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a,\(x^{10}=1^x\)
\(x^{10}=1\)
\(\Rightarrow x=\orbr{\begin{cases}1\\-1\end{cases}}\)
b)\(x^{10}=x\)
\(\Rightarrow x\left(x^9-1\right)=0\)
\(\Rightarrow x=0;1\)
-3x+(-9)+5x-5=-10
(-3x+5x)+(-9-5)=-10
-2x+(-14)=-10
-2x=-10-(-14)
-2x=24
x=24:(-2)
x=-12. chúc bạn học tối nha
a,/x+2/-12=-1
/x+2/=-1+12
/x+2/=11
x+2=11hoacx+2=-11
x+2=11 x+2=-11
x=11-2 x=-11-2
x=9 x=-13 vay x=9;x=-13
b,10-/x+3/=-4-(-10)
10-/x+3/=-4+10
10-/x+3/=6
/x+3/=10-6
/x+3/=4
x+3=4hoacx+3=-4
x+3=4 x+3=-4
x=4-3 x=-4-3
x=1 x=-7
vay x=1;x=-7
\(\frac{1}{3}+....+\frac{2}{x.\left(x+1\right)}=\frac{1999}{2001}\)
=>\(\frac{1}{2}.\left(\frac{1}{3}+...+\frac{2}{x.\left(x+1\right)}\right)=\frac{1999}{2001}.\frac{1}{2}\)
\(\Rightarrow\frac{1}{6}+\frac{1}{12}+...+\frac{1}{x.\left(x+1\right)}=\frac{1999}{4002}\)
\(\Rightarrow\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x.\left(x+1\right)}=\frac{1999}{4002}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{1999}{4002}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2001}\)
=> x=2000
Tìm stn biết: 1/3 + 1/6 + 1/10 + ...+2/x(x+1)=1999/2001
Bài giải: Gọi x là số tự nhiên cần tìm
Cho S= 1/3 + 1/6 +1/10 +...+ 1/x(x+1)
\(\Rightarrow\)S= 2/6 + 2/12+ 2/20 +...+ 2/2[x(x+1)]
\(\Rightarrow\)1/2S= 1/2.3 + 1/3.4 + 1/ 4.5 +...+1/2[x(x+1)]
\(\Rightarrow\)1/2S=1/2-1/3+1/3-1/4+...+1/(x-1) .(x+1)
\(\Leftrightarrow\)1/2S=1/2-1/x+1
Vì S = 1999 / 2001\(\Rightarrow\)1/2S=1/2-1 . (x+1)=1999/2001-1998-2001=1/2001
\(\Rightarrow\)1/x+1=1/2001
\(\Leftrightarrow\)x+1=2001
x =2001-1 =2000
Vậy số tự nhiên đó là: 2000
\(\dfrac{1}{2}\) \(\times\) ( \(x\) - \(\dfrac{2}{3}\)) - \(\dfrac{1}{3}\) \(\times\) ( 2\(x\) - 3) = \(x\)
\(\dfrac{1}{2}\) \(\times\) \(\dfrac{3x-2}{3}\) - \(\dfrac{2x-3}{3}\) = \(x\)
\(\dfrac{3x-2}{6}\) - \(\dfrac{4x-6}{6}\) = \(\dfrac{6x}{6}\)
3\(x-2-4x\) + 6 = 6\(x\)
-\(x\) + 4 - 6\(x\) = 0
7\(x\) = 4
\(x\) = \(\dfrac{4}{7}\)
x+(x+1)+(x+2)+...+(x+10)=88
=>11x+(1+2+...+10)=88
=>11x+55=88
=>x+5=8
=>x=3