x+x-1+x-2+x-3+x-4+...+x-50 = 225

<=> 51x-(1+2+...+50) = 225

<=> 51x - 1275 = 225

<=> 51x = 1500

<=> x = 500/17

\(x+x-1+x-2+...+x-50=225\)

=> \(\left(x+x+x+...+x\right)-\left(1+2+...+50\right)=225\) ( có 51 hạng tử x)

=> \(51x-\left(1+2+...+50\right)=225\)   (*)

Xét \(1+2+...+50\)

Có \(\left(50-1\right)+1=50\)  hạng tử

=> \(1+2+...+50= \left(50+1\right).50 :2 = 1275\)

Thay vào (*) : \(51x-1275=225\)

=> \(x=\frac{500}{17}\)

x+x-1+x-2+...+X-50=225

=> (x+x+...+x)-(1+2+3+...+50)=225

=> 51x-1275=225

=> 51x=1500

=> x=30

Vậy x=30

\(\text{x+x-1+x-2+x-3+x-4+...+x-50 = 225}\)

\(\Leftrightarrow\text{51x-(1+2+...+50) = 225}\)

\(\Leftrightarrow\text{51x - 1275 = 225}\)

\(\Leftrightarrow\text{51x = 1500}\)

\(\Leftrightarrow x=\frac{500}{17}\)

giai giup minh vs

Địt bố đéo biết

=15 nha bn

giúp minh vs

a: \(\left(x+5\right)^2-\left(x-5\right)^2-2x+1=0\)

=>\(x^2+10x+25-\left(x^2-10x+25\right)-2x+1=0\)

=>\(x^2+8x+26-x^2+10x-25=0\)

=>18x+1=0

=>\(x=-\dfrac{1}{18}\)

b: \(\left(2x-7\right)^2-\left(x+3\right)^2=3x^2+6\)

=>\(4x^2-28x+49-\left(x^2+6x+9\right)-3x^2-6=0\)

=>\(x^2-28x+43-x^2-6x-9=0\)

=>34-34x=0

=>34x=34

=>x=1

c: \(\left(3x+2\right)^2-9\left(x-5\right)\left(x+5\right)=225-5x\)

=>\(9x^2+12x+4-9\left(x^2-25\right)-225+5x=0\)

=>\(9x^2+17x+4-225-9x^2+225=0\)

=>17x+4=0

=>x=-4/17

\(\left(x+3\right)+\left(x+4\right)+\left(x+5\right)+...+\left(x+22\right)=450\)

\(\Rightarrow\left(x+x+x+...+x\right)+\left(3+4+5+...+22\right)=450\) ( 20 số x )

\(\Rightarrow20x+250=450\) 

\(\Rightarrow20x=200\)

\(\Rightarrow x=10\)

Vậy \(x=10\)

tìm x hả bạn