Tìm số nguyên x biết : A) 7x. ( x- 10 ) = 0 B) 17 . ( 3x – 6 ) . ( 2x – 8 ) = 0 C) ( 4 – 2x ) . ( x – 3 ) = 0 D ) – x . ( x + 7 ) . ( x – 4 ) = 0 Giú

Tìm số nguyên x biết :
A) 7x. ( x- 10 ) = 0
B) 17 . ( 3x – 6 ) . ( 2x – 8 ) = 0
C) ( 4 – 2x ) . ( x – 3 ) = 0
D ) – x . ( x + 7 ) . ( x – 4 ) = 0
Giúp mik nha thanks

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  1. a, `7x(x – 10) = 0`

    `⇒ x(x – 10) = 0`

    `⇒` \(\left[ \begin{array}{l}x = 0\\x – 10 = 0\end{array} \right.\) 

    `⇒` \(\left[ \begin{array}{l}x = 0\\x = 10\end{array} \right.\) 

    b, `17(3x – 6)(2x – 8) = 0`

    `⇒ (3x – 6)(2x – 8) = 0`

    `⇒ 3(x – 2)2(x – 4) = 0`

    `⇒ (x – 2)(x – 4) = 0`

    `⇒` \(\left[ \begin{array}{l}x – 2 = 0\\x – 4 = 0\end{array} \right.\) 

    `⇒` \(\left[ \begin{array}{l}x = 2\\x = 4\end{array} \right.\) 

    c, `(4 – 2x)(x – 3) = 0`

    `⇒` \(\left[ \begin{array}{l}4 – 2x = 0\\x – 3 = 0\end{array} \right.\) 

    `⇒` \(\left[ \begin{array}{l}2x = 4\\x = 3\end{array} \right.\) 

    `⇒` \(\left[ \begin{array}{l}x = 2\\x = 3\end{array} \right.\) 

    d, `-x(x + 7)(x – 4) = 0`

    `⇒` \(\left[ \begin{array}{l}-x(x + 7) = 0\\x – 4 = 0\end{array} \right.\) 

    `⇒` $\left[ \begin{array}{l}\left[ \begin{array}{l}x = 0\\x + 7 = 0\end{array} \right.\\x = 4\end{array} \right.$

    `⇒` $\left[ \begin{array}{l}\left[ \begin{array}{l}x = 0\\x = -7\end{array} \right.\\x = 4\end{array} \right.$

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  2. `a) 7x(x-10)=0`

    `<=>` \(\left[ \begin{array}{l}x=0\\x-10=0\end{array} \right.\) 

    `<=>`\(\left[ \begin{array}{l}x=0\\x=10\end{array} \right.\) 

    Vậy `x∈{0;10}`

    `b) 17(3x-6)(2x-8)=0`

    `<=> 17. 3(x-2).2(x-4)=0`

    `<=> 102(x-2)(x-4)=0`

    `<=>`\(\left[ \begin{array}{l}x-2=0\\x-4=0\end{array} \right.\) 

    `<=>`\(\left[ \begin{array}{l}x=2\\x=4\end{array} \right.\) 

    Vậy `x∈{2;4}`

    `c) (4-2x)(x-3)=0`

    `<=> 2(2-x)(x-3)=0`

    `<=>`\(\left[ \begin{array}{l}2-x=0\\x-3=0\end{array} \right.\) 

    `<=>`\(\left[ \begin{array}{l}x=2\\x=3\end{array} \right.\) 

    Vậy `x∈{2;3}`

    `d) -x(x+7)(x-4)=0`

    `<=> x(x+7)(4-x)=0`

    `<=>`\(\left[ \begin{array}{l}x=0\\x+7=0\\4-x=0\end{array} \right.\) 

    `<=>`\(\left[ \begin{array}{l}x=0\\x=-7\\x=4\end{array} \right.\) 

    Vậy `x∈{0;-7;4}`

     

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