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Câu 1:
a: \(\Leftrightarrow\left(x-1\right)^2\cdot\left[\left(x-1\right)^2-1\right]=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x-2\right)=0\)
hay \(x\in\left\{0;1;2\right\}\)
b: \(\Leftrightarrow\left(x-1\right)\cdot x\cdot\left(x-2\right)=0\)
hay \(x\in\left\{1;0;2\right\}\)
Câu a )
S = 5 + 52 +..... + 52012
=> S \(⋮5\)
S = 5 + 52 +..... + 52012
S = ( 5 + 53 ) + ( 52 + 54 ) + ........ + ( 52010 + 52012 )
S = 5 ( 1 + 52 ) + 52 ( 1 + 52 ) + ......... + 52010 ( 1 + 52 )
S = 5 x 26 + 52 x 26 + ................ + 52010 x 26
S = 26 ( 5 + 52 + .... + 52010 )
=> S\(⋮26\)
=>\(S⋮13\)( do 26 = 13 x 2 )
Do ( 5 , 13 ) = 1
=> \(S⋮5x13\)
=> \(S⋮65\)
=> 3(x - 2) = 4(8 + x)
=> 3x - 6 = 32 + 4x
=> -6 = 32 + 4x - 3x
=> -6 = 32 + x
=> x = -6 - 32
=> x = -38
b) Ta có : \(\left(x-\frac{1}{3}\right)^2-\frac{1}{4}=0\)
\(\Rightarrow\left(x-\frac{1}{3}\right)^2=\frac{1}{4}\)
\(\Rightarrow\orbr{\begin{cases}\left(x-\frac{1}{3}\right)^2=\left(\frac{1}{2}\right)^2\\\left(x-\frac{1}{3}\right)^2=\left(-\frac{1}{2}\right)^2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x-\frac{1}{3}=\frac{1}{2}\\x-\frac{1}{3}=-\frac{1}{2}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{6}\\x=-\frac{1}{6}\end{cases}}\)
b) \(\left(x-\frac{1}{3}\right)^2-\frac{1}{4}=0\)
\(\Leftrightarrow\left(x-\frac{1}{3}\right)^2=\frac{1}{4}\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{3}=\frac{1}{4}\\x-\frac{1}{3}=-\frac{1}{4}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{7}{12}\\x=\frac{1}{12}\end{cases}}\)
d) \(\frac{x+5}{2}=\frac{8}{x+5}\)
\(\Rightarrow\left(x+5\right)^2=16\)
\(\Rightarrow\orbr{\begin{cases}x+5=16\\x+5=-16\end{cases}\Rightarrow\orbr{\begin{cases}x=11\\x=-21\end{cases}}}\)
bài 8
c) chứng minh \(\overline{aaa}⋮37\)
ta có: \(aaa=a\cdot111\)
\(=a\cdot37\cdot3⋮37\)
\(\Rightarrow aaa⋮37\)
k mk nha
k mk nha.
#mon
\(\left(\frac{x}{10}-\frac{2}{3}\right)^2-\frac{1}{25}=0\)
\(\Leftrightarrow\left(\frac{x}{10}-\frac{2}{3}\right)^2=\frac{1}{25}\)
\(\Leftrightarrow\left(\frac{x}{10}-\frac{2}{3}\right)^2=\left(\pm\frac{1}{5}\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{x}{10}-\frac{2}{3}=\frac{1}{5}\\\frac{x}{10}-\frac{2}{3}=-\frac{1}{5}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{26}{3}\\x=\frac{14}{3}\end{cases}}\)
3) \(n^2+n+2=n\left(n+1\right)+2\)
Ta thấy \(n\left(n+1\right)⋮2\)
Số chia hết cho 2 cộng 2 không bao giờ chia hết cho 5
\(=>n^2+n+2⋮5̸\)
1) \(\overline{1abc}:2=\overline{abc8}\\ =>\overline{1abc}=\overline{abc8}.2\\ =>1000+\overline{abc}=\left(\overline{abc}.10+8\right).2\\ =>1000+\overline{abc}=\overline{abc}^2.100+16\)