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30 tháng 9 2018
a, (a, (x + 2)2 - 9 = 0
⇒ (x + 2)2 = 0 + 9 = 9
⇒ (x + 2)2 = \(\left(\pm3\right)^2\)
⇒ x + 2 = \(\pm3\)
\(\Rightarrow\left\{{}\begin{matrix}x+2=3\\x+2=-3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=3-2\\x=-3-2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)
Vậy x ∈ {1; -5}
b, \(\left(x+2\right)^2-x^2+4=0\)
⇒ x2 + 4x + 4 - x2 + 4 =0
⇒ 4x + 8 = 0
⇒ 4 (x + 2) = 0
⇒ x + 2 = 0
⇒ x = 0 - 2
⇒ x = -2
Vậy x = -2
c, (x - 3)2 = (2 - 3x)2
⇒ (x - 3)2 - (2 - 3x)2 = 0
⇒ x2 - 6x + 9 - 4 + 12x - 9x2 = 0
⇒ 6x - 8x2 + 5 = 0
⇒2 \(\left(3x-4x^2+\dfrac{5}{2}\right)\)= 0
⇒ 3x - 4x2 + \(\dfrac{5}{2}\) = 0
⇒ - (4x2- 3x + \(\dfrac{9}{16}+\dfrac{31}{16}\)) = 0
⇒ - (4x2 - 3x + \(\dfrac{9}{16}\)) - \(\dfrac{31}{16}\) = 0
⇒ - (2x - \(\dfrac{3}{4}\))2 = \(\dfrac{31}{16}\) (vô lí)
Vậy x ∈ ∅
31 tháng 7 2018
1,x^2-(x+1)(x-1)=0
x^2-x^2+1+0
1=0(vô lý)
2,5x^3+3x^2+3x+1=4x^2
x^3+3x^2+3x+1=0
(x+1)=0
x=-1
3,x^3+x^2=0
x^2(x+1)=0
x=0 or x=-1
4,2x^3-12x^2+18x=0
x^3-6x^2+9x=0
x(x^2-6x+9)=0
x(x-3)^2=0
x=0 or x=3
5,5x^2-4(x^2-2x+1)+20=0
5x^2-4x^2+8x-4+20=0
x^2+8x+16=0
(x+4)^2=0
x=-4
6,5x(x-3)+7x-21=0
5x(x-3)+7(x-3)=0
(5x+7)(x-3)=0
5x-7=0 or x-3=0
x=7/5 or x=3
7,2x^3-50x=0
2x(x^2-25)=0
2x(x-5)(x+5)=0
x=0 or x=5 or x=-5
8,(4x-1)^2-9(x+3)^2=0
(4x-1)^2-3^2*(x+3)^2=0
(4x-1)^2-(3x+9)^2=0
(4x-1-3x-9)(4x-1+3x+9)=0
(x-10)(7x+8)=0
x=10 or x=-8/7
9,3(x-2)^2-x+2=0
3*(x-2)*(x-2)-(x-2)=0
(3x-6)(x-2)-(x-2)=0
(x-2)(3x-6-1)=0
(x-2)(3x-7)=0
x=2 or x=7/3
10,9x^2+6x-8=0
9x^2+12x-6x-8=0
3x(3x-2)+4(3x-2)=0
(3x+4)(3x-2)=0
3x+4=0 or 3x-2=0
x=-4/3 or x=2/3
NV
Nguyễn Việt Lâm
Giáo viên
26 tháng 2 2020
1. \(x^2\left(x+1\right)+x+1=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow x+1=0\Rightarrow x=-1\)
2. \(\left(x-2\right)\left(6x+2\right)+\left(x-2\right)^2=0\)
\(\Leftrightarrow\left(x-2\right)\left(6x+2+x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right).7x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\7x=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
26 tháng 2 2020
3.
\(x^2-5x+6=0\)
\(\Leftrightarrow x^2-2x-3x+6=0\)
\(\Leftrightarrow x\left(x-2\right)-3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
4.
\(x^2-x-6=0\)
\(\Leftrightarrow x^2+2x-3x-6=0\)
\(\Leftrightarrow x\left(x+2\right)-3\left(x+2\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
PT
20 tháng 9 2018
a) \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-2\right)\left(x+2\right)=27\)
\(\Rightarrow x^3+3^3-x\left(x^2-4\right)=27\)
\(\Rightarrow x^3+27-x^3+4x=27\)
\(\Rightarrow27+4x=27\)
\(\Rightarrow4x=0\)
\(\Rightarrow x=0\)
PT
20 tháng 9 2018
b) \(2x^2+7x+3=0\)
\(\Rightarrow2x^2+x+6x+3=0\)
\(\Rightarrow x\left(2x+1\right)+3\left(2x+1\right)=0\)
\(\Rightarrow\left(2x+1\right)\left(x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x+1=0\\x+3=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=-1\\x=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=-3\end{matrix}\right.\)
17 tháng 7 2018
Lần sau đăng thì chia thành nhiều câu hỏi nhé
\(16^2-9.\left(x+1\right)^2=0\)
\(16^2-\text{ }\left[3.\left(x+1\right)\right]^2=0\)
\(\left[16-3.\left(x+1\right)\right].\left[16+3\left(x+1\right)\right]=0\)
\(\left[16-3x-3\right]\left[16+3x+3\right]=0\)
\(\left[13-3x\right].\left[19+3x\right]=0\)
\(\Rightarrow\orbr{\begin{cases}13-3x=0\\19+3x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=13\\3x=-19\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{13}{3}\\x=-\frac{19}{3}\end{cases}}}\)
KL:..............................
