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\(a.a\ne\pm1\)
\(b.K=\dfrac{1}{a+1}+\dfrac{2}{a^2-1}=\dfrac{a-1}{\left(a-1\right)\left(a+1\right)}+\dfrac{2}{\left(a-1\right)\left(a+1\right)}=\dfrac{a+1}{\left(a-1\right)\left(a+1\right)}=\dfrac{1}{a-1}\)
\(c.K=\dfrac{1}{1-\dfrac{1}{2}}=\dfrac{1}{\dfrac{1}{2}}=2\)
a: ĐKXĐ: \(x\notin\left\{1;-1;0\right\}\)
b: \(K=\dfrac{x^2+2x+1-x^2+2x-1+x^2-4x-1}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x+2003}{x}\)
\(=\dfrac{x^2-1}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x+2003}{x}=\dfrac{x+2003}{x}\)
c: Để K là số nguyên thì \(x\inƯ\left(2003\right)\)
hay \(x\in\left\{2003;-2003\right\}\)
\(\left(\frac{a}{a-1}-\frac{1}{a^2-a}\right)=\frac{a^2-1}{a^2-a}=\frac{a+1}{a}\)
ở phàn a+/a thiếu số 1 nhé
\(\frac{1}{a+1}+\frac{2}{a^2-1}=\frac{a-1+2}{a^2-1}=\frac{1}{a-1}\)
=> K =\(\frac{a^2-1}{a}\)
đkxđ: a khác +-1
b, thay vào mà tình
a/ \(K=\left(\frac{a}{a-1}-\frac{1}{a^2-a}\right):\left(\frac{1}{a+1}+\frac{2}{a^2-1}\right)\)
\(=\left(\frac{a}{a-1}-\frac{1}{a\left(a-1\right)}\right):\left(\frac{1}{a+1}+\frac{2}{\left(a-1\right)\left(a+1\right)}\right)\)
\(=\frac{a^2-1}{a\left(a-1\right)}:\frac{a-1+2}{\left(a-1\right)\left(a+1\right)}\)
\(=\frac{\left(a-1\right)\left(a+1\right)}{a\left(a-1\right)}.\frac{\left(a-1\right)\left(a+1\right)}{a-1}\)
\(=\frac{a+1}{a}.a+1\)
\(=\frac{\left(a+1\right)^2}{a}\)
b, Thay a=1/2
\(\Rightarrow\frac{\left(\frac{1}{2}+1\right)^2}{\frac{1}{2}}=\frac{\frac{9}{4}}{\frac{1}{2}}=\frac{9}{2}\)
a)\(A=\frac{\left(x+2\right)}{\left(x+3\right)}-\frac{5}{x^2+x-6}+\frac{1}{2-x}\)
A xác định
\(\Leftrightarrow\hept{\begin{cases}x+3\ne0\\x^2+x-6\ne0\\2-x\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne-3\\\left(x+3\right)\left(x-2\right)\ne0\\x\ne2\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne-3\\x\ne2\end{cases}}\)
Vậy A xác định \(\Leftrightarrow\hept{\begin{cases}x\ne-3\\x\ne2\end{cases}}\)
b) \(A=\frac{\left(x+2\right)}{\left(x+3\right)}-\frac{5}{\left(x^2-2x\right)+\left(3x-6\right)}+\frac{1}{2-x}\)
\(A=\frac{\left(x+2\right)}{\left(x+3\right)}-\frac{5}{x.\left(x-2\right)+3.\left(x-2\right)}+\frac{1}{2-x}\)
\(A=\frac{\left(x+2\right)}{\left(x+3\right)}-\frac{5}{\left(x-2\right)\left(x+3\right)}+\frac{1}{2-x}\)
\(A=\frac{\left(x+2\right)}{\left(x+3\right)}-\frac{5}{\left(x-2\right)\left(x+3\right)}-\frac{1}{x-2}\)
\(A=\frac{\left(x+2\right)\left(x-2\right)}{\left(x+3\right)\left(x-2\right)}-\frac{5}{\left(x-2\right)\left(x+3\right)}-\frac{\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}\)
\(A=\frac{x^2-4-5-x-3}{\left(x+3\right)\left(x-2\right)}\)
\(A=\frac{x^2-x-12}{\left(x+3\right)\left(x-2\right)}\)
\(A=\frac{\left(x^2+3x\right)-\left(4x+12\right)}{\left(x+3\right)\left(x-2\right)}\)
\(A=\frac{x.\left(x+3\right)-4.\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}\)
