x+1 chia hết 2x-1

2(x+1) chia hết 2x-1

2x+2 chia hết 2x-1

2x-1+3 chia hết 2x-1

3 chia hết 2x-1

Do 2x-1 là số lẻ nên 2x-1=-3;-1;1;3

2x=-2;0;2;4

x=-1;0;1;2

\(\left(x-2\right)\left(x^2+2x+4\right)-x\left(x-3\right)\left(x+3\right)=10\\ \Leftrightarrow x^3-8-x\left(x^2-9\right)=10\\ \Leftrightarrow x^3-8-x^3-9x=10\\ \Leftrightarrow-9x=18\\ \Leftrightarrow x=-2\)

x=2

ko bít

ko bít thì ko cần trả lời

\(x+2x+3x+...+100x=2200\)

=>\(x\left(1+2+3+...+100\right)=2200\)

=>\(x.\frac{100.101}{2}=2200\)

=>\(x.5050=2200\)

=>x=2200:5050

=>x=\(\frac{44}{101}\)

\(x+2x+3x+...+100x=220\)

\(\Rightarrow x\left(1+2+3+....+100\right)=2200\)

\(\Rightarrow5050x=2200\)

\(\Rightarrow x=\frac{44}{101}\)

a) |2x-3|+x=21

|2x-3|=21-x

\(\Rightarrow\)\(\orbr{\begin{cases}2x-3=21-x\\2x-3=-\left(21-x\right)\end{cases}}\)

TH1: 2x-3=21-x

2x-x=21+3

x=24

TH2: 2x-3=-(21-x)

2x-3 = -21+x

2x-x=-21+3

x=-18

Vậy x \(\varepsilon\){-18;24}

\(\frac{-6}{3}\left[x-\frac{1}{4}\right]=2x-1\)

\(-2x-\left[\frac{1}{4}.-2\right]=2x-1\)\

\(-2x-\frac{-1}{2}=2x-1\)

\(2x--2x=1-\frac{-1}{2}\)

\(\)\(4x=\frac{3}{2}\)

\(x=\frac{3}{2}:4\)

\(x=\frac{3}{8}\)

a)/x-2/+/x-5/=3
TH1:   

x-2+x-5=3
x+x-2-5=3
     2x-7=3
        2x=3+7
        2x=10
          x=10:2
          x=5
TH2

x-2+x-5= -3
x+x-2-5=-3
     2x-7=-3
        2x=-3+7
        2x=4
          x=4:2
          x=2
Vậy x\(\in\){5;2}

a) \(A=\left(2x-1\right)\left(x+3\right)-\left(x-2\right)\left(3x-4\right)+5x\)

\(=\left(2x^2+6x-x-3\right)-\left(3x^2-4x-6x+8\right)+5x\)

\(=\left(2x^2+5x-3\right)-\left(3x^2-10x+8\right)+5x\)

\(=2x^2+5x-3-3x^2+10x-8+5x\)

\(=x^2+20x-11\)

b) \(5x\left(2x^2-3x+1\right)-2x\left(x+1\right)\left(x-2\right)\)

\(=10x^3-15x^2+5x-2x\left(x^2-2x+x-2\right)\)

\(=10x^3-15x^2+5x-2x^3+4x^2-2x^2+4x\)

\(=8x^3-13x^2+9x\)

c) \(\left(3x+2\right)\left(x+1\right)-2x\left(x+3\right)-2x+1\)

\(=3x^2+3x+2x+2-2x^2-6x-2x+1\)

\(=x^2-3x+3\)

đkxđ:xx>3 

\(\left|5-2x\right|=x-4\) 

=>TH1:

\(5-2x=x-4\)

-x-2x=-5-4

-3x=-9

x=3(loại)

TH2:

5-2x=-x+4

x-2x=-5+4

-x=-1

x=1(loại)

vậy ko tìm đc x thỏa mãn đề bài

\(\left|5-2x\right|-3=x-7\)

\(\left|5-2x\right|=x-7+3\)

\(\left|5-2x\right|=x-4\)

Đk: \(x-4\ge0\)\(\Rightarrow x\ge4\)

Ta có: \(\left|5-2x\right|=x-4\)

\(\Rightarrow\orbr{\begin{cases}5-2x=x-4\\5-2x=-x+4\end{cases}\Rightarrow}\orbr{\begin{cases}-2x-x=-4-5\\-2x+x=4-5\end{cases}\Rightarrow}\orbr{\begin{cases}3x=9\\-x=1\end{cases}\Rightarrow}\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)( cả 2 trường hợp x ko thỏa mãn )

Vậy \(x\in\varnothing\)