\(I.\\ 1,C\\ 2,C\\ II.\\ 1,\\ a,=x^2+5x+6-x^2-5x=6\\ b,=x^3+1+x^3-1-2x^3=0\\ 2,\\ a,\Rightarrow3x^2-6x+3-3x^2+15x=1\\ \Rightarrow9x=-2\Rightarrow x=-\dfrac{2}{9}\\ b,\Rightarrow x^3+3x^2+3x+1-x\left(x^2-4x+4\right)+x-1=0\\ \Rightarrow x^3+3x^2+4x-x^3+4x^2-4x=0\\ \Rightarrow7x^2=0\\ \Rightarrow x=0\\ 3,\\ A=\left(x^2+2\cdot\dfrac{11}{2}x+\dfrac{121}{4}\right)-\dfrac{121}{4}-7\\ A=\left(x+\dfrac{11}{2}\right)^2-\dfrac{189}{4}\ge-\dfrac{189}{4}\\ A_{min}=-\dfrac{189}{4}\Leftrightarrow x=-\dfrac{11}{2}\)
I.Trắc nghiệm:
Câu 1:C
Câu 2:C
II.Tự luận:
Bài 1:
a)(x+2)(x+3)-x(x+5)=x2+5x+6-(x2+5x)=x2+5x+6-x2-5x=6
b)(x+1)(x2-x+1)+(x-1)(x2+x+1)-2x3=x3+1+x3-1-2x3=0
Bài 2:
a)3(x-1)2-3x(x-5)=1
<=>3(x2-2x+1)-(3x2-15x)=1
<=>3x2-6x+3-3x2+15x=1
<=>9x+3=1
<=>9x=-2
<=>x=\(-\dfrac{2}{9}\)
b)(x+1)3-x(x-2)2+x-1=0
<=>x3+1-x(x2-4x+4)+x-1=0
<=>x3+1-(x3-4x2+4x)+x-1=0
<=>x3+1-x3+4x2-4x+x-1=0
<=>4x2-3x=0
<=>x(4x-3)=0
<=>x=0 hay 4x-3=0
<=>x=0 hay 4x=3
<=>x=0 hay x=\(\dfrac{3}{4}\)
I.Trắc nghiệm:
Câu 1:C
Câu 2:B
II.Tự luận:
Bài 1:
a)(x+2)(x+3)-x(x+5)=x2+5x+6-(x2+5x)=x2+5x+6-x2-5x=6
b)(x+1)(x2-x+1)+(x-1)(x2+x+1)-2x3=x3+1+x3-1-2x3=0
Bài 2:
a)3(x-1)2-3x(x-5)=1
<=>3(x2-2x+1)-(3x2-15x)=1
<=>3x2-6x+3-3x2+15x=1
<=>9x+3=1
<=>9x=-2
<=>x=34