Bài 2: Tìm x Z sao cho:
a. (x + 1).(3 – x) = 0 b. (x – 2).(2x – 1) = 0 c. (3x + 9).(1 – 3x) = 0
d. (x2 + 1).(81 – x2 ) = 0 e. (x – 5)5 = 32 f. (2 – x)4 = 81
g. (31 – 2x)3 = -27 h. (x – 2).(7 – x) > 0 i. |x – 7| ≤ 3
Đánh máy ra
Giải thích các bước giải :
`a)(x+1)(3-x)=0`
`<=>`\(\left[ \begin{array}{l}x+1=0\\3-x=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=-1\\x=3\end{array} \right.\)
Vậy : \(\left[ \begin{array}{l}x=-1\\x=3\end{array} \right.\)
`b)(x-2)(2x-1)=0`
`<=>`\(\left[ \begin{array}{l}x-2=0\\2x-1=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=2\\2x=1\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=2\\x=\frac{1}{2}\end{array} \right.\)
Vậy : \(\left[ \begin{array}{l}x=2\\x=\frac{1}{2}\end{array} \right.\)
`c)(3x+9)(1-3x)=0`
`<=>3(x+3)(1-3x)=0`
`<=>(x+3)(1-3x)=0`
`<=>`\(\left[ \begin{array}{l}x+3=0\\1-3x=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=-3\\3x=1\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=2\\x=\frac{1}{3}\end{array} \right.\)
Vậy : \(\left[ \begin{array}{l}x=2\\x=\frac{1}{3}\end{array} \right.\)
`d)(x^2+1)(81-x^2)=0`
Vì `x^2 ≥ 0 => x^2+1 >0 => x^2+1 \ne 0`
`=>81-x^2=0`
`<=>x^2=81`
`<=>x^2=(±9)^2`
`<=>x=±9`
Vậy `x=±9`
`e)(x-5)^5=32`
`<=>(x-5)^5=2^5`
`<=>x-5=2`
`<=>x=2+5`
`<=>x=7`
Vậy `x=7`
`f)(2-x)^4=81`
`<=>(2-x)^4=(±3)^4`
`<=>`\(\left[ \begin{array}{l}2-x=3\\2-x=-3\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=-1\\x=5\end{array} \right.\)
Vậy : \(\left[ \begin{array}{l}x=-1\\x=5\end{array} \right.\)
`g)(31-2x)^3=-27`
`<=>(31-2x)^3=(-3)^3`
`<=>31-2x=-3`
`<=>2x=34`
`<=>x=17`
Vậy `x=17`
`h)(x-2)(7-x)>0`
`<=>`\(\left[ \begin{array}{l}\left \{ {{x-2>0} \atop {7-x>0}} \right. \\\left \{ {{x-2<0} \atop {7-x<0}} \right. \end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}\left \{ {{x>2} \atop {x<7}}=>2<x<7 ™ \right. \\\left \{ {{x<2} \atop {x>7}}=>7<x<2(Loại) \right. \end{array} \right.\)
Vậy `2<x<7`
`i)|x-7| ≤ 3`
`<=>-3 ≤ x-7 ≤ 3`
`<=>-3+7 ≤ x-7+7 ≤ 3+7`
`<=>4 ≤ x ≤ 10`
Vậy `4≤x≤10`
Bài 2:
a, ( x + 1 ) ( 3 – x ) = 0
`=>` \(\left[ \begin{array}{l}x+1=0\\3-x=0\end{array} \right.\)
`=>` \(\left[ \begin{array}{l}x=-1\\x=3\end{array} \right.\)
Vậy x ∈ { -1 ; 3 }
b, ( x – 2 ) ( 2x – 1 ) = 0
`=>` \(\left[ \begin{array}{l}x-2=0\\2x-1=0\end{array} \right.\)
`=>` \(\left[ \begin{array}{l}x=2\\x =\frac{1}{2}\end{array} \right.\)
Vậy x ∈ { 2 ; `1/2` }
c, ( 3x + 9 ) ( 1 – 3x ) = 0
`=>` \(\left[ \begin{array}{l}3x+9=0\\1-3x=0\end{array} \right.\)
`=>` \(\left[ \begin{array}{l}x=-3\\x=\frac{1}{3}\end{array} \right.\)
d, ( x² + 1 ) ( 81 – x² ) = 0
`=>` \(\left[ \begin{array}{l}x^2+1=0\\81-x^2=0\end{array} \right.\)
`=>` \(\left[ \begin{array}{l}x=-1(vô lí)\\x=9\end{array} \right.\)
`=>` \(\left[ \begin{array}{l}x^2=-1(loại)\\x=±9\end{array} \right.\)
Vậy x ∈ { ±9 }
e, `(x-5)^5` = 32
`(x-5)^5` = `2^5`
`=>` x – 5 = 5
`=>` x = 10
Vậy x = 10
f, `(2-x)^4` = 81
`=>` `(2-x)^4` = `3^4`
`=>` 2 – x = 3
`=>` x = -1
Vậy x = -1
g, `(31-2x)^3` = -27
`(31-2x)^3` = `(-3)^3`
`=>` 31 – 2x = -3
`=>` 2x = 34
`=>` x = 17
Vậy x = 17
h, ( x – 2 ) ( 7 – x ) > 0
`=>` x – 2 và 7 – x cùng dấu
+, Với x – 2 > 0 `=>` 7 – x > 0
`=>` x > 2 và x < 7 ( t/m )
+, Với x – 2 < 0 `=>` 7 – x < 0
`=>` x < 2 và x > 7 ( vô lí )
Vậy 2 < x < 7
i, | x – 7 | ≤ 3
`=>` x – 7 ∈ { 0 ; ±1; ±2 ; ±3 }
`=>` x ∈ { 7; 8; 6; 9; 5; 10; 4 }
Vậy x ∈ { 7; 8; 6; 9; 5; 10; 4 }