Tìm \(x\) biết \(\left( {x + 1} \right)\left( {x + 3} \right) - x\left( {x + 2} \right) = 7\).

\(5x{y^3}\)

\( - 2{x^3}yz\)

\({x^2}y\,( - 5x)\)

\( - 2x{y^3}\)

\(2\)

\(1\)

\(0\)

\(3\)

\(\,P\left( x \right) + Q\left( x \right) = \,{x^2} + x + \dfrac{3}{4};\)\(P\left( x \right) - Q\left( x \right)\, = 8{x^4} + 8{x^3} + 13{x^2} - x + \dfrac{5}{4}\)       

\(\,P\left( x \right) + Q\left( x \right) = 8{x^4} + 8{x^3} + 13{x^2} - x + \dfrac{5}{4}\,;\)\(P\left( x \right) - Q\left( x \right)\, = {x^2} + x + \dfrac{3}{4}\)         

\(\,P\left( x \right) + Q\left( x \right) = \,{x^2} + x - \dfrac{3}{4};\)\(P\left( x \right) - Q\left( x \right)\, = 8{x^4} + 8{x^3} + 13{x^2} + x + \dfrac{5}{4}\)

\(\,P\left( x \right) + Q\left( x \right) = \,{x^2} + 2x + \dfrac{3}{4};\)\(P\left( x \right) - Q\left( x \right)\, = 8{x^4} + 8{x^3} - 13{x^2} - x + \dfrac{5}{4}\)

\(P\left( x \right) = \,4{x^4} + 4{x^3} + 7{x^2} + 1\,;\) \(Q\left( x \right) = \, - 4{x^4} + 4{x^3} + 6{x^2} + x - \dfrac{1}{4}\)

\(P\left( x \right) = \,4{x^4} + 4{x^3} + 7{x^2} + 1\,;\) \(Q\left( x \right) = \,4{x^4} - 4{x^3} - 6{x^2} + x - \dfrac{1}{4}\)

\(P\left( x \right) = \,4{x^4} + 4{x^3} + 7{x^2}\,;\) \(Q\left( x \right) = \, - 4{x^4} - 4{x^3} - 6{x^2} + x - \dfrac{1}{4}\)

\(P\left( x \right) = \,4{x^4} + 4{x^3} + 7{x^2} + 1\,;\) \(Q\left( x \right) = \, - 4{x^4} - 4{x^3} - 6{x^2} + x - \dfrac{1}{4}\)

\(P\left( x \right) = \,4{x^4} + 4{x^3} + 7{x^2} + 1\,;\) \(Q\left( x \right) = \, - 4{x^4} + 4{x^3} + 6{x^2} + x - \dfrac{1}{4}\)

\(P\left( x \right) = \,4{x^4} + 4{x^3} + 7{x^2} + 1\,;\) \(Q\left( x \right) = \,4{x^4} - 4{x^3} - 6{x^2} + x - \dfrac{1}{4}\)

\(P\left( x \right) = \,4{x^4} + 4{x^3} + 7{x^2}\,;\) \(Q\left( x \right) = \, - 4{x^4} - 4{x^3} - 6{x^2} + x - \dfrac{1}{4}\)

\(P\left( x \right) = \,4{x^4} + 4{x^3} + 7{x^2} + 1\,;\) \(Q\left( x \right) = \, - 4{x^4} - 4{x^3} - 6{x^2} + x - \dfrac{1}{4}\)

\({\rm N}\) là giao ba đường trung trực của \(\Delta BHC.\)

\({\rm N}\) là giao ba đường phân giác của \(\Delta BHC.\)

\({\rm N}\) là trực tâm của \(\Delta BHC.\)

\({\rm N}\) là trọng tâm của \(\Delta BHC.\)

\(IK = AD\)

\(DK//AI\)

\(IK > AD\)

Cả A, B đều đúng