Tô Thị Hằng

\(\left(3x-6\right)\left(4x-8\right)=0\)

\(\Rightarrow\orbr{\begin{cases}3x-6=0\\4x-8=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}3x=0+6\\4x=0+8\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}3x=6\Rightarrow x=6:3=2\\4x=8\Rightarrow x=8:4=2\end{cases}}\)

Vậy...

ko chi tiết lắm

Chữ Ben 10 xấu nhỉ !

\(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+..........+\left(x+10\right)=165\)

\(\Leftrightarrow x+1+x+2+..........+x+10=165\)

\(\Leftrightarrow\left(x+x+x+.......+x\right)+\left(1+2+....+10\right)=165\)

\(\Leftrightarrow10x+55=165\)

\(\Leftrightarrow10x=110\)

\(\Leftrightarrow x=11\)

Vậy ...

\(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+10\right)=165\)

\(\Leftrightarrow x+1+x+2+x+3+...+x+10=165\)

\(\Leftrightarrow\left(x+x+x+...+x\right)+\left(1+2+3+...+10\right)=165\)

\(\Leftrightarrow10x+55=165\)

\(\Leftrightarrow10x=165-55\)

\(\Leftrightarrow10x=110\)

\(\Leftrightarrow x=110:10\)

\(\Leftrightarrow x=11\)

Vậy \(x=11\)

[x+1]+[x+2]+[x+3]+................+[x+10] =165

(x+x+x+...+x)+(1+2+3+10) = 165

         10x      +       55       = 165

                     10x              = 165 - 55

                      10x             = 110

                       x                = 110 : 10

                       x                = 11

Vậy ...

[x + 1] + [x + 2] + [x + 3] + ..... + [x + 10] = 165

10x + (1 + 2 + 3 + ..... + 10) = 165

10x + 45 = 165

10x = 120

=> x = 12

Bằng 12 nha bn ! 

k mink nha 

Vấn đề ở đây là trong dấu ngoặc thứ nhất có bao nhiêu x hả bạn

ưn...đúng đó

(x + x + ... + x) + (1 + 2 + 3 + ... + 10) = 165

=> 10x + 55 = 165

=> 10x = 110

=> 10x = 11

(X+1)+(x+2)+......+(x+10)=165

(X+x+x+.......+x)+(1+2+3+......+10)=165

10.x                  +55                        =165

10x=165-55

10x=110

 x=110:10

X=10

de vay

a) 2y - 12y = 0

\(\Rightarrow\) y ( 2-12) = 0

\(\Rightarrow\) y . (-10) =0 

\(\Rightarrow\) y = 0 : (-10) = 0

b) (y-7)(y-8) = 0

\(\Rightarrow\orbr{\begin{cases}y-7=0\\y-8=0\end{cases}\Rightarrow\orbr{\begin{cases}y=0+7\\y=0+8\end{cases}\Rightarrow}\orbr{\begin{cases}y=7\\y=8\end{cases}}}\)

c) x + x.2+x.3+x.4+...+x.10 = 165

\(\Rightarrow\) x ( 1+2+3+.....+8+9+10) = 165

\(\Rightarrow\)x . \(\frac{\left(1+10\right).10}{2}\)=165

\(\Rightarrow\) x . 55 = 165

\(\Rightarrow x=\frac{165}{55}=3\)

Can you k for me ,Lê Thị Kim Chi!

a) \(2y-12y=0\)

\(\Leftrightarrow-10y=0\)

\(\Leftrightarrow y=0:\left(-10\right)\)

\(\Leftrightarrow y=0\)

b) \(\left(y-7\right)\left(y-8\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}y-7=0\\y-8=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}y=0+7\\y=0+8\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}y=7\\y=8\end{cases}}\)

c) \(x+x.2+x.3+......+x.10=165\)

\(\Leftrightarrow x.\left(1+2+3+.....+10\right)=165\)

\(\Leftrightarrow x.55=165\)

\(\Leftrightarrow x=165:55\)

\(\Leftrightarrow x=3\)

(=)10x+(1+2+....10)=165

(=)10x+55=165

(=)10x=110

(=)x=11

(X + 1) + (X + 2) + (X + 3) + ..... + (X + 10) = 165

( X x 10 ) + ( 1 + 2 + 4 + ... + 10 ) = 165

( X x 10 ) + 45 = 165

X x 10 = 165 - 45

X x 10 = 120

X = 120 : 10

X = 12

Ta có \(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+10\right)=165\)

\(\Rightarrow\left(x+x+x+...+x\right)+\left(1+2+3+...+10\right)=165\)

\(\Rightarrow10x+55=165\)

\(\Rightarrow10x=110\)

\(\Rightarrow x=11\)