a: ĐKXĐ: x<>2
Thay x=4 vào A, ta được:
\(A=\dfrac{4\cdot4+4}{4-2}=\dfrac{20}{2}=10\)
b: \(B=\dfrac{x}{2\left(x-1\right)}-\dfrac{x^2+1}{2\left(x-1\right)\left(x+1\right)}=\dfrac{x^2+x-x^2-1}{2\left(x-1\right)\left(x+1\right)}=\dfrac{1}{2x+2}\)
a, ĐKXĐ:\(x-2\ne0\Rightarrow x\ne2\)
Thay x=4 vào A ta có:\(A=\dfrac{4x+4}{x-2}=\dfrac{4.4+4}{4-2}=\dfrac{20}{2}=10\)
\(b,B=\dfrac{x}{2x-2}-\dfrac{x^2+1}{2\left(x^2-1\right)}\\ \Rightarrow B=\dfrac{x\left(x+1\right)}{2\left(x-1\right)\left(x+1\right)}-\dfrac{x^2+1}{2\left(x-1\right)\left(x+1\right)}\\ \Rightarrow B=\dfrac{x^2+x-x^2-1}{2\left(x-1\right)\left(x+1\right)}\\ \Rightarrow B=\dfrac{x-1}{2\left(x-1\right)\left(x+1\right)}\\ \Rightarrow B=\dfrac{1}{2\left(x+1\right)}\)
\(c,AB=\dfrac{4x+4}{x-2}.\dfrac{1}{2\left(x+1\right)}\\ \Rightarrow AB=\dfrac{4\left(x+1\right)}{x-2}.\dfrac{1}{2\left(x+1\right)}\\ \Rightarrow AB=\dfrac{2}{x-2}\)
Để AB nguyên thì\(\dfrac{2}{x-2}\in Z\Rightarrow x-2\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
Ta có bảng:
| x-2 | -1 | -2 | 1 | 2 |
| x | 1(loại) | 0(tm) | 3(tm) | 4(tm) |
Vậy \(x\in\left\{0;3;4\right\}\)