25 tháng 6 2019
Tìm x,biết:
a/ x + 5x2 =0
⇔x ( 1 + 5x ) = 0
\(\Leftrightarrow\) x = 0 hoặc 1 + 5x = 0
1) x = 0
2) 1+ 5x = 0 \(\Leftrightarrow\) x = \(\frac{-1}{5}\)
Vậy: S = \(\left\{0;\frac{-1}{5}\right\}\)
b/x+1=(x+1)2
\(\Leftrightarrow\) (x+1) - (x+1)2 = 0
\(\Leftrightarrow\) ( x+ 1)(1-x-1) = 0
\(\Leftrightarrow\) (x+1).(-x) = 0
\(\Leftrightarrow\) x+1 = 0 hoặc x = 0
\(\Leftrightarrow\) x= -1 ; 0
Vậy: S=\(\left\{-1;0\right\}\)
c/ x3+x=0
\(\Leftrightarrow\) x(x2 + 1) = 0
\(\Leftrightarrow\) x = 0 hoặc x2 + 1 = 0
Ta có : x2 + 1 \(\ge\) 0 vs mọi x
Vậy: S = \(\left\{0\right\}\)
d/5x(x−2)−(2−x)=0
\(\Leftrightarrow\) 5x(x-2) + (x - 2) = 0
\(\Leftrightarrow\) (x - 2)(5x+1) = 0
\(\Leftrightarrow\) x - 2 = 0 hoặc 5x+ 1 = 0
\(\Leftrightarrow\) x = 2 hoặc x = \(\frac{-1}{5}\)
Vậy: S = \(\left\{\frac{-1}{5};2\right\}\)
g/ x(x−4)+(x−4)2=0
⇔ (x - 4)( x+x-4) = 0
\(\Leftrightarrow\) x - 4 = 0 hoặc 2x-4=0
\(\Leftrightarrow\) x = 4 hoặc x = 2
Vậy: S= \(\left\{2;4\right\}\)
h/ x2−3x=0
⇔x (x-3) = 0
\(\Leftrightarrow\) x = 0 hoặc x = 3
Vậy: S = \(\left\{0;3\right\}\)
Vậy: S= \(\left\{0;3\right\}\)
i/4x(x+1)=8(x+1)
⇔4x(x+1)-8(x+1) = 0
\(\Leftrightarrow\) 4(x+1) (x - 2) = 0
\(\Leftrightarrow\) x+1 = 0 hoặc x - 2 = 0
\(\Leftrightarrow\) x= -1 hoặc x = 2
Vậy: S=\(\left\{-1;2\right\}\)
DN
5 tháng 9 2018
\(a,x^2-2x=0\)
\(\Rightarrow x\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
Vậy ...
\(b,\left(5-2x\right)^2-16=0\)
\(\Rightarrow\left(5-2x\right)^2=16\)
\(\Rightarrow\left(5-2x\right)^2=4^2\)
\(\Rightarrow5-2x=\pm4\)
\(\Rightarrow\left[{}\begin{matrix}5-2x=4\\5-2x=-4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=1\\2x=9\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{2}{9}\end{matrix}\right.\)
Vậy ...
\(c,x\left(x+3\right)-x^2-11=0\)
\(\Rightarrow x^2+3x-x^2-11=0\)
\(\Rightarrow3x-11=0\)
\(\Rightarrow3x=11\)
\(\Rightarrow x=\dfrac{11}{3}\)
Vậy ...
\(2\left(x+2\right)^2-x^3-8=0\)
\(\Leftrightarrow2\left(x^2+4x+4\right)-x^3-8=0\)
\(\Leftrightarrow2x^2+8x+8-x^3-8=0\)
\(\Leftrightarrow x\left(-x+2x+8\right)=0\)
\(\Leftrightarrow x\left(x-4\right)\left(x+2\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}x=0\\x-4=0\\x+2=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=0\\x=4\\x=-2\end{cases}}\)
Đổi ngoặc nhọn thành ngoặc vuông giúp mình, mình cảm ơn.
\(2x^2+8x+8-x^2-8=0\)
\(x^2+8x=0\)
\(x\left(x+8\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=-8\end{cases}}\)