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Bài 2: Vì: 2m - 2n = 256 nên m> n
Đặt m - n = d ( d > 0 )
Ta có : 2m - 2n = 2n ( 2d - 1 ) = 256 = 28.1
=> 2n = 28 và 2d - 1 = 1
=> n = 8 và d = 1
=> m = 1 + 8 = 10
Vậy n = 8 ; m = 9
![](https://rs.olm.vn/images/avt/0.png?1311)
B1: A=|x-13|+|x-2014|=|x-13|+|2014-x| \(\ge\) |x-13+2014-x| = 2001
Dấu "=" xảy ra khi \(\left(x-13\right)\left(2014-x\right)\ge0\Rightarrow13\le x\le2014\)
Vậy GTNN của A = 2001 khi 13\(\le\)x\(\le\)2014
B2
a, 3n+2-2n+2+3n-2n
=3n.32-2n.22+3n-2n
=3n(9+1)-2n(4+1)
=3n.10-2n.5
=3n.10-2n-1.10
=10(3n-2n-1) chia hết cho 10
b, \(\left(x-7\right)^{x+1}+\left(x-7\right)^{x+11}=0\)
\(\Rightarrow\left(x-7\right)^{x+1}\left[1-\left(x-7\right)^{10}\right]=0\)
\(\Rightarrow\orbr{\begin{cases}\left(x-7\right)^{x+1}=0\\1-\left(x-7\right)^{10}=0\end{cases}\Rightarrow\orbr{\begin{cases}x-7=0\\x-7=\pm1\end{cases}}\Rightarrow x\in\left\{6;7;8\right\}}\)
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\(\Leftrightarrow-\dfrac{7}{3}+\dfrac{5}{2}< \left|x-\dfrac{2}{7}\right|< -\dfrac{7}{4}+\dfrac{5}{2}\)
\(\Leftrightarrow\dfrac{1}{6}< \left|x-\dfrac{2}{7}\right|< \dfrac{3}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{6}< x-\dfrac{2}{7}< \dfrac{3}{4}\\-\dfrac{3}{4}< x-\dfrac{2}{7}< -\dfrac{1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{19}{42}< x< \dfrac{29}{28}\\-\dfrac{13}{28}< x< \dfrac{5}{42}\end{matrix}\right.\)
mà x>0
nên \(\left[{}\begin{matrix}\dfrac{19}{42}< x< \dfrac{29}{28}\\0< x< \dfrac{5}{42}\end{matrix}\right.\)
Giá trị x > 0 nguyên thỏa mãn \(-\frac{7}{3}< \left|\frac{2}{7}-x\right|-\frac{5}{2}< -\frac{7}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(-\dfrac{7}{3}< \left|x-\dfrac{2}{7}\right|-\dfrac{5}{2}< -\dfrac{7}{4}\)
=>1/6<|x-2/7|<3/4
\(\Leftrightarrow\left\{{}\begin{matrix}x\in\left(-\infty;\dfrac{5}{42}\right)\cup\left(\dfrac{19}{42};+\infty\right)\\x\in\left(-\dfrac{13}{28};\dfrac{29}{28}\right)\end{matrix}\right.\)
mà x là số nguyên
nên \(x\in\left\{0;1\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
a, \(2y.\left(y-\dfrac{1}{7}\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}2y=0\\y-\dfrac{1}{7}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y=0\\y=\dfrac{1}{7}\end{matrix}\right.\)
Vậy \(y\in\left\{0;\dfrac{1}{7}\right\}\)
b, \(\dfrac{-2}{5}+\dfrac{2}{3}y+\dfrac{1}{6}y=\dfrac{-4}{15}\)
\(\Rightarrow\dfrac{5}{6}y=\dfrac{-4}{15}+\dfrac{2}{5}\)
\(\Rightarrow\dfrac{5}{6}y=\dfrac{2}{15}\)
\(\Rightarrow y=\dfrac{4}{25}\)
Vậy \(y=\dfrac{4}{25}\)
Chúc bạn học tốt!!!
Bài 1:
a, \(2y\left(y-\dfrac{1}{7}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2y=0\\y-\dfrac{1}{7}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}y=0\\y=\dfrac{1}{7}\end{matrix}\right.\)
Vậy...
b, \(\dfrac{-2}{5}+\dfrac{2}{3}y+\dfrac{1}{6}y=\dfrac{-4}{15}\)
\(\Rightarrow\dfrac{5}{6}y=\dfrac{2}{15}\)
\(\Rightarrow y=\dfrac{4}{25}\)
Vậy...
Bài 2:
a, \(x\left(x-\dfrac{4}{7}\right)>0\)
\(\Rightarrow\left\{{}\begin{matrix}x>0\\x-\dfrac{4}{7}>0\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}x< 0\\x-\dfrac{4}{7}< 0\end{matrix}\right.\)
\(\Rightarrow x>\dfrac{4}{7}\left(x\ne0\right)\) hoặc \(x< \dfrac{4}{7}\left(x\ne0\right)\)
Vậy...
Các phần còn lại tương tự nhé
Lời giải:
$(x+7)^{n+1}-(x+7)^{n-3}=0$
$(x+7)^{n-3}[(x+7)^4-1]=0$
$\Rightarrow (x+7)^{n-3}=0$ hoặc $(x+7)^4-1=0$
Nếu $(x+7)^{n-3}=0$
$\Rightarrow x+7=0\Rightarrow x=-7$
Nếu $(x+7)^4-1=0$
$\Rightarrow (x+7)^4=1=1^4=(-1)^4$
$\Rightarrow x+7=1$ hoặc $x+7=-1$
$\Rightarrow x=-6$ hoặc $x=-8$.