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Theo đề toán ta có : 80-16 chia hết cho a và 123-24 chia hết cho a
=> 64 chia hết cho a và 99 chia hết cho a
=> a thuộc ƯC(64;99)
UCLN(64;99)=1
=> Ư(64;99)=1
=> a=1
![](https://rs.olm.vn/images/avt/0.png?1311)
a) -2x-x+17=34+x-25
<=> -2x-x-x=34-25-17
<=> -4x=-8
<=> x=2
b)17x+16x+37=2x+43
<=> 33x-2x=43-37
<=> 31x=6
<=>x=\(\frac{6}{31}\)
c) -2x-3x+51=34+2x-50
<=>-5x-2x=34-50-51
<=>-7x=-67
<=>x=\(\frac{67}{7}\)
d) [-2.(4x2-4x+1)- 1]=-106
<=> -8x2+8x-2-1=-106
<=>-8x2+8x=-103
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x\left(x-2\right)-1\left(x-2\right)=2\)
\(x^2-2x-x+2-2=0\)
\(x^2-3x=0\)
\(x\left(x-3\right)=0\)
\(Th1:x=0\)
\(Th2:x-3=0=>x=3\)
Vậy\(x\in\left\{0;3\right\}\)
Toán 8 mà
(x-1).(x-2)=2
x2-2x-x+2=2
x2-3x=0
x(x-3)=0
TH1: x=0
TH2: x-3=0 =>x=3
![](https://rs.olm.vn/images/avt/0.png?1311)
Ko cần đâu bn à mk mong bn đấy
a)\(\left(3x-1\right)\left(5-\frac{1}{2}x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x-1=0\\5-\frac{1}{2}x=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{1}{3}\\x=10\end{cases}}\)
b)\(2\left|\frac{1}{2}x-\frac{1}{3}\right|-\frac{3}{2}=\frac{1}{4}\)
\(2\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{7}{4}\)
\(\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{7}{8}\)
\(\Rightarrow\hept{\begin{cases}\frac{1}{2}x-\frac{1}{3}=\frac{7}{8}\\\frac{1}{2}x-\frac{1}{3}=-\frac{7}{8}\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{29}{12}\\x=-\frac{13}{12}\end{cases}}\)
a)\(\left(3x-1\right)\left(\frac{-1}{2}x+5\right)=0\)
\(\Leftrightarrow\)3x - 1 = 0 hay \(\frac{-1}{2}\)x + 5 = 0
\(\Leftrightarrow\)3x = 1 I\(\Leftrightarrow\)\(\frac{-1}{2}\)x = -5
\(\Leftrightarrow\) x = \(\frac{1}{3}\) I\(\Leftrightarrow\) x = 10
b) 2 I \(\frac{1}{2}x-\frac{1}{3}\)I - \(\frac{3}{2}\)=\(\frac{1}{4}\)
\(\Leftrightarrow\) 2 I\(\frac{1}{2}x-\frac{1}{3}\)I = \(\frac{7}{4}\)
\(\Leftrightarrow\) I\(\frac{1}{2}x-\frac{1}{3}\)I = \(\frac{7}{8}\)
\(\Leftrightarrow\)\(\frac{1}{2}x-\frac{1}{3}\)= \(\frac{7}{8}\) hay \(\frac{1}{2}x-\frac{1}{3}\)= \(\frac{-7}{8}\)
\(\Leftrightarrow\)\(\frac{1}{2}x\) = \(\frac{29}{24}\) I\(\Leftrightarrow\)\(\frac{1}{2}x\) = \(\frac{-13}{24}\)
\(\Leftrightarrow\) x = \(\frac{29}{12}\) I\(\Leftrightarrow\) x = \(\frac{-13}{12}\)
c) (2x +\(\frac{3}{5}\))2 - \(\frac{9}{25}\)= 0
\(\Leftrightarrow\)(2x +\(\frac{3}{5}\))2 = \(\frac{9}{25}\)
\(\Leftrightarrow\) 2x +\(\frac{3}{5}\) = \(\frac{3}{5}\) hay 2x +\(\frac{3}{5}\)= \(\frac{-3}{5}\)
\(\Leftrightarrow\) 2x = 0 I \(\Leftrightarrow\)2x = \(\frac{-6}{5}\)
\(\Leftrightarrow\) x = 0 I \(\Leftrightarrow\) x = \(\frac{-3}{5}\)
d) 3(x -\(\frac{1}{2}\)) - 5(x +\(\frac{3}{5}\)) = -x + \(\frac{1}{5}\)
\(\Leftrightarrow\)3x - \(\frac{3}{2}\)- 5x - 3 = -x + \(\frac{1}{5}\)
\(\Leftrightarrow\)-2x + x - \(\frac{9}{2}\)- \(\frac{1}{5}\)= 0
\(\Leftrightarrow\)-x = \(\frac{-47}{10}\)
\(\Leftrightarrow\) x = \(\frac{47}{10}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=2+2^2+2^3+2^4+...+2^{99}+2^{100}\)
\(\Rightarrow A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{99}+2^{100}\right)\)
\(\Rightarrow A=\left(2+2^2\right)+2^2\left(2+2^2\right)+...+2^{98}\left(2+2^2\right)\)
\(\Rightarrow A=\left(2+2^2\right)\left(1+2^2+...+2^{98}\right)\)
\(\Rightarrow A=6\left(1+2^2+...+2^{98}\right)⋮6\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1+2 . (25 + 9) - 4 MŨ 3
= 1 + 2 . 34 - 64
= 1 + 68 - 64
= 69 - 64
= 5
Cảm ơn bạn nha ^^