`a,(x+3)(x^2+2021)=0`

`x^2+2021>=2021>0`

`=>x+3=0`

`=>x=-3`

`2,x(x-3)+3(x-3)=0`

`=>(x-3)(x+3)=0`

`=>x=+-3`

`b,x^2-9+(x+3)(3-2x)=0`

`=>(x-3)(x+3)+(x+3)(3-2x)=0`

`=>(x+3)(-x)=0`

`=>` $\left[ \begin{array}{l}x=0\\x=-3\end{array} \right.$

`d,3x^2+3x=0`

`=>3x(x+1)=0`

`=>` $\left[ \begin{array}{l}x=0\\x=-1\end{array} \right.$

`e,x^2-4x+4=4`

`=>x^2-4x=0`

`=>x(x-4)=0`

`=>` $\left[ \begin{array}{l}x=0\\x=4\end{array} \right.$

1) a) \(\left(x+3\right).\left(x^2+2021\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+3=0\\x^2+2021=0\end{matrix}\right.\\\left[{}\begin{matrix}x=-3\left(nhận\right)\\x^2=-2021\left(loại\right)\end{matrix}\right. \)

=> S={-3}

Bài 1: 

a) Ta có: \(\left(x+3\right)\left(x^2+2021\right)=0\)

mà \(x^2+2021>0\forall x\)

nên x+3=0

hay x=-3

Vậy: S={-3}

Bài 2: 

b) Ta có: \(x\left(x-3\right)+3\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)

Vậy: S={3;-3}

\(\dfrac{-5}{2}-\left(\dfrac{-7}{2}-\dfrac{7}{4}\right):\left(\dfrac{-7}{3}\right)-\left(\dfrac{-7}{3}\right)-2021^0\)

\(=\dfrac{-5}{2}-\dfrac{9}{4}+\dfrac{7}{3}-1\)

\(=-\dfrac{41}{12}\)

\(\left(\dfrac{1}{2}\right)^5:\left(\dfrac{1}{3}\right)^2+2\left(-\dfrac{1}{2}\right)^2-2021^0\\ =\dfrac{1}{32}:\dfrac{1}{9}+2.\dfrac{1}{4}-1\\ =\dfrac{9}{32}+\dfrac{1}{2}-1\\ =-\dfrac{7}{32}\)

\(-\dfrac{7}{32}\)

a) \(A=3\left|2x-\dfrac{3}{2}\right|+2021^0=3\left|2x-\dfrac{3}{2}\right|+1\ge1\)

\(minA=1\Leftrightarrow2x=\dfrac{3}{2}\Leftrightarrow x=\dfrac{3}{4}\)

b) \(B=2\left|x-6\right|+3\left(2y-1\right)^2+2021^0=2\left|x-6\right|+3\left(2y-1\right)^2+1\ge1\)

\(minB=1\Leftrightarrow\) \(\left\{{}\begin{matrix}x=6\\y=\dfrac{1}{2}\end{matrix}\right.\)

\(A=3\left|2x-\dfrac{3}{2}\right|+1\ge1\\ A_{min}=1\Leftrightarrow2x-\dfrac{3}{2}=0\Leftrightarrow x=\dfrac{3}{4}\\ B=2\left|x-6\right|+3\left(2y-1\right)^2+1\ge1\\ B_{min}=1\Leftrightarrow\left\{{}\begin{matrix}x=6\\y=\dfrac{1}{2}\end{matrix}\right.\)

a) 2021 - (1/3)² . 3²

= 2021 - 1/9 . 9

= 2021 - 1

= 2020

b) 5/10 + 9 . (-3/2)

