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\(A=\left(\frac{1}{1-x}-1\right):\left(x+1-\frac{1-2x}{1-x}\right)\) \(\left(ĐK:x\ne1;x\ne2\right)\)
\(=\frac{1-1+x}{1-x}:\frac{\left(1-x\right)\left(x+1\right)-\left(1-2x\right)}{1-x}\)
\(=\frac{x}{1-x}\cdot\frac{1-x}{1-x^2-1+2x}\)
\(=\frac{x}{-x^2+2x}\)
\(=\frac{x}{-x\left(x-2\right)}=-\frac{1}{x-2}=\frac{1}{2-x}\)
b) Để A=\(\frac{1}{2}\) \(\Leftrightarrow\)\(\frac{1}{2-x}=\frac{1}{2}\)
\(\Leftrightarrow2-x=2\)
\(\Leftrightarrow-x=0\Leftrightarrow x=0\)
c) Để A>1 \(\Leftrightarrow\)\(\frac{1}{2-x}>1\)
\(\Leftrightarrow\)\(\frac{1}{2-x}-1>0\)
\(\Leftrightarrow\)\(\frac{1-2+x}{2-x}>0\)
\(\Leftrightarrow\)\(\frac{x-1}{2-x}>0\)
\(\Leftrightarrow\begin{cases}x-1>0\\2-x>0\end{cases}\) hoặc \(\begin{cases}x-1< 0\\2-x< 0\end{cases}\)
\(\Leftrightarrow\begin{cases}x>1\\x< 2\end{cases}\) hoặc \(\begin{cases}x< 1\\x>2\end{cases}\)(vô nghiệm)
\(\Leftrightarrow1< x< 2\)
Vậy \(1< x< 2\) thì A<1
\(x^2-x+1=x^2-2.x.\frac{1}{2}+\left(\frac{1}{2}\right)^2+\frac{3}{4}=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}>0\forall x\)
\(-x^2+4x-5=-\left(x^2-2.x.2+2^2\right)-1=-\left(x-2\right)^2-1< 0\forall x\)
\(a\left(2a-3\right)-2a\left(a+1\right)=a\left(2a-3-2a-2\right)=-5a⋮5\forall a\inℤ\)
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
\(\frac{3}{\frac{4001}{2000}-2}=\frac{3}{\frac{4001}{2000}-\frac{4000}{2000}}=\frac{3}{\frac{1}{2000}}=6000\)
a) \(ĐKXĐ:\hept{\begin{cases}x\ne-2\\x\ne3\\x\ne2\end{cases}}\)
\(A=\left(1-\frac{4}{x+2}\right):\left(1+\frac{1}{x-3}\right)\)
\(\Leftrightarrow A=\frac{x-2}{x+2}:\frac{x-2}{x-3}\)
\(\Leftrightarrow A=\frac{x-3}{x+2}\)
b) Để A nguyên
\(\Leftrightarrow x-3⋮x+2\)
\(\Leftrightarrow x+2-5⋮x+2\)
\(\Leftrightarrow5⋮x+2\)
\(\Leftrightarrow x+2\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)
\(\Leftrightarrow x\in\left\{-3;-1;-7;3\right\}\)
Vậy để A nguyên \(\Leftrightarrow x\in\left\{-3;1;-7;3\right\}\)
c) Để A > 0
\(\Leftrightarrow\frac{x-3}{x+2}>0\)
\(\Leftrightarrow1-\frac{5}{x+2}>0\)
\(\Leftrightarrow\frac{5}{x+2}< 0\)
\(\Leftrightarrow x+2< 0\)(vì 5 > 0)
\(\Leftrightarrow x< -2\)
Vậy để A > 0 \(\Leftrightarrow x< -2\)