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a) \(ĐKXĐ:\hept{\begin{cases}x\ne2\\x\ne3\end{cases}}\)
\(A=\frac{2x-12}{x^2-5x+6}-\frac{x+3}{x-2}+\frac{2x}{x-3}\)
\(\Leftrightarrow A=\frac{2x-12-x^2+9+2x^2-4x}{\left(x-2\right)\left(x-3\right)}\)
\(\Leftrightarrow A=\frac{x^2-2x-3}{\left(x-2\right)\left(x-3\right)}\)
\(\Leftrightarrow A=\frac{\left(x-3\right)\left(x+1\right)}{\left(x-2\right)\left(x-3\right)}\)
\(\Leftrightarrow A=\frac{x+1}{x-2}\)
b) Thay \(x=5\)vào A ta được :
\(A=\frac{5+1}{5-2}=2\)
c) Để \(A\inℤ\)
\(\Leftrightarrow x+1⋮x-2\)
\(\Leftrightarrow3⋮x-2\)
\(\Leftrightarrow x-2\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
\(\Leftrightarrow x\in\left\{1;3;-1;5\right\}\)
Vì \(x\ne3\)
Vậy để \(A\inℤ\Leftrightarrow x\in\left\{1;-1;5\right\}\)
\(ĐKXĐ:x\ne\pm1\)
a) \(P=\frac{2x+3}{x+1}-\frac{x+2}{x-1}+\frac{3x+5}{x^2-1}\)
\(\Leftrightarrow P=\frac{\left(2x+3\right)\left(x-1\right)-\left(x+2\right)\left(x+1\right)+3x+5}{\left(x-1\right)\left(x+1\right)}\)
\(\Leftrightarrow P=\frac{2x^2+x-3-x^2-3x-2+3x+5}{\left(x-1\right)\left(x+1\right)}\)
\(\Leftrightarrow P=\frac{x^2+x}{\left(x-1\right)\left(x+1\right)}\)
\(\Leftrightarrow P=\frac{x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\)
\(\Leftrightarrow P=\frac{x}{x-1}\)
b) Để \(P\inℤ\)
\(\Leftrightarrow x⋮x-1\)
\(\Leftrightarrow x-1+1⋮x-1\)
\(\Leftrightarrow1⋮x-1\)
\(\Leftrightarrow x-1\inƯ\left(1\right)=\left\{\pm1\right\}\)
\(\Leftrightarrow x\in\left\{0;2\right\}\)
Vậy để \(A\inℤ\Leftrightarrow x\in\left\{0;2\right\}\)
a) Điều kiện : \(x\ne2;x\ne3\)
\(B=\frac{2x-9}{x^2-5x+6}-\frac{x+3}{x-2}-\frac{2x+4}{3-x}=\frac{2x-9}{\left(x-2\right)\left(x-3\right)}-\frac{x+3}{x-2}+\frac{2x+4}{x-3}\)
\(=\frac{2x-9-\left(x-3\right)\left(x+3\right)+2\left(x+2\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}=\frac{2x-9-x^2+9+2x^2-8}{\left(x-2\right)\left(x-3\right)}=\frac{x^2+2x-8}{\left(x-2\right)\left(x-3\right)}\)
\(=\frac{\left(x-2\right)\left(x+4\right)}{\left(x-2\right)\left(x-3\right)}=\frac{x+4}{x-3}\)
b) Điều kiện \(x\in Z;x\ne2;x\ne3\)
Có \(B=\frac{x+4}{x-3}\in Z\), mà x+4 và x-3 nguyên do x nguyên, nên
\(x+4⋮x-3\Leftrightarrow7⋮x-3\), do đó \(x-3\inƯ\left(7\right)=\left\{1;7;-1;-7\right\}\Rightarrow x\in\left\{4;10;2;-4\right\}\)
mà do x khác 2 (điều kiện) nên ta kết luận \(x\in\left\{4;10;-4\right\}\)