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Câu d là dấu " ) "  đúng không bạn ?

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

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

\(-1+3x=3\)

\(3x=3-\left(-1\right)=3+1\)

\(3x=4\)

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

8/15 - 2/15 : x = 0,2

2/15 : x = 8/15 - 0,2

2/15 : x = 1/3

x = 2/15 : 1/3

x = 2/5

em làm ngược lại thoi 

2/15 : x = 8 /15 - 0,2

x = 2/15:( 8 /15 - 0,2)

a)4x(x+1)+(3-2x)=15

  4x² + 4x + 3 - 2x = 15

  4x² + 2x -12 = 0

  4x² + 8x - 6x - 12 = 0

  4x(x+2) - 6(x+2) = 0

   (x+2)(4x - 6)= 0 => TH1: x = -2 . TH2: x = 3/2

b)9x(x-2014)-x+2014=0

9x(x-2014)-(x-2014) = 0

(x-2014) (9x -1) = 0 => TH1: x = 2014. TH2: x=1/9

a)x=4

các bạn ơi nhanh lên mình k cho

\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{\left(2x+1\right).\left(2x+3\right)}=\frac{15}{93}\)

\(2.\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{\left(2x+1\right).\left(2x+3\right)}\right)=2.\frac{15}{93}\)

\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{\left(2x+1\right).\left(2x+3\right)}=\frac{30}{93}\)

\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{2x+1}-\frac{1}{2x+3}=\frac{10}{31}\)

\(\frac{1}{3}-\frac{1}{2x+3}=\frac{10}{31}\)

\(\frac{1}{2x+3}=\frac{1}{3}-\frac{10}{31}\)

\(\frac{1}{2x+3}=\frac{1}{93}\)

=> 2x + 3 = 93

=> 2x = 93 - 3

=> 2x = 90

=> x = 90 : 2

=> x = 45

Vậy x = 45

sai rồi

\(a,2x-1-3x\left(2x-1\right)=0\)

\(\Leftrightarrow2x-1-6x^2+3x=0\)

\(\Leftrightarrow5x-1-6x^2=0\)

\(\Leftrightarrow6x^2-5x+1=0\)

\(\Leftrightarrow6x^2-2x-3x+1=0\)

\(\Leftrightarrow2x\left(3x-1\right)-\left(3x-1\right)=0\)

\(\Leftrightarrow\left(2x-1\right)\left(3x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x-1=0\\3x-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x=1\\3x=1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{1}{3}\end{cases}}\)

\(b,2x^2+4x=0\)

\(\Leftrightarrow2x\left(x+4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x=0\\x+4=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-4\end{cases}}\)

a) \(2x\left(x-3\right)+6\left(3-x\right)=0\)

\(\Leftrightarrow2\left[x\left(x-3\right)+3\left(3-x\right)\right]=0\)

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

\(\Leftrightarrow x-3=0\)

\(\Rightarrow x=3\)

b) \(3x\left(2x-5\right)-15\left(5-2x\right)=0\)

\(\Leftrightarrow3\left[x\left(2x-5\right)-5\left(5-2x\right)\right]=0\)

\(\Leftrightarrow x\left(2x-5\right)-5\left(5-2x\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(2x-5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+5=0\\2x-5=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-5\\x=\frac{5}{2}\end{cases}}\)

bạn cho mình cách giải đc ko

tổng trên có số số hạng là:

(x-1):1+1=x (số)

tổng trên là:

(x+1)x:2=820

=> (x+1)x=820.2=1640=41.40

=>x=40

hinh nhu de bai sai