a)
\(\left|x\right|=2=>\left[{}\begin{matrix}x=2\\x=-2\left(loaividieukien\right)\end{matrix}\right.\)
thay x=2 vào biểu thức B ta có
\(\dfrac{2\cdot2+2}{2+2}=\dfrac{6}{4}=1,5\)
b)
\(\dfrac{x+1}{2x-2}+\dfrac{1}{2-2x^2}\\ =\dfrac{x+1}{2x-2}-\dfrac{1}{2x^2-2}\\ =\dfrac{x+1}{2\left(x-1\right)}-\dfrac{1}{2\left(x-1\right)\left(x+1\right)}\\ =\dfrac{\left(x+1\right)^2}{2\left(x-1\right)\left(x+1\right)}-\dfrac{1}{2\left(x-1\right)\left(x+1\right)}\\ =\dfrac{x^2+2x+1-1}{2\left(x-1\right)\left(x+1\right)}\\ =\dfrac{x^2+2x}{2\left(x-1\right)\left(x+1\right)}\\ =\dfrac{x\left(x+2\right)}{2\left(x-1\right)\left(x+1\right)}\)
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