1) 1 + 2 + 3 + ... + x = 1225
=> (1 + x).x:2 = 1225
=> (1 + x).x = 1225.2
=> (1 + x).x = 2450
=> (1 + x).x = 50.49 = (-49).(-50)
Vậy \(x\in\left\{50;-49\right\}\)
2) 2 + 4 + 6 + ... + 2x = 210
=> 2.(1 + 2 + 3 + ... + x) = 210
=> 2.(1 + x).x:2 = 210
=> (1 + x).x = 15.14 = (-14).(-15)
Vậy \(x\in\left\{15;-14\right\}\)
a)
\(1+2+......+x=1225\)
\(\Rightarrow\frac{\left(x+1\right)x}{2}=1225\)
\(\Rightarrow\left(x+1\right)x=2450\)
\(\Rightarrow\left(x+1\right)x=49.50\)
=> x = 49
Vậy x = 49
b)
\(2+4+....+2x=210\)
\(\Rightarrow2\left(1+2+....+x\right)=210\)
\(\Rightarrow2.\frac{\left(x+1\right)x}{2}=210\)
\(\Rightarrow x\left(x+1\right)=210\)
\(\Rightarrow x\left(x+1\right)=210\)
\(\Rightarrow x\left(x+1\right)=14.15\)
=> x = 14
Vậy x = 14
1) 1 + 2 + 3 +..... + X = 1225
Đặt A= 1 + 2 + 3 +..... + X
Tổng A có số số hạng theo x là:
\(\left(x-1\right):1+1=x\) (số)
Tổng A theo x là:
\(\left(x+1\right)\cdot x:2=\frac{x^2+x}{2}\)
Thay A vào ta đc:
\(\frac{x^2+x}{2}=1225\)\(\Leftrightarrow x^2+x-2450=0\)
\(\Leftrightarrow x^2-49x+50x-2450=0\)
\(\Leftrightarrow x\left(x-49\right)+50\left(x-49\right)=0\)
\(\Leftrightarrow\left(x-49\right)\left(x+50\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=49\left(tm\right)\\x=-50\left(loai\right)\end{array}\right.\)
2)Đặt B= 2 + 4 + 6 +........ + 2X
Tổng B có số số hạng theo x là:
\(\left(2x-2\right):2+1=x\) (số)
Tổng B theo x là:
\(\left(2x+2\right)\cdot x:2=x^2+x\)
Thay B vào ta dc:
\(x^2+x=210\)\(\Leftrightarrow x^2+x-210=0\)
\(\Leftrightarrow x^2-14x+15x-210=0\)
\(\Leftrightarrow x\left(x-14\right)+15\left(x-14\right)=0\)
\(\Leftrightarrow\left(x+15\right)\left(x-14\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=14\left(tm\right)\\x=-15\left(loai\right)\end{array}\right.\)
2+4+6+....+2x=210
\(\Leftrightarrow\)2.1+2.2.+2.2+....+2.x=210
2.( 1+2+3+4+5+....+x)=210
2.( x+1).x:2=210
(x+1).x=210
(x+1) .x =14.15
xay x =14
