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

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

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

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

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

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

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-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`

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

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

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

-Từ số 4! đến số 10! đều chia hết cho 20 do có thừa số 4.5=20.

-Mà 1!+2!+3!=1+2+6=91!+2!+3!=1+2+6=9 chia 20 dư 9 nên tổng đó chia 20 dư 9. 

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

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

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)\(\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=6.\)

\(\Leftrightarrow x^2-4x+4-x^2+9-6=0\)

\(\Leftrightarrow-4x+7=0\)

\(\Leftrightarrow4x=7\Leftrightarrow x=1,75\)

\(b,4\left(x-3\right)^2-\left(2x-1\right)\left(2x+1\right)=10.\)

\(\Leftrightarrow4\left(x^2-6x+9\right)-4x^2+1-10=0\)

\(\Leftrightarrow-24x+27=0\)

\(\Leftrightarrow24x=27\Leftrightarrow x=1,125\)

Trả lời :.....................................

\(\Leftrightarrow24x=27\Leftrightarrow x=1,125..........................\)

Hk tốt,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,

Học sinh giỏi 6A