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TP

Tìm \(x\in Z\) biết :

\(a.\left(x-4\right).\left(x-7\right)=0\)

\(b:x.\left(x+3\right)=0\)

\(c:\left(x-2\right)\left(5-x\right)=0\)

\(d:\left(x-1\right).\left(x^2+1\right)=0\)

TL
11 tháng 9 2016 lúc 19:50

a) \(\left(x-4\right)\left(x-7\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x-4=0\\x-7=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=4\\x=7\end{array}\right.\)

b) \(x\left(x+3\right)=0\)

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

c) \(\left(x-2\right)\left(5-x\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x-2=0\\5-x=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=2\\x=5\end{array}\right.\)

d) \(\left(x-1\right)\left(x^2+1\right)=0\)

\(\Leftrightarrow x-1=0\) ( Vì \(x^2+1>0\) )

\(\Leftrightarrow x=1\)

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IM
11 tháng 9 2016 lúc 19:50

a)

\(\left(x-4\right)\left(x-7\right)=0\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=4\\x=7\end{array}\right.\)

Vậy x = 4 ; x = 7

b)

\(x\left(x+3\right)=0\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=0\\x=-3\end{array}\right.\)

Vậy x = 0 ; x = - 3

c)

\(\left(x-2\right)\left(5-x\right)=0\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=2\\x=5\end{array}\right.\)

Vậy x = 2 ; x = 5

d)

\(\left(x-1\right)\left(x^2+1\right)=0\)

Mà \(x^2+1\ge1\)

=> x = - 1

Vậy x = - 1

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NT
11 tháng 9 2016 lúc 19:53

a, \(\left(x-4\right).\left(x-7\right)=0\)

\(\Rightarrow x-4=0\) hoặc \(x-7=0\)

+) \(x-4=0\Rightarrow x=4\)

+) \(x-7=0\Rightarrow x=7\)

Vậy x = 4 hoặc x = 7

b, \(x.\left(x+3\right)=0\)

\(\Rightarrow x=0\) hoặc \(x+3=0\)

+) \(x+3=0\Rightarrow x=-3\)

Vậy x = 0 hoặc x = -3

c, \(\left(x-2\right).\left(5-x\right)=0\)

\(\Rightarrow x-2=0\) hoặc \(5-x=0\)

+) \(x-2=0\Rightarrow x=2\)

+) \(5-x=0\Rightarrow x=5\)

Vậy x = 2 hoặc x = 5

d, \(\left(x-1\right).\left(x^2+1\right)=0\)

\(\Rightarrow x-1=0\) hoặc \(x^2+1=0\)

+) \(x-1=0\Rightarrow x=1\)

+) \(x^2+1=0\Rightarrow x^2=-1\Rightarrow\) không có giá trị x thỏa mãn đề bài.

Vậy x = 1 hoặc không có giá trị x thỏa mãn đề bài

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PU
11 tháng 9 2016 lúc 19:55

\(\text{a) (x-4)(x-7)=0}\)

\(\Rightarrow\left[\begin{array}{nghiempt}x-4=0\\x-7=0\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=0+4\\x=0+7\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=4\\x=7\end{array}\right.\)

\(\text{b) x.(x+3)=0}\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=0\\x+3=0\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=0\\x=0-3=-3\end{array}\right.\)

\(\text{Vậy}\left[\begin{array}{nghiempt}x=0\\x=-3\end{array}\right.\)

\(\text{c)(x-2)(5-x)=0}\)

\(\Rightarrow\left[\begin{array}{nghiempt}x-2=0\\5-x=0\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=0+2\\x=5-0\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=2\\x=5\end{array}\right.\)

\(\text{d) (x-1).(x^2 +1)=0}\)

\(\Rightarrow\left[\begin{array}{nghiempt}x-1=0\\x^2+1=0\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=0+1=1\\x^2=0-1=-1\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=1\\x\in\varnothing\end{array}\right.\)

\(\text{Vậy x=1}\)

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