B nha em

Chọn B

\(x:\frac{1}{2}+x:\frac{1}{4}+x:\frac{1}{100}=\frac{106}{10}\)

\(Xx2+Xx4+Xx100=\frac{106}{10}\)

\(Xx\left(2+4+100\right)=\frac{106}{10}\)

\(Xx106=\frac{106}{10}\)

\(X=\frac{106}{10}:106\)

\(\Rightarrow X=\frac{1}{10}\)

Ta có :

\(x:\frac{1}{2}+x:\frac{1}{4}+x:\frac{1}{100}=\frac{106}{10}\)

\(\Leftrightarrow\)\(2x+4x+100x=\frac{106}{10}\)

\(\Leftrightarrow\)\(x\left(2+4+100\right)=\frac{106}{10}\)

\(\Leftrightarrow\)\(106x=\frac{106}{10}\)

\(\Leftrightarrow\)\(x=\frac{106}{10}:106\)

\(\Leftrightarrow\)\(x=\frac{106}{10}.\frac{1}{106}\)

\(\Leftrightarrow\)\(x=\frac{1}{10}\)

Vậy \(x=\frac{1}{10}\)

x :1/2 +x :1/4+ x: 1/100=106/10

x : ( 1/2 + 1/4 + 1/100 ) = 106/10

x : 301/400 = 106/10

     x            = 106/10 * 301/400

     x            = 15503/2000

(x+1)+(x+4)+(x+7)+(x+10)=62

=>4x+22=62

=>4x=62-22

=>4x=40

=>x=10

t tôi nha bn

x+(1+4+7+10)=62

x+22=62

x=62-22

x=40

( x + 1 ) + ( x + 4 ) + ( x + 7 ) + ( x + 10 ) = 62

4x+ (1+4+7+10)=62

4x+22=62

4x=62-22

4x=40

x=40:4

x=10

`x-3/5 =-4/10`

`=>x=-4/10 + 3/5`

`=>x=-4/10 + 6/10`

`=>x= 2/10`

`=>x=1/5`

`----`

`(x-1)/(-2) = -8/(x-1)`

`=> (x-1).(x-1) = (-2).(-8)`

`=> (x-1)^2 = 16`

`=> (x-1)^2 = +- 4^2`

\(\Rightarrow\left[{}\begin{matrix}x-1=4\\x-1=-4\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=4+1\\x=-4+1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)

Lời giải:

$2+4+6+8+10+...+2500=2(1+2+3+4+...+1250)$

$=2.1250(1250+1):2=1250(1250+1)=x(x+1)$

$\Rightarrow x=1250$ (điều kiện $x$ là stn)

\(\left(x+1\right)+\left(x+4\right)+\left(x+7\right)+\left(x+10\right)=62\)

\(\Rightarrow x+1+x+4+x+7+x+10=62\)

\(\Rightarrow4x+22=62\)

\(\Rightarrow4x=40\)

\(\Rightarrow x=10\)

x+(1+4+7+10)=62

x+22=62

x=62-22

x=40

khoan cach giua 2 so la :2-1=1

co tat ca so so hang hay x la :(10-1):1+1=10

(x+1)+(x+2)+.....................(x+10)=2015

x*10+[(1+10)*10/2]=2015

x*10+55               =2015

x*10                     =2015-55

x*10                     =1960

        x                  =1960/10

       x                    =196   

10x+55=2015

10x=1960

x=196

tìm x biết:( x+1) + ( x+ 2) + ( x+3)+ ( x+4)+.....+( x+9)+ ( x+10) = 2015

Giải:Ta có:\(\left(x+1\right)+\left(x+2\right)+......+\left(x+10\right)=2015\)

\(\Rightarrow x+1+x+2+.....+x+10=2015\)

\(\Rightarrow10x+\left(1+2+......+10\right)=2015\)

\(\Rightarrow10x+55=2015\Rightarrow10x=2015-55\)

\(\Rightarrow10x=1960\Rightarrow x=196\)

Vậy .....................

\(a,x+5x^2=0\\ \Rightarrow a,x\left(1+5x\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{5}\end{matrix}\right.\\ b,\left(x+3\right)^2+\left(4+x\right)\left(4-x\right)=0\\ \Rightarrow x^2+6x+9+16-x^2=0\\ \Rightarrow6x+25=0\\ \Rightarrow6x=-25\\ \Rightarrow x=-\dfrac{25}{6}\)

\(c,5x\left(x-1\right)=x-1\\ \Rightarrow c,5x\left(x-1\right)-\left(x-1\right)\\ \Rightarrow\left(x-1\right)\left(5x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\\ d,x^2-2x-3=0\\ \Rightarrow\left(x^2-3x\right)+\left(x-3\right)=0\\ \Rightarrow x\left(x-3\right)+\left(x-3\right)=0\\ \Rightarrow\left(x+1\right)\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\)