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\(2x+1⋮x-1\)
=>\(2x-2+3⋮x-1\)
=>\(3⋮x-1\)
=>\(x-1\in\left\{1;-1;3;-3\right\}\)
=>\(x\in\left\{2;0;4;-2\right\}\)
2x+1⋮x−12x+1⋮x-1
⇔(2x−2)+3⋮x−1⇔(2x-2)+3⋮x-1
⇔2(x−1)+3⋮x−1⇔2(x-1)+3⋮x-1
Mà x−1⋮x−1x-1⋮x-1
⇒2(x−1)⋮x−1⇒2(x-1)⋮x-1
⇒3⋮x−1⇒3⋮x-1
⇔x−1∈Ư(3)={±1;±3}⇔x-1∈Ư(3)={±1;±3}
⇔x∈{0;2;4;−2}⇔x ∈{0;2;4;-2}
Vậy x∈{0;±2;4}x ∈{0;±2;4} thì 2x+1⋮x−1
\(a,x^3-1=-28\\ \Leftrightarrow x^3=-27\\ \Leftrightarrow x^3=\left(-3\right)^3\\ \Leftrightarrow x=-3\\ b,\left(y-1\right)^2-32=-23\\ \Leftrightarrow\left(y-1\right)^2=9\\ \Leftrightarrow\left[{}\begin{matrix}y-1=3\\y-1=-3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}y=4\\y=-2\end{matrix}\right.\\ c,15-16:\left|x\right|=-1\\ \Leftrightarrow16:\left|x\right|=16\\ \Leftrightarrow\left|x\right|=1\\ \Leftrightarrow x=\pm1\)
a: \(x\in\left\{-6;-5;-4;-3;-2;-1;0;1;2;3\right\}\)
c: \(x\in\left\{-4;-3;-2;-1\right\}\)
\(\left(n-4\right)⋮\left(n-1\right)\Rightarrow\left(n-1-3\right)⋮\left(n-1\right)\)
\(Mà\left(n-1\right)⋮\left(n-1\right)\Rightarrow-3⋮\left(n-1\right)\Rightarrow n-1\inƯ\left(-3\right)=\left\{\pm1;\pm3\right\}\Rightarrow n\in\left\{-2;0;2;4\right\}\)
A<1/1*2+1/2*3+...+1/2021*2022
=>A<1-1/2+1/2-1/3+...+1/2021-1/2022<1
Đặt \(A=\dfrac{10^{1987}+1}{10^{1988}+1};B=\dfrac{10^{1989}+1}{10^{1990}+1}\)
Ta có: \(A=\dfrac{10^{1987}+1}{10^{1988}+1}\)
\(\Leftrightarrow10A=\dfrac{10^{1988}+10}{10^{1988}+1}=1+\dfrac{9}{10^{1988}+1}\)
Ta có: \(B=\dfrac{10^{1989}+1}{10^{1990}+1}\)
\(\Leftrightarrow10B=\dfrac{10^{1990}+10}{10^{1990}+1}=1+\dfrac{9}{10^{1990}+1}\)
Ta có: \(10^{1988}+1< 10^{1990}+1\)
\(\Leftrightarrow\dfrac{9}{10^{1988}+1}>\dfrac{9}{10^{1990}+1}\)
\(\Leftrightarrow1+\dfrac{9}{10^{1988}+1}>1+\dfrac{9}{10^{1990}+1}\)
hay A>B
| x + 4 | = 1
=> x + 4 = 1 hoặc x + 4 = -1
Ta xét 2 trường hợp :
TH1 : x + 4 = 1
x = 1 - 4
x = -3
TH2 : x + 4 = -1
x = -1 - 4
x = - 5
Vậy x \(\in\){ - 3 ; - 1 }
|x + 4| = 1
=> x + 4 = \(\pm1\)
TH1 : x + 4 = 1 TH2 : x + 4 = -1
x = 1 - 4 x = -1 - 4
x = -3 x= -5
~HT~