1+2+3+4+...+x=120
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a/ \(\frac{x}{2}+\frac{x}{3}+\frac{x}{4}=180\)
\(\Leftrightarrow x\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)=180\)
\(\Leftrightarrow\frac{13}{12}x=180\)
\(\Leftrightarrow x=\frac{2160}{13}\)
b/ \(\frac{1}{x.2}+\frac{2}{x.3}+\frac{3}{x.4}=120\)
\(\Leftrightarrow\frac{1}{x}\left(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}\right)=120\)
\(\Leftrightarrow\frac{1}{x}.\frac{23}{12}=120\)
\(\Leftrightarrow\frac{1}{x}=\frac{1440}{23}\)
\(\Leftrightarrow x=\frac{23}{1440}\)
# Tính lũy thừa
a = 3
n = 2
power = a ** n
print(power) # Kết quả: 9
# Tính giai thừa
n = 5
factorial = 1
for i in range(1, n+1):
factorial *= i
print(factorial) # Kết quả: 120
(10 - 4.x) + 120 : 8 = 5 mũ 2
(10-4.x) +120 : 8 = 25
(10-4.x)+120 = 25.8
(10-4.x)=200
4.x=200+10
x=210:4
x=52,5
\(\dfrac{1}{120}\cdot120+x:\dfrac{1}{3}=-4\)
\(\Leftrightarrow1+x\cdot3=-4\)
\(\Leftrightarrow3x=-5\)
\(\Leftrightarrow x=-\dfrac{5}{3}\)
\(\dfrac{1}{120}.120+x:\dfrac{1}{3}=-4\)
\(1+x:\dfrac{1}{3}=-4\)
\(x:\dfrac{1}{3}=-4-1\)
\(x:\dfrac{1}{3}=-5\)
\(x=-5.\dfrac{1}{3}\)
\(x=\dfrac{-5}{3}\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-120\)
\(=\left(x^2+5x\right)^2+10\left(x^2+5x\right)-96\)
\(=\left(x^2+5x+16\right)\left(x^2+5x-6\right)\)
\(=\left(x^2+5x+16\right)\left(x+6\right)\left(x-1\right)\)
Ta có:
10 - 4x + 120:23 = 1 + 42
=> 10 - 4x + 15 = 1 + 16
=> 4x - 5 = 17
=> 4x = 17 + 5 = 22
=> x = 22/4
Vậy: x = 22/4
\(1+2+3+4+...+x=120\)
\(\Rightarrow\frac{\left(x+1\right).x}{2}=120\)
\(\Rightarrow\left(x+1\right).x=120.2\)
\(\Rightarrow\left(x+1\right).x=240\)
\(\Rightarrow\left(x+1\right).x=16.15\)
\(\Rightarrow x=15\)