(X+3)×(x-3) - (x+3)^2 = 0
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KV
9 tháng 1 2024
1) Do x ∈ Z và 0 < x < 3
⇒ x ∈ {1; 2}
2) Do x ∈ Z và 0 < x ≤ 3
⇒ x ∈ {1; 2; 3}
3) Do x ∈ Z và -1 < x ≤ 4
⇒ x ∈ {0; 1; 2; 3; 4}
10 tháng 8 2017
\(\left(x-3\right)^3+\left(x+3\right)^3=0\)
\(\Leftrightarrow x^3-9x^2+27x-27+x^3+9x^2+27x+27=0\)\(\Leftrightarrow2x^3+54x^2=0\)
\(\Leftrightarrow x^2\left(2x+54\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\2x+54=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-27\end{matrix}\right.\)
\(b,\left(x+1\right)^3-\left(x-1\right)^3=0\)
\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1=0\)\(\Leftrightarrow6x^2+2=0\)
\(\Leftrightarrow6x^2=-2\)
\(\Leftrightarrow x^2=-3\) ( vô lí)
Vậy pt vô nghiệm
\(c,x^2-4x+3=0\)
\(\Leftrightarrow x^2-3x-x+3=0\)
\(\Leftrightarrow x\left(x-3\right)-\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)
\(d,4x^2+4x+1=0\)
\(\Leftrightarrow\left(2x+1\right)^2=0\)
\(\Rightarrow2x+1=0\)
\(\Leftrightarrow2x=-1\Rightarrow x=-\dfrac{1}{2}\)
\(e,\left(x+2\right)^2-\left(x+3\right)^2=0\)
\(\Leftrightarrow\left(x+2-x-3\right)\left(x+2+x+3\right)=0\)
\(\Leftrightarrow-\left(2x+5\right)=0\)
\(\Leftrightarrow-2x-5=0\)
\(\Leftrightarrow-2x=5\Rightarrow x=-\dfrac{5}{2}\)
Học tốt nha you <3
10 tháng 8 2017
\(\left(x-3\right)^3+\left(x+3\right)^3=0\)
\(\Leftrightarrow x^3-9x^2+27x-27+x^3+9x^2+27x+27=0\)\(\Leftrightarrow2x^3+54x^2=0\)
\(\Leftrightarrow x^2\left(2x+54\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\2x+54=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-27\end{matrix}\right.\)
\(b,\left(x+1\right)^3-\left(x-1\right)^3=0\)
\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1=0\)\(\Leftrightarrow6x^2+2=0\)
\(\Leftrightarrow6x^2=-2\)
\(\Leftrightarrow x^2=-3\) ( vô lí)
Vậy pt vô nghiệm
\(c,x^2-4x+3=0\)
\(\Leftrightarrow x^2-3x-x+3=0\)
\(\Leftrightarrow x\left(x-3\right)-\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)
\(d,4x^2+4x+1=0\)
\(\Leftrightarrow\left(2x+1\right)^2=0\)
\(\Rightarrow2x+1=0\)
\(\Leftrightarrow2x=-1\Rightarrow x=-\dfrac{1}{2}\)
\(e,\left(x+2\right)^2-\left(x+3\right)^2=0\)
\(\Leftrightarrow\left(x+2-x-3\right)\left(x+2+x+3\right)=0\)
\(\Leftrightarrow-\left(2x+5\right)=0\)
\(\Leftrightarrow-2x-5=0\)
\(\Leftrightarrow-2x=5\Rightarrow x=-\dfrac{5}{2}\)
Học tốt nha you <3
10 tháng 8 2017
\(\left(x-3\right)^3+\left(x+3\right)^3=0\)
\(\Leftrightarrow x^3-9x^2+27x-27+x^3+9x^2+27x+27=0\)\(\Leftrightarrow2x^3+54x^2=0\)
\(\Leftrightarrow x^2\left(2x+54\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\2x+54=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-27\end{matrix}\right.\)
\(b,\left(x+1\right)^3-\left(x-1\right)^3=0\)
\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1=0\)\(\Leftrightarrow6x^2+2=0\)
\(\Leftrightarrow6x^2=-2\)
\(\Leftrightarrow x^2=-3\) ( vô lí)
Vậy pt vô nghiệm
\(c,x^2-4x+3=0\)
\(\Leftrightarrow x^2-3x-x+3=0\)
\(\Leftrightarrow x\left(x-3\right)-\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)
\(d,4x^2+4x+1=0\)
\(\Leftrightarrow\left(2x+1\right)^2=0\)
\(\Rightarrow2x+1=0\)
\(\Leftrightarrow2x=-1\Rightarrow x=-\dfrac{1}{2}\)
\(e,\left(x+2\right)^2-\left(x+3\right)^2=0\)
\(\Leftrightarrow\left(x+2-x-3\right)\left(x+2+x+3\right)=0\)
\(\Leftrightarrow-\left(2x+5\right)=0\)
\(\Leftrightarrow-2x-5=0\)
\(\Leftrightarrow-2x=5\Rightarrow x=-\dfrac{5}{2}\)
Học tốt nha you <3
VM
25 tháng 10 2019
\(2x^2-6x=0\)
\(\Rightarrow2x.\left(x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0:2\\x=0+3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
Vậy \(x\in\left\{0;3\right\}.\)
\(2x.\left(x+2\right)-3.\left(x+2\right)=0\)
\(\Rightarrow\left(x+2\right).\left(2x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+2=0\\2x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0-2\\2x=3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=3:2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=\frac{3}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{-2;\frac{3}{2}\right\}.\)
\(x^3-16x=0\)
\(\Rightarrow x.\left(x^2-16\right)=0\)
\(\Rightarrow x.\left(x^2-4^2\right)=0\)
\(\Rightarrow x.\left(x-4\right).\left(x+4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-4=0\\x+4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=0+4\\x=0-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)
Vậy \(x\in\left\{0;4;-4\right\}.\)
Chúc bạn học tốt!