= 1/2 - 27/2

= -26/2

= -13

c) -10 . (-2021/2022)⁰ + (2/5)² : 2

= -10 . 1 + 4/25 . 2

= -10 + 8/25

= -68/7

\(a,2021-\left(\dfrac{1}{3}\right)^2\cdot3^2\\ =2021-\dfrac{1}{9}\cdot9\\ =2021-\dfrac{9}{9}\\ =2021-1=2020\\ b,\dfrac{5}{10}+9\cdot\dfrac{-3}{2}\\ =\dfrac{5}{10}+\dfrac{-27}{2}\\ =\dfrac{5}{10}+\dfrac{-135}{10}\\ =-\dfrac{130}{10}\\ =-13\\ c,-10\cdot\left(-\dfrac{2021}{2022}\right)^0+\left(\dfrac{2}{5}\right)^2:2\\ =-10\cdot1+\dfrac{4}{25}\cdot\dfrac{1}{2}\\ =-10+\dfrac{4}{50}\\ =-10+\dfrac{2}{25}\\ =-\dfrac{248}{25}\)

1 

e) E >= 2021 

dấu = xảy ra khi x=1/2

g) G = |x-1|+ |2-x| >= |x-1+2-x|=1

Dấu = xảy ra khi (x-1)(2-x)>=0 <=> 1<=x<=2

h) H = |x-1|+|x-2| + |x-3| 

Ta có : |x-1| + |x-3| = |x-1| + |3-x| >= |x-1+3-x| = 2

|x-2| >=0

=> H>=2

Dấu = xảy ra khi (x-1)(3-x) >=0 ; x-2=0

<=> x=2

k) K = |x-1| + |2x-1| 

2K = |2x-2| + |2x-1| + |2x-1|

Ta có : |2x-2| + |2x-1|  = |2x-2| + |1-2x| >= |2x-2+1-2x|=1

|2x-1| >=0 

Dấu = xảy ra (2x-2)(1-2x) >=0; 2x-1=0

<=> x=1/2

e)Vì \(\left|x-\dfrac{1}{2}\right|\ge0\forall x\)

\(\Leftrightarrow2\left|x-\dfrac{1}{2}\right|\ge0\forall x\\ \Rightarrow2\left|x-\dfrac{1}{2}\right|+2012\ge2012\forall x\)

Dấu "=" xảy ra khi x=\(\dfrac{1}{2}\)

Vậy...

b)G=|x-1|+ |2-x|\(\)

áp dụng bđt |a+b|+ |c+d|\(\ge\left|a+b+c+d\right|\forall x\)

\(\Rightarrow\)ta có |x-1|+ |2-x|\(\ge\) \(\left|x-1+2-x\right|\forall x\)

\(\Leftrightarrow\text{|x-1|+ |2-x| }\ge1\forall x\)

Dấu "=" xảy ra khi 1\(\le x\le2\) \(\forall x\)

Vậy...

h)H= |x-1|+|x-2| + |x-3| 

Ta có |x-1| + |x-3|         

=|x-1| + |3-x| ( trong giá trị tuyệt đối đổi dấu không cần đặt dấu trừ ở ngoài)       

 =>|x-1| + |3-x|\(\ge\left|x-1+3-x\right|\forall x\)          

<=>|x-1| + |3-x|\(\ge2\forall x\) (1)

Mà |x-2|\(\ge0\forall x\) (2)

Từ (1) và (2)=> ta có |x-1|+|x-2| + |x-3| \(\ge2\forall x\)

Dấu "=" xảy ra khi x-2=0

<=>x=2

Vậy...

k) K = |x-1| + |2x-1| 

2K = |2x-2| + |2x-1| + |2x-1|

Mà : |2x-2| + |2x-1| 

=|2x-2| + |1-2x|\(\ge\text{|2x-2+1-2x|}\) \(\forall x\)

Lại có |2x-1| \(\ge\)0 \(\forall x\)

Dấu "=" xảy ra 2x-1=0

<=>x=\(\dfrac{1}{2}\)

Vậy....

e) Ta có: \(2\left|x-\dfrac{1}{2}\right|\ge0\forall x\)

\(\Leftrightarrow2\left|x-\dfrac{1}{2}\right|+2021\ge2021\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)

1:

a: =7/5(40+1/4-25-1/4)-1/2021

=21-1/2021=42440/2021

b: =5/9*9-1*16/25=5-16/25=109/25