x2 + x = 0
Những câu hỏi liên quan
a) x2(x - 5) + 5 - x = 0; b) 3x4 - 9x3 = -9x2 + 27x;
c) x2(x + 8) + x2 = -8x; d) (x + 3)(x2 -3x + 5) = x2 + 3x.
e) 3x(x - 1) + x - 1 = 0;
f) (x - 2)(x2 + 2x + 7) + 2(x2 - 4) - 5(x - 2) = 0;
g) (2x - 1)2 - 25 = 0;
h) x3 + 27 + (x + 3)(x - 9) = 0.
i)8x3 - 50x = 0; k) 2(x + 3)-x2 - 3x = 0;
m)6x2 - 15x - (2x - 5)(2x + 5) =
a: \(\Leftrightarrow\left(x-5\right)\left(x+1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\\x=1\end{matrix}\right.\)
d: \(\Leftrightarrow\left(x+3\right)\left(x^2-4x+5\right)=0\)
\(\Leftrightarrow x+3=0\)
hay x=-3
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Tìm x biết:
a/ 5x( x- 3) = x – 3 b/ x3 - x = 0 c/ x2 – 7x + 6 = 0
d/ x2 – 4 + ( x – 2)2 = 0 e/ x2 – 16 –( x +4) = 0 f/ x2 + x – 2 = 0
a: \(\Leftrightarrow\left(x-3\right)\left(5x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{5}\end{matrix}\right.\)
b: \(\Leftrightarrow x\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
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c) C x(y2 +z2)+y(z2 +x2)+z(x2 +y2)+2xyz.
d) D x3(y−z)+y3(z−x)+z3(x−y).
e) E (x+y)(x2 −y2)+(y+z)(y2 −z2)+(z+x)(z2 −x2).
b) x2 +2x−24 0.
d) 3x(x+4)−x2 −4x 0.
f) (x−1)(x−3)(x+5)(x+7)−297 0.
(2x−1)2 −(x+3)2 0.
c) x3 −x2 +x+3 0.
e) (x2 +x+1)(x2 +x)−2 0.
a) A x2(y−2z)+y2(z−x)+2z2(x−y)+xyz.
b) B x(y3 +z3)+y(z3 +x3)+z(x3 +y3)+xyz(x+y+z). c) C x(y2 −z2)−y(z2 −x2)+z(x2 −y2).
Đọc tiếp
c) C = x(y2 +z2)+y(z2 +x2)+z(x2 +y2)+2xyz.
d) D = x3(y−z)+y3(z−x)+z3(x−y).
e) E = (x+y)(x2 −y2)+(y+z)(y2 −z2)+(z+x)(z2 −x2).
b) x2 +2x−24 = 0.
d) 3x(x+4)−x2 −4x = 0.
f) (x−1)(x−3)(x+5)(x+7)−297 = 0.
(2x−1)2 −(x+3)2 = 0.
c) x3 −x2 +x+3 = 0.
e) (x2 +x+1)(x2 +x)−2 = 0.
a) A = x2(y−2z)+y2(z−x)+2z2(x−y)+xyz.
b) B = x(y3 +z3)+y(z3 +x3)+z(x3 +y3)+xyz(x+y+z). c) C = x(y2 −z2)−y(z2 −x2)+z(x2 −y2).
Đề bài yêu cầu gì vậy em.
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Bài 11: Tìm x biết:
a) (x+2)(x2-2x+4) - x(x2+2) =15
b) (x+3)2 –x(3x+1)2 +(2x+1)(4x2 -2x+1) =28
c) (x2-1)3 - (x4+x2+1)(x2-1) = 0
d) (x-2)3 –(x-3)(x2 + 3x+9) +6(x+1)2 = 49
Cho các tập hợp A {x ∈ R : (x2 - 4) (x2 - 1) 0}; B {x ∈ R : (x2 - 4) (x2 + 1) 0}; C {-1; 0; 1; 2}; D {x ∈ R :
x
4
-
5
x
2
+
4
x
0}. Khẳng định nào sau đây đúng? A. A B. B. C A. C. D B. D. D A.
