\ne1\\x\ne-1\end{matrix}\right.\)
b)Ta có: \(A=\dfrac{x^2+2x+1}{x^2-1}=\dfrac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}=\dfrac{\left(x+1\right)}{\left(x-1\right)}\)
Vậy \(A=\dfrac{x+1}{x-1}\Leftrightarrow\left\{{}\begin{matrix}x\ne1\\x\ne-1\end{matrix}\right.\)
c) Ta có A=2 <-> \(\dfrac{x+1}{x-1}=2\Leftrightarrow x+1=2\left(x-1\right)\Leftrightarrow x+1=2x-2\)
\(\Leftrightarrow x+1-2x+2=0\Leftrightarrow3-x=0\Rightarrow x=3\)
Vậy khi x=3 thì A=2
a) Vì \(x^3-1=\left(x-1\right)\left(x^2+x+1\right)\) nên điều kiện xác định của A là \(x^3-1\ne0\)
=> \(x\ne1\)
b) Rút gọn A:
\(A=\frac{5x+1+\left(1-2x\right)\left(x-1\right)+2\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\frac{5x+1+x-1-2x^2+2x+2x^2+2x+2}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\frac{10x+2}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{2\left(5x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
c) Vì \(x^2+x+1=\left(x^2+2.x.\frac{1}{2}+\frac{1}{4}\right)+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\)
Nên để A > 0 thì \(5x+1\) và \(x-1\) phải cùng dấu.
TH1: \(\hept{\begin{cases}5x+1>0\\x-1>0\end{cases}}\) => \(x>1\)
TH2: \(\hept{\begin{cases}5x+1< 0\\x-1< 0\end{cases}}\) => \(x< -\frac{1}{5}\)
Vậy để A > 0 thì \(x>1\) hoặc \(x< -\frac{1}{5}\)
a) Biểu thức A xác định `<=>x^2-1 ne 0 <=> (x-1)(x+1) ne 0 <=> x ne +-1`
b) `A=(x^2-3x-4)/(x^2 -1) = (x^2+x-4x-4)/(x^2-1) = (x(x+1)-4(x+1))/(x^2-1)`
`= ((x+1)(x-4))/((x+1)(x-1))=(x-4)/(x-1)`
c) `A` là số nguyên `<=> (x-4) vdots\ (x-1)`
`<=>[(x-1)-3] vdots\ (x-1)`
`<=> -3\ vdots\ (x-1)`
`<=> (x-1)\ in\ Ư(-3)`
`<=>(x-1)\ in\ {-3;-1;3;1}`
`<=>x\ in\ {-2;0;4;2}`
Vậy...
a: ĐKXĐ: x<>1; x<>-1
b: \(A=\dfrac{\left(x-4\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{x-4}{x-1}\)
c: Để A là số nguyên thì x-1-3 chia hết cho x-1
=>\(x-1\in\left\{1;-1;3;-3\right\}\)
=>\(x\in\left\{2;0;4;-2\right\}\)
a) A xác định \(\Leftrightarrow\hept{\begin{cases}3x\ne0\\x+1\ne0\\2-4x\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne0\\x\ne-1\\x\ne\frac{1}{2}\end{cases}}}\)
\(A=\left(\frac{x+2}{3x}+\frac{2}{x+1}-3\right):\frac{2-4x}{x+1}-\frac{3x+1-x^2}{3x}\)
\(A=\left[\frac{\left(x+2\right)\left(x+1\right)}{3x\left(x+1\right)}+\frac{2\cdot3x}{3x\left(x+1\right)}-\frac{3\cdot3x\left(x+1\right)}{3x\left(x+1\right)}\right]\cdot\frac{x+1}{2\left(1-2x\right)}-\frac{3x+1-x^2}{3x}\)
\(A=\frac{x^2+3x+2+6x-9x^2-9x}{3x\left(x+1\right)}\cdot\frac{x+1}{2\cdot\left(1-2x\right)}-\frac{3x+1-x^2}{3x}\)
\(A=\frac{\left(-8x^2+2\right)\left(x+1\right)}{3x\left(x+1\right)2\left(1-2x\right)}-\frac{3x+1-x^2}{3x}\)
\(A=\frac{2\left(1-4x^2\right)}{3x\cdot2\left(1-2x\right)}-\frac{3x+1-x^2}{3x}\)
\(A=\frac{2\left(1-2x\right)\left(1-2x\right)}{3x\cdot2\left(1-2x\right)}-\frac{3x+1-x^2}{3x}\)
\(A=\frac{1+2x}{3x}-\frac{3x+1-x^2}{3x}\)
\(A=\frac{2x+1-3x-1+x^2}{3x}\)
\(A=\frac{x^2-x}{3x}\)
\(A=\frac{x\left(x-1\right)}{3x}\)
\(A=\frac{x-1}{3}\)
b) Thay x = 4 ta có :
\(A=\frac{4-1}{3}=\frac{3}{3}=1\)
c) Để A thuộc Z thì \(x-1⋮3\)
\(\Rightarrow x-1\in B\left(3\right)=\left\{0;3;6;...\right\}\)
\(\Rightarrow x\in\left\{1;4;7;...\right\}\)
Vậy.....
a) \(M=\frac{x^4+2}{x^6+1}+\frac{x^2-1}{x^4-x^2+1}+\frac{x^2+3}{x^4+4x^2+3}\)
\(M=\frac{x^4+2}{\left(x^2+1\right)\left(x^4-x^2+1\right)}+\frac{x^2-1}{x^4-x^2+1}-\frac{x^2+3}{x^4+3x^2+x^2+3}\)
\(M=\frac{x^4+2}{\left(x^2+1\right)\left(x^4-x^2+1\right)}+\frac{x^2-1}{x^4-x^2+1}-\frac{x^2+3}{x^2\left(x^2+3\right)+x^2+3}\)
\(M=\frac{x^4+2}{\left(x^2+1\right)\left(x^4-x^2+1\right)}+\frac{x^2-1}{x^4-x^2+1}-\frac{x^2+3}{\left(x^2+3\right)\left(x^2+1\right)}\)
\(M=\frac{x^4+2}{\left(x^2+1\right)\left(x^4-x^2+1\right)}+\frac{x^2-1}{x^4-x^2+1}-\frac{1}{x^2+1}\)
\(M=\frac{x^4+2+x^4-1-x^4+x^2-1}{\left(x^2+1\right)\left(x^4-x^2+1\right)}\)
\(M=\frac{0+x^4+x^2}{\left(x^2+1\right)\left(x^4-x^2+1\right)}\)
\(M=\frac{x^2\left(x^2+1\right)}{\left(x^2+1\right)\left(x^4-x^2+1\right)}\)
\(M=\frac{x^2}{x^4-x^2+1}\)
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