\(A=\frac{\left(x+3\right)\left(x-4\right)}{\left(x+3\right)\left(x-2\right)}\)
\(A=\frac{x-4}{x-2}\left(x+3\ne0\right)\)
c) \(A=-\frac{3}{4}\)
\(\Leftrightarrow\frac{x-4}{x-2}=-\frac{3}{4}\)
\(\Leftrightarrow4.\left(x-4\right)=-3.\left(x-2\right)\)
\(\Leftrightarrow4x-16=-3x+6\)
\(\Leftrightarrow7x=22\)
\(\Leftrightarrow x=\frac{22}{7}\)
Vậy \(x=\frac{22}{7}\)
Tham khảo nhé~
\(K=\left(\frac{a}{a-1}-\frac{1}{a^2-a}\right):\left(\frac{1}{a+1}+\frac{2}{a^2+1}\right)\)
a/ K xác định khi \(\hept{\begin{cases}a-1\ne0\\a^2-a=a\left(a-1\right)\ne0\\a+1\ne0\end{cases}}\) <=> \(\hept{\begin{cases}a\ne\pm1\\a\ne0\end{cases}}\)
b/ \(K=\left(\frac{a}{a-1}-\frac{1}{a^2-a}\right):\left(\frac{1}{a+1}+\frac{2}{a^2+1}\right)=\left(\frac{a}{a-1}-\frac{1}{a\left(a-1\right)}\right):\left(\frac{1}{a+1}+\frac{2}{a^2+1}\right)\)
=> \(K=\frac{a^2-1}{a\left(a-1\right)}:\frac{a^2+1+2a+2}{\left(a+1\right)\left(a^2+1\right)}\)
=> \(K=\frac{\left(a-1\right)\left(a+1\right)}{a\left(a-1\right)}.\frac{\left(a+1\right)\left(a^2+1\right)}{a^2+2a+3}\)
=> \(K=\frac{\left(a+1\right)^2\left(a^2+1\right)}{a\left(a^2+2a+3\right)}\)
c/ a=1/2
=> \(K=\frac{\left(\frac{1}{2}+1\right)^2\left(\frac{1}{4}+1\right)}{\frac{1}{2}\left(\frac{1}{4}+1+3\right)}=\frac{\frac{9}{4}.\frac{5}{4}}{\frac{17}{8}}=\frac{45}{16}.\frac{8}{17}=\frac{45}{2.17}\)
=> \(K=\frac{45}{34}\)
a, ĐKXĐ: \(a\ne1;a\ne-1\)
Ta có:
\(P=\frac{2a^2}{a^2-1}+\frac{a}{a+1}-\frac{a}{a-1}=\frac{2a^2}{\left(a-1\right)\left(a+1\right)}\) \(+\frac{a\left(a-1\right)}{\left(a+1\right)\left(a-1\right)}-\frac{a\left(a+1\right)}{\left(a-1\right)\left(a+1\right)}\)
\(\Rightarrow P=\frac{2a^2+a^2-a-a^2-a}{\left(a-1\right)\left(a+1\right)}=\frac{2a^2-2a}{\left(a-1\right)\left(a+1\right)}=\frac{2a\left(a-1\right)}{\left(a+1\right)\left(a-1\right)}\)
\(\Rightarrow P=\frac{2a}{a+1}\)
b. Để P có giá trị nguyên \(\Rightarrow2a⋮a+1\Rightarrow2\left(a+1\right)-2a⋮a+1\Rightarrow2a+2-2a⋮a+1\)
\(\Rightarrow2⋮a+1\) vì \(a\in Z\Rightarrow a+1\in\left\{-2;-1;1;2\right\}\Rightarrow a\in\left\{-3;-2;0;1\right\}\)
Vậy \(a\in\left\{-3;-2;0;1\right\}\)
a,ĐK : \(a\ne\pm1\)
\(K=\left(\frac{a}{a-1}-\frac{1}{a^2-a}\right):\left(\frac{1}{a+1}+\frac{2}{a^2-1}\right)\)
\(=\left(\frac{a}{a-1}-\frac{1}{a\left(a-1\right)}\right):\left(\frac{1}{a+1}+\frac{2}{\left(a-1\right)\left(a+1\right)}\right)\)
\(=\left(\frac{a^2}{a\left(a-1\right)}-\frac{1}{a\left(a-1\right)}\right):\left(\frac{a-1}{\left(a+1\right)\left(a-1\right)}+\frac{2}{\left(a+1\right)\left(a-1\right)}\right)\)
\(=\left(\frac{\left(a-1\right)\left(a+1\right)}{a\left(a-1\right)}\right):\left(\frac{a+1}{\left(a+1\right)\left(a-1\right)}\right)\)
\(=\frac{a+1}{a}.\frac{a-1}{1}=\frac{a^2-1}{a}\)
b, Thay a = 1/2 ta được :
\(K=\frac{\left(\frac{1}{2}\right)^2-1}{\frac{1}{2}}=\frac{\frac{1}{4}-1}{\frac{1}{2}}=\frac{-\frac{3}{4}}{\frac{1}{2}}=-\frac{3}{8}\)