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\(2+4+6+...+2x=1010\)\(0\)

\(\Rightarrow2.\left(1+2+3+...+x\right)=10100\)

\(\Rightarrow1+2+3+...+n=10100:2=5050\)

\(\Rightarrow n.\left(n+1\right):2=5050\)

\(\Rightarrow n.\left(n+1\right)=5050.2=10100\)

\(\Rightarrow n.\left(n+1\right)=100.101\)

\(\Rightarrow n=100\)

\(2+4+6+.......+2x=10100.\)

\(2\cdot1+2\cdot2+2\cdot3+........+2x=10100\)

\(2\cdot\left(1+2+3+.....+x\right)=10100\)

\(1+2+3+....+x=10100:2\)

\(1+2+3+...+x=5050\)

\(\Rightarrow\left(x+1\right).\left(x-1+1\right):2=5050\)

\(\left(x+1\right).x:2=5050\)

\(\Rightarrow x.\left(x+1\right)=5050.2=10100\)

\(\text{Mà:}10100=2^2\cdot5^2\cdot101=100\cdot101\)

\(\Rightarrow x=100\)

A=2(1+2+3+.........+x)=10100

suy ra 1+2+3+.................+x=5050

\(\frac{x\left(x+1\right)}{2}\)=5050 \(\Rightarrow\)x²+x=10100\(\Rightarrow\)x²+x-10100=0\(\Rightarrow\)x=100 hoac x=-101 ma x>0 nen x=100

100

tick cho mình tròn 110 đi

huhuhu

Ai tick cho mình tròn 40 với 

Cảm ơn các bạn nhiều lắm

n=100

tick cho mình nha

chắt chắn 100%

\(2+4+6+.....+2n=10100\)

\(\Rightarrow\frac{\left(\left(2n-2\right):2+1\right).\left(2+2n\right)}{2}=10100\)

\(\Rightarrow\frac{\left(n-1+1\right).\left(2.\left(n+1\right)\right)}{2}=10100\)

\(\Rightarrow\frac{\left(n-0\right).2\left(n+1\right)}{2}=10100\)

\(\Rightarrow n.2\left(n+1\right)=10100.2=20200\)

\(\Rightarrow n.\left(n+1\right)=20200:2=10100\)

\(\Rightarrow n.\left(n+1\right)=101.100\)

\(\Rightarrow n=100\)

Vậy n = 100

Nhớ k cho mình nhé! Thank you!!!

Có : 

 2 + 4 + 6 +...+ 2.x = 10100

=>  2.( 1 + 2 + 3 +...+ n ) = 10100

=> 1 + 2 + 3 +...+ n = 10100 : 2 = 5050

=> n.(n+1) : 2 = 5050

=> n. ( n + 1 ) = 5050.2 = 10100

=> n. ( n + 1 ) = 100 . 101

=> n = 100

2+4+6+8+.....+2n=10100

2(1+2+3+4+....+n)=10100

1+2+3+4+....+n=10100:2

1+2+3+4+...+n=5050

\(\frac{\left(n+1\right)n}{2}\)=5050

(n+1)n=5050.2

(n+1)n=10100

(n+1)n=101.100

Vậy n=100

Chúc bạn học tốt

n=100

các bạn cho mk vài li-ke cho tròn 600 với 

Ta có : 2 + 4 + 6 +... + 2n = 10100

=> 2(1 + 2 + 3 + ... + n) = 10100

=> 2n(n + 1) : 2              = 10100

=> n(n + 1) = 100.101

=> n = 100

Vậy n = 100

                                       Lời giải:

   2 + 4 + 6 + ....+ 2n = 10100

= 2 . ( 1 + 2 + 3 + .... + n) = 10100

= 2n. (n + 1) :2                    = 10100

= n . (n + 1)  = 100 . 101

=> n = 100

        Vậy n=100

Chúc học tốt!!!

\(2+4+6+.....+2n=10100\)

\(\Leftrightarrow2\left(1+2+3+.....+n\right)=10100\)

\(\Leftrightarrow2.\frac{n\left(n+1\right)}{2}=10100\)\(\Leftrightarrow n\left(n+1\right)=10100=100.101\)

\(\Leftrightarrow n=100\)

Vậy \(n=100\)

a: Ta có: \(3\left(2x-3\right)+2\left(2-x\right)=-3\)

\(\Leftrightarrow6x-9+4-2x=-3\)

\(\Leftrightarrow4x=2\)

hay \(x=\dfrac{1}{2}\)

Bài 1: (2x+6)+2014=2017

      => 2x+6          = 2017-2014

      => 2x+6          = 3

      => 2x              = -3

      => x                = -3/2

Bài 2: x+2x+3x+.....+100x = 10100

=> x.(1+2+3+......+99+100) = 10100

=> x. 5050 = 10100

=> x          = 10100 : 5050

=> x          = 2

Vậy x = 2

Bài 1 :

\(48:\left(2x+6\right)+2014=2017\)

\(48:\left(2x+6\right)=2017-2014\)

\(48:\left(2x+6\right)=3\)

\(2x+6=48:3\)

\(2x+6=16\)

\(2x=16-6\)

\(2x=10\)

\(x=10:5\)

\(x=2\)

b) \(x+2x+3x+...+100x=10100\)

\(\Rightarrow1x+2x+3x+...+100x=10100\)

\(\Rightarrow x\cdot\left(1+2+3+...+100\right)=10100\)

\(\Rightarrow x\cdot\frac{\left(100+1\right)\cdot100}{2}=10100\)

\(\Rightarrow x\cdot5050=10100\)

\(\Rightarrow x=10100:5050\)

\(\Rightarrow x=2\)

1.

 \(x^2-5x+6=0\\ \Rightarrow x^2-2x-3x+6=0\\ \Rightarrow\left(x^2-2x\right)-\left(3x-6\right)=0\\ \Rightarrow x\left(x-2\right)-3\left(x-2\right)=0\\ \Rightarrow\left(x-2\right)\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)

2.

\(\left(x+4\right)^2-\left(3x-1\right)^2=0\\ \Rightarrow\left(x+4-3x+1\right)\left(x+4+3x-1\right)=0\\ \Rightarrow\left(-2x+5\right)\left(4x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}-2x+5=0\\4x+3=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{3}{4}\end{matrix}\right.\)

3.

\(x^2-2x+24=0\\ \Rightarrow\left(x^2-2x+1\right)+23=0\\ \Rightarrow\left(x-1\right)^2+23=0\)

Vì (x-1)²≥0

23>0

\(\Rightarrow\left(x-1\right)^2+23>0\)

Vậy x vô nghiệm

4.

\(9x^2-4=0\\ \Rightarrow\left(3x-4\right)\left(3x+4\right)=0\\ \Rightarrow\left[{}\begin{matrix}3x-4=0\\3x+4=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=-\dfrac{4}{3}\end{matrix}\right.\)

5.

\(x^2+2x-8=0\\ \Rightarrow\left(x^2+2x+1\right)-9=0\\ \Rightarrow\left(x+1\right)^2-3^2=0\\ \Rightarrow\left(x-2\right)\left(x+4\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2=0\\x+4=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)