Tìm x biết : ( x - 1 ) ^20 = ( x - 1 ) ^30
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x : 4dư 1⇒ x+3 ⋮ 4
x: 7 dư 4⇒x+3 ⋮ 7
⇒ (x+3) ϵ BC(4;7)
4=4 TSNT chung: không có
7=7 riêng : 4;7
BCNN(4;7)= 4.7=28
⇒ x+3 ϵ B(28)= (0;28;56;...)
mà 25<x<30⇒ 22< x+3 <27
⇒ x+3 = 28
x = 28-3=25
vậy x=25
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1/2 + 1/6 + 1/12 + 1/20 + 1/30 1/x = 41/42
5/6 + 1/x = 41/42
1/x = 41/42 - 5/6
1/x = 1/7
vậy x = 7
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\(\left(4x-1\right)^{30}=\left(4x-1\right)^{20}\)
\(\Leftrightarrow\left(4x-1\right)^{30}-\left(4x-1\right)^{20}=0\)
\(\Leftrightarrow\left(4x-1\right)^{20}\left[\left(4x-1\right)^{10}-1\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(4x-1\right)^{20}=0\\\left(4x-1\right)^{10}-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}4x-1=0\\\left(4x-1\right)^{10}=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}4x=1\\4x-1=-1;1\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=\frac{1}{4}\\x=0\\x=\frac{1}{2}\end{cases}}\)
Vậy \(x=\frac{1}{4};0;\frac{1}{2}\)
P/s : phần \(\Leftrightarrow\hept{\begin{cases}x=\frac{1}{4}\\x=0\\x=\frac{1}{2}\end{cases}}\) thay dấu \(\hept{\begin{cases}\\\\\end{cases}}\) thành dấu \(\orbr{\begin{cases}\\\end{cases}}\) nhé!
\(\Leftrightarrow\orbr{\begin{cases}4x=1\\4x-1=\pm1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{4}\\\orbr{\begin{cases}x=0\\x=\frac{1}{2}\end{cases}}\end{cases}}\)
\(\left(4x-1\right)^{30}=\left(4x-1\right)^{20}\)
\(\Leftrightarrow\left(4x-1\right)^{30}-\left(4x-1\right)^{20}=0\)
\(\Leftrightarrow\left(4x-1\right)^{20}.\left[\left(4x-1\right)^{10}-1\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(4x-1\right)^{20}=0\\\left(4x-1\right)^{10}-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}4x-1=0\\\left(4x-1\right)^{10}=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{4}\\4x-1=\pm1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=0\end{cases}}\)
Vậy x = 1/4 hoặc 1/2 hoặc 0
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\(\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+7\right)}\)
\(\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+7}=\frac{1}{18}\)
\(\frac{1}{x+4}-\frac{1}{x+7}=\frac{1}{18}\)
( x - 1 ) 20 = ( x - 1 ) 30
<=> (x-1)20 [ (x-1)10-1]=0
\(\Leftrightarrow\orbr{\begin{cases}\left(x-1\right)^{20}=0\\\left(x-1\right)^{10}=1\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=1\\x=2\end{cases}}\)hoặc x=0
Bài giải
\(\left(x-1\right)^{20}=\left(x-1\right)^{30}\)
\(\left(x-1\right)^{30}-\left(x-1\right)^{20}=0\)
\(\left(x-1\right)^{20} [\left(x-1\right)^{10}-1 ] =0\)
\(\Rightarrow\orbr{\begin{cases}\left(x-1\right)^{20}=0\\\left(x-1\right)^{10}-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x-1=0\\\left(x-1\right)^{10}=1\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x-1=1\text{ hoặc }x-1=-1\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=2\text{ hoặc }x=0\end{cases}}\)
Vậy \(x\in\left\{0\text{ ; }1\text{ ; }2\right\}\)