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(\(x\) + 2)n+1 = ( \(x\) + 2)n+11
(\(x+2\))n+1 - ( \(x\) + 2)n+11 = 0
(\(x\) + 2)n+1.( 1 + (\(x\) + 2)10) = 0
(\(x\) + 2)10 + 1 > 0 ∀ \(x\)
=> (\(x\) + 2)n+1 = 0 ⇒ \(x\) + 2 = 0 ⇒ \(x\) = -2
vậy \(x\) = -2
b) 3x - 6 - (8x + 4) - (10x + 15) = 50
=> 3x - 6 - 8x - 4 - 10x - 15 = 50
=> (3x - 8x - 10x) = 6+ 4 + 15 + 50
=> -15x = 75 => x = 75 : (-15) = -5
c) => 2x - 3 = 2 - x hoặc 2x - 3 = - (2 - x) (Vì 2 số có giá trị tuyệt đối bằng nhau thì chings bằng nhau hoặc đối nhau)
+) nếu 2x - 3 = 2 - x => 2x+ x = 2 + 3 => 3x = 5 => x = 5/3
+) nếu 2x - 3 = -(2 - x) => 2x - 3 = -2 + x => 2x - x = -2 + 3 => x = 1
Vậy x = 5/3 hoặc x = 1
a) (n-1)n+11-(n-1)n=0
(n-1)n(n-1)11-(n-1)n=0
(n-1)n[(n-1)11-1]=0
(n-1)n=0 hoặc (n-1)11-1=0
n-1=0 hoặc (n-1)11 =1
n=1 hoặc n-1 =1
n=1 hoặc n =2
\(\left(x+2\right)^{n+1}=\left(x+2\right)^{n+11}\)
\(\Leftrightarrow\orbr{\begin{cases}x+2=1\\x+2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-1\\x=-2\end{cases}}}\)
\(\left(x+2\right)^{n+1}-\left(x+2\right)^{x+11}=0\\ \Leftrightarrow\left(x+2\right)^{n+1}\left(1-\left(x+2\right)^{10}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(x+2\right)^{n+1}=0\\1-\left(x+2\right)^{10}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\\\left(x+2\right)^{10}=1\Rightarrow x+2=-1\Rightarrow x=-3\end{cases}}}\)
\(\left(x+2\right)^{n+1}=\left(x+2\right)^{n+11}\)
\(\Leftrightarrow\left(x+2\right)^{n+1}-\left(x+2\right)^{n+11}=0\)
\(\Leftrightarrow\left(x+2\right)^{n+1}-\left(x+2\right)^{n+1}\cdot\left(x+2\right)^{10}=0\)
\(\Leftrightarrow\left(x+2\right)^{n+1}\left[1-\left(x+2\right)^{10}\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(x+2\right)^{n+1}=0\\1-\left(x+2\right)^{10}=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\x+2\in\left\{\pm1\right\}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-2\\x\in\left\{-1;-3\right\}\end{cases}}\)
Vậy....
=> (x+2)n+11:(x+2)n+1=1
<=> (x+2)10=1
th1:x+2=1
<=>x=-1
th2:x+2=-1
<=>x=-3
vậy x=-1 hoặc x=-3
\(\left(x+2\right)^{n+1}=\left(x+2\right)^{n+11}\)
\(\Rightarrow\left(x+2\right)^{n+11}-\left(x+2\right)^{n+1}=0\)
\(\Rightarrow\left(x+2\right)^{n+1}\left[\left(x+2\right)^{10}-1\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x+2\right)^{n+1}=0\\\left(x+2\right)^{10}-1=0\end{matrix}\right.\)
+) \(\left(x+2\right)^{n+1}=0\Rightarrow x+2=0\Rightarrow x=-2\)
+) \(\left(x+2\right)^{10}-1=0\Rightarrow\left(x+2\right)^{10}=1\)
\(\Rightarrow\left[{}\begin{matrix}x+2=1\\x+2=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=-3\end{matrix}\right.\)
Vậy \(x\in\left\{-2;-1;-3\right\}\)
\(c,\Rightarrow\left[{}\begin{matrix}-2\left(x+2\right)+\left(4-x\right)=11\left(x< -2\right)\\2\left(x+2\right)+\left(4-x\right)=11\left(-2\le x\le4\right)\\2\left(x+2\right)+\left(x-4\right)=11\left(x>4\right)\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{11}{3}\left(tm\right)\\x=3\left(tm\right)\\x=\dfrac{11}{3}\left(ktm\right)\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{11}{3}\end{matrix}\right.\)
\(a,\Rightarrow\left[{}\begin{matrix}x+\dfrac{5}{2}=3x+1\\x+\dfrac{5}{2}=-3x-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=-\dfrac{7}{8}\end{matrix}\right.\)
\(M\left(x\right)+N\left(x\right)\)
\(=5x^3-x^2-4+2x^4-2x^2+2x+1\)
\(=2x^4+5x^3-3x^2+2x-3\)
\(M\left(x\right)-N\left(x\right)\)
\(=5x^3-x^2-4-\left(2x^4-2x^2+2x+1\right)\)
\(=5x^3-x^2-4-2x^4+2x^2-2x-1\)
\(=-2x^4+5x^3+x^2-2x-5\)
\(M\left(x\right)+P\left(x\right)=N\left(x\right)\)
\(\Rightarrow P\left(x\right)=N\left(x\right)-M\left(x\right)\)
\(\Rightarrow P\left(x\right)=2x^4-2x^2+2x+1-\left(5x^3-x^2-4\right)\)
\(\Rightarrow P\left(x\right)=2x^4-2x^2+2x+1-5x^3+x^2+4\)
\(\Rightarrow P\left(x\right)=2x^4-5x^3-x^2+2x+5\)