Question

Tìm x∈Z, biết:

a)x.(x-6)=0

b)(-7-x).(-x+5)=0

c)(x+3).(x-7)=0

d)(x-3).(x²+12)=0

e)(x+1).(2-x) ≥0

f)(x-3).(x-5) ≤0

Answer 1

a) \(x\left(x-6\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x-6=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)

b) \(\left(-7-x\right)\left(-x+5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}-7-x=0\\-x+5=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-7\\x=-5\end{matrix}\right.\)

c) \(\left(x+3\right)\left(x-7\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+3=0\\x-7=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=7\end{matrix}\right.\)

d) \(\left(x-3\right)\left(x^2+12\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-3=0\\x^2+12=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=3\\x^2=-12\text{(vô lý)}\end{matrix}\right.\)

\(\Rightarrow x=3\)

e) \(\left(x+1\right)\left(2-x\right)\ge0\)

\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x+1\ge0\\2-x\ge0\end{matrix}\right.\\\left[{}\begin{matrix}x+1\le0\\2-x\le0\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x\ge-1\\x\le2\end{matrix}\right.\\\left[{}\begin{matrix}x\le-1\\x\ge2\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}-1\le x\le2\\x\in\varnothing\end{matrix}\right.\)

\(\Rightarrow-1\le x\le2\)

f) \(\left(x-3\right)\left(x-5\right)\le0\)

\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x-3\le0\\x-5\ge0\end{matrix}\right.\\\left[{}\begin{matrix}x-3\ge0\\x-5\le0\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x\le3\\x\ge5\end{matrix}\right.\\\left[{}\begin{matrix}x\ge3\\x\le5\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow3\le x\le5\)

Answer 2

a) =>\(\left[{}\begin{matrix}x=0\\x-6=0\end{matrix}\right.=>\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)

b => \(\left[{}\begin{matrix}-7-x=0\\-x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-7\\x=5\end{matrix}\right.\)

d) => \(\left[{}\begin{matrix}x-3=0\\x^2+12=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x^2=-12\end{matrix}\right.\)(vô lí) => x=3

Answer 3

c) => \(\left[{}\begin{matrix}x+3=0\\x-7=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-3\\x=7\end{matrix}\right.\)

Answer 4

a)  x.(x - 6) = 0 \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)

b) (-7 - x)(-x + 5) = 0

\(\Leftrightarrow\left[{}\begin{matrix}-7-x=0\\-x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-7\\x=5\end{matrix}\right.\)

c) (x + 3)(x - 7) = 0 

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=7\end{matrix}\right.\)

d) (x - 3)(x² + 12) = 0 

Ta có: x² ≥ 0 ∀ x => x² + 12 ≥ 12 ∀ x ∈ Z

mà (x - 3)(x² + 12) = 0 nên x - 3 = 0 \(\Leftrightarrow\)x = 3

e) (x + 1)(2 - x) ≥ 0

+) Xét: (x + 1)(2 - x) = 0 \(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\2-x=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)

+) Xét: (x + 1)(2 - x) > 0

=> x + 1 và 2 - x cùng dấu

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+1>0\\2-x>0\end{matrix}\right.\\\left\{{}\begin{matrix}x+1< 0\\2-x< 0\end{matrix}\right.\end{matrix}\right.\)                  \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>-1\\x< 2\end{matrix}\right.\\\left\{{}\begin{matrix}x< -1\\x>2\end{matrix}\right.\end{matrix}\right.\)

 \(\Rightarrow\left[{}\begin{matrix}-1< x< 2\\2< x< -1\end{matrix}\right.\)

=> -1 < x < 2 ( do 2 < x < -1 là vô lí) 

mà x ∈ Z nên x ∈ {0; 1}

f) (x - 3)(x - 5) = x(x - 5) - 3(x - 5)

= x² - 5x - 3x + 15 = (x² - 2.x.4 + 4²) - 1

= (x - 1)² - 1

Có: (x - 1)² ≥ 0 ∀ x ∈ Z => (x - 1)² - 1 ≥ -1

=> (x - 3)(x - 5) ≥ -1

mà (x - 3)(x - 5) ≤ 0 nên -1 ≤ (x - 3)(x - 5) ≤ 0

Mặt khác: (x - 3)(x - 5) ∈ Z (do x ∈ Z)

=> (x - 3)(x - 5) = -1 hoặc (x - 3)(x - 5) = 0

Xét: (x - 3)(x - 5) = -1

=> x - 3 và x - 5 trái dấu mà x ∈ Z

=> x - 3 = -1 ; x - 5 = 1 hoặc x - 3 = 1 ; x - 5 = -1

=> x = 2 ; x = 6 hoặc x = 4 ; x = 4

Xét: (x - 3)(x - 5) = 0 \(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x-5=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=5\end{matrix}\right.\)