Câu 1:
a: \(A=\dfrac{x^3+x^2}{x^3-2x^2+x}:\left(\dfrac{x^2+x}{x^2}+\dfrac{1}{x-1}+\dfrac{2-x^2}{x^2-x}\right)\)
\(=\dfrac{x^2\left(x+1\right)}{x\left(x^2-2x+1\right)}:\left(\dfrac{x\left(x+1\right)}{x^2}+\dfrac{1}{x-1}+\dfrac{2-x^2}{x\left(x-1\right)}\right)\)
\(=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\left(\dfrac{x+1}{x}+\dfrac{1}{x-1}+\dfrac{2-x^2}{x\left(x-1\right)}\right)\)
\(=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\dfrac{\left(x+1\right)\left(x-1\right)+x+2-x^2}{x\left(x-1\right)}\)
\(=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}\cdot\dfrac{x\left(x-1\right)}{x^2-1+x+2-x^2}\)
\(=\dfrac{x^2\cdot\left(x+1\right)}{x-1}\cdot\dfrac{1}{x+1}=\dfrac{x^2}{x-1}\)
b: \(A=\dfrac{x^2}{x-1}=\dfrac{x^2-1+1}{x-1}=x+1+\dfrac{1}{x-1}=x-1+\dfrac{1}{x-1}+2\)
=>\(A>=2\cdot\sqrt{\left(x-1\right)\cdot\dfrac{1}{x-1}}+2=2+2=4\)
Dấu '=' xảy ra khi x-1=1
=>x=2(nhận)
Câu 2:
1: ĐKXĐ: \(x\notin\left\{2;-2;-3\right\}\)
\(\dfrac{4}{x^2-4}+\dfrac{1}{x^2+5x+6}=\dfrac{-5}{4}\)
=>\(\dfrac{4}{\left(x-2\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}=\dfrac{-5}{4}\)
=>\(\dfrac{4\left(x+3\right)+x-2}{\left(x-2\right)\left(x+2\right)\left(x+3\right)}=\dfrac{-5}{4}\)
=>\(\dfrac{4x+12+x-2}{\left(x^2-4\right)\left(x+3\right)}=\dfrac{-5}{4}\)
=>\(\dfrac{5\left(x+2\right)}{\left(x+2\right)\left(x-2\right)\left(x+3\right)}=\dfrac{-5}{4}\)
=>\(\dfrac{1}{\left(x-2\right)\left(x+3\right)}=\dfrac{-1}{4}\)
=>(x-2)(x+3)=-4
=>\(x^2+x-6+4=0\)
=>\(x^2+x-2=0\)
=>(x+2)(x-1)=0
=>\(\left[{}\begin{matrix}x=-2\left(loại\right)\\x=1\left(nhận\right)\end{matrix}\right.\)
2: \(P=\dfrac{a^3+b^3+c^3-3bac}{a^2+b^2+c^2-ab-bc-ac}\)
\(=\dfrac{\left(a+b\right)^3+c^3-3ba\left(a+b\right)-3abc}{a^2+b^2+c^2-ab-bc-ac}\)
\(=\dfrac{\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)}{a^2+b^2+c^2-ab-bc-ac}\)
\(=\dfrac{\left(a+b+c\right)\left[a^2+2ab+b^2-ac-bc+c^2-3ab\right]}{a^2+b^2+c^2-ab-bc-ac}\)
\(=a+b+c=2025\)
3: \(P\left(x\right)⋮Q\left(x\right)\)
=>\(ax^4+bx^3+1⋮x^2-2x+1\)
=>\(ax^4-2a\cdot x^3+a\cdot x^2+\left(2a+b\right)x^3-\left(4a+2b\right)x^2+\left(2a+b\right)x+\left(3a+2b\right)x^2-2\cdot\left(3a+2b\right)x+\left(3a+2b\right)+\left(-2a-b+6a+4b\right)x-3a-2b+1⋮x^2-2x+1\)
=>\(\left\{{}\begin{matrix}3a-2b+1=0\\-2a-b+6a+4b=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}3a-2b=-1\\4a+3b=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}12a-8b=-4\\12a+9b=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}12a-8b-12a-9b=-4-0=-4\\4a=-3b\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-17b=-4\\a=-\dfrac{3}{4}b\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{4}{17}\\a=-\dfrac{3}{4}\cdot\dfrac{4}{17}=-\dfrac{3}{17}\end{matrix}\right.\)
