1: \(\dfrac{2x^3+11x^2+18x-3}{2x+3}\)

\(=\dfrac{2x^3+3x^2+8x^2+12x+6x+9-12}{2x+3}\)

\(=x^2+4x+3-\dfrac{12}{2x+3}\)

\(a,\Rightarrow x=\dfrac{7}{3}+\dfrac{6}{7}=\dfrac{49+18}{21}=\dfrac{67}{21}\\ b,\Rightarrow x=\dfrac{15}{7}-\dfrac{13}{14}=\dfrac{30-13}{14}=\dfrac{17}{14}\\ c,\Rightarrow3x=\dfrac{9}{4}-\dfrac{5}{6}=\dfrac{27-10}{12}=\dfrac{17}{12}\\ \Rightarrow x=\dfrac{17}{12}\cdot\dfrac{1}{3}=\dfrac{17}{36}\)

\(x=\dfrac{7}{3}+\dfrac{6}{7}=\dfrac{67}{21}\)

\(x=\dfrac{15}{7}-\dfrac{13}{14}=\dfrac{17}{14}\)

\(x=\left(\dfrac{9}{4}-\dfrac{5}{6}\right):3=\dfrac{17}{36}\)

\(x-\dfrac{6}{7}=\dfrac{7}{3}\\ \Rightarrow x=\dfrac{7}{3}+\dfrac{6}{7}\\ \Rightarrow x=\dfrac{67}{21}\)

\(x+\dfrac{13}{14}=\dfrac{15}{7}\\ \Rightarrow x=\dfrac{15}{7}-\dfrac{13}{14}\\ \Rightarrow x=\dfrac{17}{14}\)

\(\dfrac{9}{4}-3\times x=\dfrac{5}{6}\\ \Rightarrow3\times x=\dfrac{9}{4}-\dfrac{5}{6}\\ \Rightarrow3\times x=\dfrac{17}{12}\\ \Rightarrow x=\dfrac{17}{12}:3\\ \Rightarrow x=\dfrac{17}{36}\)

3x^2 +18x +33

3(x^2 +6x +11)

3(x^2 +2.x.3 +3^2) +11- 3^2

3(x+3)^ 2- 2

vì 3(x+3)^2 lớn hơn hoặc bằngg 0 => 3(x+3)^2 -2 lớn  hơn hoặc = -2 

=>biếu thức trên nhỏ nhất = - 2 khi x+3 = 0 => x= -3 

\(a,\left(3x+1\right)\left(3x-1\right)-\left(18x^3+5x^2-2x\right):2x\\ =\left(9x^2-1\right)-\left(9x^2+\dfrac{5}{2}x-1\right)\\ =9x^2-1-9x^2-\dfrac{5}{2}x+1=\dfrac{5}{2}x\)

\(b,3x\left(x-2021\right)-x+2021=0\\ \Rightarrow b,3x\left(x-2021\right)-\left(x-2021\right)=0\\ \Rightarrow\left(x-2021\right)\left(3x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=2021\\x=\dfrac{1}{3}\end{matrix}\right.\)

1​023 chia hết cho 3 không chia hết cho 9

​vt: Phải chia hết cho 3 => x=3t khi x=3t thì vế trái chia hết cho 9 => đpcm

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Chọn A