Tìm x€Z : (x²-9).(3x+15)=0 (4x-8).(x²+1)=0 (x+1).(3-x)=0 (x-2).(2x-1)=0 (3x+9).(1-3x)=0 (x²+1).(81-x²)=0 27/10/2021 Bởi Rose Tìm x€Z : (x²-9).(3x+15)=0 (4x-8).(x²+1)=0 (x+1).(3-x)=0 (x-2).(2x-1)=0 (3x+9).(1-3x)=0 (x²+1).(81-x²)=0
`a)(x^2-9)(3x+15)=0` `→` \(\left[ \begin{array}{l}x^2-9=0\\3x+15=0\end{array} \right.\) `→` \(\left[ \begin{array}{l}x^2=9\\3x=-15\end{array} \right.\) `→` \(\left[ \begin{array}{l}x=±3\\x=-5\end{array} \right.\) Vậy `x∈{3;-3;-5}` `b)(4x-8)(x^2+1)=0` `→4x-8=0` (vì `x^2+1≥1`) `→4x=8` `→x=2` Vậy `x=2` `c)(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}` `d)(x-2)(2x-1)=0` `→` \(\left[ \begin{array}{l}x-2=0\\2x-1=0\end{array} \right.\) `→` \(\left[ \begin{array}{l}x=2\\x=\dfrac{1}{2}\end{array} \right.\) Vậy `x∈{2;1/2}` `e)(3x+9)(1-3x)=0` `→` \(\left[ \begin{array}{l}3x+9=0\\1-3x=0\end{array} \right.\) `→` \(\left[ \begin{array}{l}x=-3\\x=\dfrac{1}{3}\end{array} \right.\) Vậy `x∈{-3;1/3}` `f)(x^2+1)(81-x^2)=0` `→81-x^2=0` (vì `x^2+1≥1`) `→x^2=81` `→x^2=9^2` `→` \(\left[ \begin{array}{l}x=9\\x=-9\end{array} \right.\) Vậy `x∈{9;-9}` Bình luận
(x²-9).(3x+15)=0 => x² – 9 = 0 hoặc 3x + 15 =0 Xét x² – 9 = 0 => x² = 9 => x = ±3 Xét 3x + 15 = 0 => 3x = -15 => x = -5 (4x-8).(x²+1)=0 => 4x – 8 = 0 hoặc x² – 1 =0 Xét : 4x – 8 = 0 => 4x = 8 => x = 2 Xét x² – 1 = 0 => x² = 1 => x = 1 (x+1).(3-x)=0 => x + 1 =0 hoặc 3 – x = 0 Xét x + 1 = 0 => x = -1 Xét : 3 – x = 0 => -x = -3 => x = 3 (x-2).(2x-1)=0 => x – 2 = 0 hoặc 2x – 1 = 0 Xét : x – 2 = 0 => x = 2 Xét 2x – 1 = 0 => 2x = 1 => x = 0,5 (3x+9).(1-3x)=0 => 3x + 9 = 0 hoặc 1 – 3x = 0 Xét 3x + 9 = 0 => 3x = -9 => x = -3 Xét : 1 – 3x = 0 => 3x = 1 => x = $\frac{1}{3}$ (x²+1).(81-x²)=0 => x² + 1 = 0 hoặc 81 – x² = 0 Xét : x² + 1 =0 => x² = -1 (Vô lý ) Xét 81 – x² = 0 => x² = 81 => x = ± 9 Bình luận
`a)(x^2-9)(3x+15)=0`
`→` \(\left[ \begin{array}{l}x^2-9=0\\3x+15=0\end{array} \right.\)
`→` \(\left[ \begin{array}{l}x^2=9\\3x=-15\end{array} \right.\)
`→` \(\left[ \begin{array}{l}x=±3\\x=-5\end{array} \right.\)
Vậy `x∈{3;-3;-5}`
`b)(4x-8)(x^2+1)=0`
`→4x-8=0` (vì `x^2+1≥1`)
`→4x=8`
`→x=2`
Vậy `x=2`
`c)(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}`
`d)(x-2)(2x-1)=0`
`→` \(\left[ \begin{array}{l}x-2=0\\2x-1=0\end{array} \right.\)
`→` \(\left[ \begin{array}{l}x=2\\x=\dfrac{1}{2}\end{array} \right.\)
Vậy `x∈{2;1/2}`
`e)(3x+9)(1-3x)=0`
`→` \(\left[ \begin{array}{l}3x+9=0\\1-3x=0\end{array} \right.\)
`→` \(\left[ \begin{array}{l}x=-3\\x=\dfrac{1}{3}\end{array} \right.\)
Vậy `x∈{-3;1/3}`
`f)(x^2+1)(81-x^2)=0`
`→81-x^2=0` (vì `x^2+1≥1`)
`→x^2=81`
`→x^2=9^2`
`→` \(\left[ \begin{array}{l}x=9\\x=-9\end{array} \right.\)
Vậy `x∈{9;-9}`
(x²-9).(3x+15)=0
=> x² – 9 = 0 hoặc 3x + 15 =0
Xét x² – 9 = 0
=> x² = 9
=> x = ±3
Xét 3x + 15 = 0
=> 3x = -15
=> x = -5
(4x-8).(x²+1)=0
=> 4x – 8 = 0 hoặc x² – 1 =0
Xét : 4x – 8 = 0
=> 4x = 8
=> x = 2
Xét x² – 1 = 0
=> x² = 1
=> x = 1
(x+1).(3-x)=0
=> x + 1 =0 hoặc 3 – x = 0
Xét x + 1 = 0
=> x = -1
Xét : 3 – x = 0
=> -x = -3
=> x = 3
(x-2).(2x-1)=0
=> x – 2 = 0 hoặc 2x – 1 = 0
Xét : x – 2 = 0
=> x = 2
Xét 2x – 1 = 0
=> 2x = 1
=> x = 0,5
(3x+9).(1-3x)=0
=> 3x + 9 = 0 hoặc 1 – 3x = 0
Xét 3x + 9 = 0
=> 3x = -9
=> x = -3
Xét : 1 – 3x = 0
=> 3x = 1
=> x = $\frac{1}{3}$
(x²+1).(81-x²)=0
=> x² + 1 = 0 hoặc 81 – x² = 0
Xét : x² + 1 =0
=> x² = -1 (Vô lý )
Xét 81 – x² = 0
=> x² = 81
=> x = ± 9