NV
Nguyễn Việt Lâm
Giáo viên
8 tháng 1 2024
a.
\(x\left(x+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+7=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-7\end{matrix}\right.\)
b.
\(\left(x+12\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+12=0\\x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-12\\x=3\end{matrix}\right.\)
c.
\(\left(-x+5\right)\left(3-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-x+5=0\\3-x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=3\end{matrix}\right.\)
d.
Em kiểm tra lại ngoặc cuối của câu này
29 tháng 9 2018
\(2x^3-50x=0\)
<=> \(2x\left(x^2-25\right)=0\)
<=> \(2x\left(x-5\right)\left(x+5\right)=0\)
đến đây
bạn tự giải nhé
hk tốt
KN
28 tháng 7 2015
1) (2x-1)(x+3)(2-x)=0
=>2x-1 =0 hoặc x+3=0 hoặc 2-x=0
=>x=1/2 hoặc x=-3 hoặc x=2
2)x^3 + x^2 + x + 1 = 0
=>.x^2(x+1)+(x+1)=0
=>(x^2+1)(x+1)=0
=>x^2+1=0 hoặc x+1=0
=> x =-1
3) 2x(x-3)+5(x-3) =0
=>(2x+5)(x-3)=0
=>2x+5=0 hoặc x-3=0
=>x=-5/2 hoặc x=3
4)x(2x-7)-(4x-14)=0
=> (x-2)(2x-7)=0
=> x-2 =0 hoặc 2x-7=0
=>x=2 hoặc x=7/2
5)2x^3+3x^2+2x+3=0
=>x^2(2x+3)+2x+3=0
=>(x^2+1)(2x+3)=0
=>x^2+1=0 hoặc 2x+3=0
=> x =-3/2
19 tháng 6 2021
a, 4x2 - 49 = 0
⇔⇔ (2x)2 - 72 = 0
⇔⇔ (2x - 7)(2x + 7) = 0
⇔{2x−7=02x+7=0⇔⎧⎪ ⎪⎨⎪ ⎪⎩x=72x=−72⇔{2x−7=02x+7=0⇔{x=72x=−72
b, x2 + 36 = 12x
⇔⇔ x2 + 36 - 12x = 0
⇔⇔ x2 - 2.x.6 + 62 = 0
⇔⇔ (x - 6)2 = 0
⇔⇔ x = 6
19 tháng 6 2021
e, (x - 2)2 - 16 = 0
⇔⇔ (x - 2)2 - 42 = 0
⇔⇔ (x - 2 - 4)(x - 2 + 4) = 0
⇔⇔ (x - 6)(x + 2) = 0
⇔{x−6=0x+2=0⇔{x=6x=−2⇔{x−6=0x+2=0⇔{x=6x=−2
f, x2 - 5x -14 = 0
⇔⇔ x2 + 2x - 7x -14 = 0
⇔⇔ x(x + 2) - 7(x + 2) = 0
⇔⇔ (x + 2)(x - 7) = 0
⇔{x+2=0x−7=0⇔{x=−2x=7
bn co muon choi surviv.io ko
\(\left(x+3\right)\left(x-3\right)-\left(x+3\right)^2=0\)
\(\Rightarrow\left(x+3\right)\cdot\left(x-3-x-3\right)=0\)
\(\Rightarrow\left(x+3\right)\cdot\left(-6\right)=0\)
\(\Rightarrow x+3=0\)
\(\Rightarrow x=-3\)