Câu 1:
1: \(A=\left(\dfrac{x}{x^2-4}-\dfrac{2}{x-2}+\dfrac{1}{x+2}\right):\left(x-2+\dfrac{10-x^2}{x+2}\right)\)
\(=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x-2}+\dfrac{1}{x+2}\right):\dfrac{\left(x-2\right)\left(x+2\right)+10-x^2}{x+2}\)
\(=\dfrac{x-2\left(x+2\right)+x-2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x+2}{x^2-4+10-x^2}\)
\(=\dfrac{2x-2-2x-4}{\left(x-2\right)}\cdot\dfrac{1}{6}=\dfrac{-1}{x-2}\)
|x-1|=3
=>\(\left[{}\begin{matrix}x-1=3\\x-1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\left(nhận\right)\\x=-2\left(loại\right)\end{matrix}\right.\)
Thay x=4 vào A, ta được:
\(A=\dfrac{-1}{4-2}=-\dfrac{1}{2}\)
2: \(5a^2+b^2+c^2-4ab-2a+2c+2=0\)
=>\(\left(4a^2-4ab+b^2\right)+\left(a^2-2a+1\right)+\left(c^2+2c+1\right)=0\)
=>\(\left(2a-b\right)^2+\left(a-1\right)^2+\left(c+1\right)^2=0\)
=>\(\left\{{}\begin{matrix}a-1=0\\c+1=0\\2a-b=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=1\\c=-1\\b=2a=2\end{matrix}\right.\)
\(B=\dfrac{19}{a^{19}}-\dfrac{5}{\left(b-3\right)^5}+\dfrac{1890}{c^{2024}}\)
ư\(=\dfrac{19}{1^{19}}-\dfrac{5}{\left(2-3\right)^5}+\dfrac{1890}{\left(-1\right)^{2024}}\)
\(=19+5+1890=1890+24=1914\)
