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\(\left(3x-7\right)^{2015}=\left(3x-7\right)^{2017}\Rightarrow\left(3x-7\right)^{2017}-\left(3x-7\right)^{2015}=0\Leftrightarrow\left(3x-7\right)^{2015}\left[\left(3x-7\right)^2-1\right]=0\Leftrightarrow\orbr{\begin{cases}3x-7=0\\\left(3x-7\right)^2=1\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}3x=7\\3x-7=1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{3}\\x=\frac{1+7}{3}=\frac{8}{3}\end{cases}}\)
Vậy phương trình có hai nghiệm là \(x=\frac{7}{3}\)và \(x=\frac{8}{3}\)
Vì \(\left(3x-7\right)^{2015}=\left(3x-7\right)^{2017}\) =>3x-7=0 hoặc 3x-7=1
- Nếu 3x-7=0=>x=\(\frac{7}{3}\)
- Nếu 3x-7=1=>x=\(\frac{8}{3}\)
Vậy \(x=\orbr{\begin{cases}\frac{7}{3}\\\frac{8}{3}\end{cases}}\)
(3x - 7)2015 = (3x - 7)2017
(3x - 7)2017 - (3x - 7)2015 = 0
(3x - 7)2017[(3x - 7)2 - 1] = 0
=> (3x - 7)2017 = 0 hoặc (3x - 7)2 = 1
=> 3x - 7 = 0 hoặc 3x - 7 = ± 1
=> x = 7/3 hoặc x = { 8/3 ; 2 }
Vậy x = { 2; 7/3; 8/3 }
\(y\left(y^2-1\right)=0\Leftrightarrow\orbr{\begin{cases}y=0\\y^2-1=0\end{cases}}\)
\(\frac{5x+7}{4}+\frac{3x+5}{8}>\frac{9x+4}{5}\)
\(\frac{10\cdot\left(5x+7\right)}{40}+\frac{5\cdot\left(3x+5\right)}{40}>\frac{8\cdot\left(9x+4\right)}{40}\)
10.(5x + 7) + 5.(3x + 5) > 8.(9x + 4)
10.(5x + 7) + 5.(3x + 5) - 8.(9x + 4) > 0
50x + 70 + 15x + 25 - 72x - 32 > 0
- 7x + 63 > 0
- 7.(x - 9) > 0
\(\Rightarrow x-9
Ta có : |2x - 1| + 1 = x
=> |2x - 1| = x - 1
\(\Leftrightarrow\orbr{\begin{cases}2x-1=x-1\\2x-1=1-x\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x-x=-1+1\\2x+x=1+1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\3x=2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{2}{3}\end{cases}}\)
1) x (x-2016) + 2015 (2016-x) = 0
x (x-2016) - 2015 (x- 2016) = 0
(x-2015)(x-2016) =0
\(\Rightarrow\orbr{\begin{cases}x-2015=0\\x-2016=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2015\\x=2016\end{cases}}}\)
Vậy x= 2015; 2016
2) -5x (x-15) + (15-x) = 0
-5x (x-15) - (x-15) =0
(-5x -1) (x-15) =0
\(\Rightarrow\orbr{\begin{cases}-5x-1=0\\x-15=0\end{cases}\Rightarrow\orbr{\begin{cases}-5x=1\\x=15\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{1}{5}\\x=15\end{cases}}}\)
Vậy x= -1/5; 15
3) 3x (3x-7) - (7-3x) =0
3x(3x-7) + (3x -7) =0
(3x+1) (3x-7) =0
\(\Rightarrow\orbr{\begin{cases}3x+1=0\\3x-7=0\end{cases}\Rightarrow\orbr{\begin{cases}3x=-1\\3x=7\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{1}{3}\\x=\frac{7}{3}\end{cases}}}\)
Vậy x= -1/3 ; 7/3
a) \(\left(3x^2+5x-3\right)+\left(x-3x^2-3\right)=0\)
\(\Leftrightarrow6x-6=0\)
\(\Leftrightarrow6x=6\Leftrightarrow x=1\)
b) \(\left(3x^2-5x\right)-\left(3x^2+x-12\right)=0\)
\(\Leftrightarrow3x^2-5x-3x^2-x+12=0\)
\(\Leftrightarrow-6x=-12\Leftrightarrow x=2\)
\(\left(3x-7\right)^{2015}=\left(3x-7\right)^{2017}\)
\(\Leftrightarrow\left(3x-7\right)^{2017}-\left(3x-7\right)^{2015}=0\)
\(\Leftrightarrow\left(3x-7\right)^2=0\)
\(\Leftrightarrow3x-7=0\)
\(\Leftrightarrow3x=7\Leftrightarrow x=\frac{7}{3}\)
Vậy \(x=\frac{7}{3}\)
\(\left(3x-7\right)^{2015}=\left(3x-7\right)^{2017}\)
\(\Rightarrow\left(3x-7\right)^{2017}-\left(3x-7\right)^{2015}=0\)
\(\Rightarrow\left(3x-7\right)^{2015}\left[\left(3x-7\right)^2-1\right]=0\)
\(\Rightarrow\left(3x-7\right)^{2015}=0\) hoặc \(\left(3x-7\right)^2-1=0\)
+) \(\left(3x-7\right)^{2015}=0\Rightarrow3x-7=0\Rightarrow x=\frac{7}{3}\)
+) \(\left(3x-7\right)^2-1=0\Rightarrow\left(3x-7\right)^2=1\)
\(\Rightarrow\left[\begin{matrix}3x-7=1\\3x-7=-1\end{matrix}\right.\Rightarrow\left[\begin{matrix}x=\frac{8}{3}\\x=2\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{7}{3};\frac{8}{3};2\right\}\)