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Bài này giống tìm nghiệm quá :
1) \(5\left(x-7\right)=0\)
\(\left(x-7\right)=0\div5\)
\(\left(x-7\right)=0\)
\(x=0+7\)
\(x=7\)
2) \(25\left(x-4\right)=0\)
\(\left(x-4\right)=0\div25\)
\(\left(x-4\right)=0\)
\(x=0+4\)
\(x=4\)
3) \(34\left(2x-6\right)=0\)
\(\left(2x-6\right)=0\div34\)
\(\left(2x-6\right)=0\)
\(2x=0+6\)
\(2x=6\)
\(x=6\div2\)
\(x=3\)
4) \(2007\left(3x-12\right)=0\)
\(\left(3x-12\right)=0\div2007\)
\(\left(3x-12\right)=0\)
\(3x=0+12\)
\(3x=12\)
\(x=12\div3\)
\(x=4\)
5) \(47\left(5x-15\right)=0\)
\(\left(5x-15\right)=0\div47\)
\(\left(5x-15\right)=0\)
\(5x=0+15\)
\(5x=15\)
\(x=15\div5\)
\(x=3\)
6) \(13\left(4x-24\right)=0\)
\(\left(4x-24\right)=0\div13\)
\(\left(4x-24\right)=0\)
\(4x=0+24\)
\(4x=24\)
\(x=24\div4\)
\(x=6\)
![](https://rs.olm.vn/images/avt/0.png?1311)
2007. (3x -12)=0
=> 3x - 12 = 0
=> 3x = 0 + 12
=> 3x = 12
=> x = 12 : 3
=> x = 4
Vậy x = 4
trả lời câu hỏi của quá khứ
x=4
k mk đi
hok trong quá khứ tot
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(6^{2009}-6^{2007}\right)\div6^{2007}\)
\(=\left(6^{2009}-6^{2007}\right)\cdot\frac{1}{6^{2007}}\)
\(=6^{2009}\cdot\frac{1}{6^{2007}}-6^{2007}\cdot\frac{1}{6^{2007}}\)
\(=6^2-1\)
\(=36-1=35\)
![](https://rs.olm.vn/images/avt/0.png?1311)
gọi tổng đó là A:
A=1+22+23+24+...+298
2A=22+23+24+...+298+299
2A-A=299-1
A=299-1
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\left(-2+x\right).\left(3x-6\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}-2+x=0\\3x-6=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=2\end{cases}\Leftrightarrow}x=2}\)
Vậy ................
\(b,\left(3x+9\right)\left(2x+6\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x+9=0\\2x+6=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-3\\x=-3\end{cases}\Leftrightarrow x=-3}\)
Vậy ......................
a)\(\(\left(-2+x\right)\left(3x-6\right)=0\)\)
\(\(\Leftrightarrow\orbr{\begin{cases}-2+x=0\\3x-6=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=2\end{cases}\Rightarrow}x=2}\)\)
b) cmtt
_Minh ngụy_
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(3x-6\right)^2+\left(3x-6\right)^3=0\)
\(\left(3x-6\right)^2\cdot\left(1+3x-6\right)=0\)
\(\left(3x-6\right)^2\cdot\left(3x-5\right)=0\)
\(\orbr{\begin{cases}3x-6=0\\3x-5=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=\frac{5}{3}\end{cases}}}\)
(3x-6)2+(3x-6)3=0
2.(3x-6)2=0
(3x-6)2=0:2=0
(3x-6)2=02
3x-6=0
3x=6
x=2
![](https://rs.olm.vn/images/avt/0.png?1311)
1/3 + 1/6 + 1/10 + ... + 2/x(x+1) = 2005/2007
=> 2/6 + 2/12 + 2/20 + ... + 2/x(x+1) = 2005/2007
=> 2(1/2*3 + 1/3*4 + 1/4*5 + ... + 1/x*(x+1) = 2005/2007
=> 2(1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/x - 1/x+1) = 2005/2007
=> 2(1/2 - 1/x + 1) = 2005/2007
=> 1/2 - 1/x + 1 = 2005/4014
=> 1/x+1 = 1/2007
=> x + 1 = 2007
=> x = 2006
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x.\left(x+1\right)}=\frac{2005}{2007}\)
\(\rightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2005}{2007}\)
\(\rightarrow2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2005}{2007}\)
\(\rightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2005}{2007}\)
\(\rightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2005}{2007}\)
\(\rightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2005}{2007}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2005}{2007}:2\)
\(\rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2005}{2007}:2\) \(\Rightarrow\frac{1}{x+1}=\frac{1}{2007}\)
\(\Rightarrow x+1=2007\rightarrow x=2006\)
Vậy x = 2006.
\(2007\cdot\left(3x-6\right)=0\)
\(\Rightarrow3x-6=0\) \(\Rightarrow3x=6\) \(\Rightarrow x=6:3=2\)
~~ Chúc bạn học tốt ~~
X=2
học tốt !