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\(a-b⋮7\Rightarrow a⋮6,b⋮7\)
\(\Rightarrow4a⋮7;3b⋮7\)
\(\Rightarrow4a+3b⋮7\) (đpcm)
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Đặt \(x=\frac{y}{2}=\frac{z}{3}=k\left(k\in Q\right)\)\(\Rightarrow x=k;y=2k;z=3k\)
Thế (1) vào biểu thức trên
\(\Rightarrow2\left(x^2+y^2\right)-z^2=9\)
\(\Leftrightarrow2\left[\left(k\right)^2+\left(2k\right)^2\right]-\left(3k\right)^2=9\)
\(\Rightarrow2\left(k^2+4k^2\right)-9k^2=9\)
\(\Rightarrow2k^2+8k^2-9k^2=9\)
\(\Rightarrow k^2=9\)
\(\Rightarrow k=\hept{\begin{cases}3\\-3\end{cases}}\)
Với k = 3
\(\Rightarrow x=3;y=3.2=6;z=3.3=9\)
Với k = -3
\(\Rightarrow x=-3;y=-3.2=-6;z=-3.3=-9\)
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\(\Rightarrow S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+..+\frac{1}{n^2}-\frac{1}{n+1^2}\)
\(\Rightarrow S=1-\frac{1}{n+1}\)
\(\Rightarrow S+\frac{n}{n+1}\)
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\(n^2+1⋮n-1\)
Ta có : \(n^2+1=n^2-n+n-1+2=n\left(n-1\right)+n-1+2=\left(n+1\right)\left(n-1\right)+2\)
\(\Rightarrow n^2+1⋮n-1\Leftrightarrow\left(n+1\right)\left(n-1\right)+2⋮n-1\)
\(\Leftrightarrow2⋮n-1\Leftrightarrow n-1\inƯ\left(2\right)=\left\{1;-1;2;-2\right\}\)
\(\Leftrightarrow n\in\left\{2;0;3;-1\right\}\)
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[ 212 + ( 48 - 2 . x . x2 ) : 40 - 3 = 3
[ 212 + ( 48 - 2 . x . x2 ) : 40 = 3 + 3 = 6
[ 212 + ( 48 - 2 . x . x2 ) = 6 . 40 = 240
=> 48 - 2 . x . x2 = 240 - 212 = 28
2 . x . x2 = 48 - 28 = 20
x. x2 = 20 : 2 = 10
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\(\begin{cases}3\left(x-7\right)=4\left(y-5\right)\\4x-3y+8=0\end{cases}\)
\(\Leftrightarrow\begin{cases}3x-4y=1\\4x-3y=-8\end{cases}\Leftrightarrow\begin{cases}9x-12y=3\\-16x+12y=32\end{cases}\Leftrightarrow}\begin{cases}-7x=35\\3x-4y=1\end{cases}\Leftrightarrow\begin{cases}x=-5\\y=-4\end{cases}}}\)
\(\begin{cases}3\left(x-7\right)=4\left(y-5\right)\\4x-3y+8=0\end{cases}\)
<=> \(\begin{cases}3x-4y=1\\4x-3y=-8\end{cases}\)<=> \(\begin{cases}9x-12y=3\\-16x+12y=32\end{cases}\)<=> \(\begin{cases}-7x=35\\3x-4y-1\end{cases}\)<=> \(\begin{cases}x=-5\\y=-4\end{cases}\)
3n chia hết cho 3 - n àh