(2x-3)^6 = (2x -3)^8
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13 tháng 11 2021
\(x\left(x-2\right)+x-2=0\\ \Leftrightarrow x\left(x-2\right)+\left(x-2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
\(x^2-2x+1=9\\ \Leftrightarrow\left(x-1\right)^2=9\\ \Leftrightarrow\left[{}\begin{matrix}x-1=-3\\x-1=3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x=4\end{matrix}\right.\)
\(7x^2=2x\\ \Leftrightarrow7x^2-2x=0\\ \Leftrightarrow x\left(7x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\7x-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{7}\end{matrix}\right.\)
\(x^2-6x=8\\ \Leftrightarrow x^2-6x-8=0\\ \left(x^2-6x+9\right)-17=0\\ \Leftrightarrow\left(x-3\right)^2-\sqrt{17^2}=0\\ \Leftrightarrow\left(x-3-\sqrt{17}\right)\left(x-3+\sqrt{17}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-3-\sqrt{17}=0\\x-3+\sqrt{17}=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=3+\sqrt{17}\\x=3-\sqrt{17}\end{matrix}\right.\)
3 tháng 10 2021
\(a,\left(-\dfrac{2}{3}\right)^8+\left(-\dfrac{2}{3}\right)^8=2\left(-\dfrac{2}{3}\right)^8=2\cdot\dfrac{\left(-2\right)^8}{3^8}=\dfrac{2\cdot2^8}{3^8}=\dfrac{2^9}{3^8}=\dfrac{512}{6561}\)
12 tháng 8 2021
1: Ta có: \(2x\left(x+3\right)-6\left(x-3\right)=0\)
\(\Leftrightarrow2x^2+6x-6x+18=0\)
\(\Leftrightarrow2x^2+18=0\left(loại\right)\)
2: Ta có: \(2x^2\left(2x+3\right)+\left(2x+3\right)=0\)
\(\Leftrightarrow2x+3=0\)
hay \(x=-\dfrac{3}{2}\)
3: Ta có: \(\left(x-2\right)\left(x+1\right)-4x\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(1-3x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)
12 tháng 8 2021
4: Ta có: \(2x\left(x-5\right)-3x+15=0\)
\(\Leftrightarrow\left(x-5\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{3}{2}\end{matrix}\right.\)
5: Ta có: \(3x\left(x+4\right)-2x-8=0\)
\(\Leftrightarrow\left(x+4\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=\dfrac{2}{3}\end{matrix}\right.\)
6: Ta có: \(x^2\left(2x-6\right)+2x-6=0\)
\(\Leftrightarrow2x-6=0\)
hay x=3
8 tháng 4 2023
1: =>x+1/2=0 hoặc 2/3-2x=0
=>x=-1/2 hoặc x=1/3
2: =>7/6x=5/2:3,75=2/3
=>x=2/3:7/6=2/3*6/7=12/21=4/7
3: =>2x-3=0 hoặc 6-2x=0
=>x=3 hoặc x=3/2
4: =>-5x-1-1/2x+1/3=3/2x-5/6
=>-11/2x-3/2x=-5/6-1/3+1
=>-7x=-1/6
=>x=1/42
23 tháng 4 2023
cho A=1/101+1/102+1/103+...+1/199+1/200 chứng minh 1/2 <A<1
11 tháng 8 2021
g: Ta có: \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)=0\)
\(\Leftrightarrow3\left(6x^2-5x+1\right)-\left(18x^2-29x+3\right)=0\)
\(\Leftrightarrow18x^2-15x+3-18x^2+29x-3=0\)
\(\Leftrightarrow14x=0\)
hay x=0
30 tháng 9 2021
a: Ta có: \(3\left(2x-3\right)+2\left(2-x\right)=-3\)
\(\Leftrightarrow6x-9+4-2x=-3\)
\(\Leftrightarrow4x=2\)
hay \(x=\dfrac{1}{2}\)
Bài giải
\(\left(2x-3\right)^6=\left(2x-3\right)^8\)
\(\left(2x-8\right)^8-\left(2x-3\right)^6=0\)
\(\left(2x-3\right)^6\left[\left(2x-3\right)^2-1\right]=0\)
\(\Rightarrow\orbr{\begin{cases}\left(2x-3\right)^6=0\\\left(2x-3\right)^2-1=0\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}2x-3=0\\\left(2x-3\right)^2=1\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}2x=3\\2x-3=-1\text{ hoặc }2x-3=1\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=1\text{ hoặc }x=2\end{cases}}\)
\(\Rightarrow\text{ }x\in\left\{\frac{3}{2}\text{ ; }1\text{ ; }2\right\}\)
\(\left(2x-3\right)^6=\left(2x-3\right)^8\)
\(\Rightarrow\left(2x-3\right)^8-\left(2x-3\right)^6=0\)
\(\left(2x-3\right)^6\left[\left(2x-3\right)^2-1\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(2x-3\right)^6=0\\\left(2x-3\right)^2-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}2x-3=0\\\left(2x-3\right)^2=1\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{3}{2}\\x=1;x=2\end{cases}}}\)