Đọc tiếp
Cho các tập hợp A = {x ∈ R : (x2 - 4) (x2 - 1) = 0}; B = {x ∈ R : (x2 - 4) (x2 + 1) = 0}; C = {-1; 0; 1; 2}; D = {x ∈ R : x 4 - 5 x 2 + 4 x = 0}. Khẳng định nào sau đây đúng?
A. A = B.
B. C = A.
C. D = B.
D. D = A.
Đáp án: D
(x2 - 4) (x2 - 1) = 0 ⇔ x = ±2; x = ±1 nên A = {-2; -1; 1; 2}
(x2 - 4) (x2 + 1) = 0 ⇔ x2 - 4 = 0 ⇔ x = ±2 nên B = {-2; 2}
x4 - 5x2 + 4)/x = 0 ⇔ x4 - 5x2 + 4 = 0 ⇔ x = ±2; x = ±1 nên D = {-2; -1; 1; 2}
=> A = D
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(x + 3)(1 – x) > 0
(x2 – 1)(x2 – 4) < 0
(x2 – 20)(x2 – 30) < 0
Tui đang cần gấp, giúp tui nhaa
\(\left(x+3\right)\left(1-x\right)>0.\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+3>0.\\1-x>0.\end{matrix}\right.\\\left\{{}\begin{matrix}x+3< 0.\\1-x< 0.\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>-3.\\x< 1.\end{matrix}\right.\\\left\{{}\begin{matrix}x< -3.\\x>1.\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow-3< x< 1.\)
\(\left(x^2-1\right)\left(x^2-4\right)< 0.\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x^2-1< 0.\\x^2-4>0.\end{matrix}\right.\\\left\{{}\begin{matrix}x^2-1>0.\\x^2-4< 0.\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x^2< 1.\\x^2>4.\end{matrix}\right.\\\left\{{}\begin{matrix}x^2>1.\\x^2< 4.\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\left[{}\begin{matrix}x< 1.\\x>-1.\end{matrix}\right.\\\left[{}\begin{matrix}x>2.\\x< -2.\end{matrix}\right.\end{matrix}\right.\\\left\{{}\begin{matrix}\left[{}\begin{matrix}x>1.\\x< -1.\end{matrix}\right.\\\left[{}\begin{matrix}x< 2.\\x>-2.\end{matrix}\right.\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}-1< x< 1.\\\left[{}\begin{matrix}x>2.\\x< -2.\end{matrix}\right.\end{matrix}\right.\\\left\{{}\begin{matrix}\left[{}\begin{matrix}x>1.\\x< -1.\end{matrix}\right.\\-2< x< 2.\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x>2.\\x< -2.\\-2< x< -1.\\1< x< 2.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x< -2.\\x>2.\end{matrix}\right.\)
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h*) (x + 3)(1 – x) > 0
i*) (x2 – 1)(x2 – 4) < 0
k*) (x2 – 20)(x2 – 30) < 0
Bài 4: Tìm các số nguyên x sao cho
a) –3 ⋮ (x – 2)
b) (3x + 7) ⋮ (x – 2)
c*) (x2 + 7x + 2) ⋮ (x + 7)
a, \(\Rightarrow x-2\inƯ\left(-3\right)=\left\{\pm1;\pm3\right\}\)
| x-2 | 1 | -1 | 3 | -3 |
| x | 3 | 1 | 5 | -1 |
b, \(3\left(x-2\right)+13⋮x-2\Rightarrow x-2\inƯ\left(13\right)=\left\{\pm1;\pm13\right\}\)
| x-2 | 1 | -1 | 13 | -13 |
| x | 3 | 1 | 15 | -11 |
c, \(x\left(x+7\right)+2⋮x+7\Rightarrow x+7\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
| x+7 | 1 | -1 | 2 | -2 |
| x | -6 | -8 | -5 | -9 |
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Bài 1: Giải các pt sau: 1) x2 + 5x + 6 = 0 2)
x2 - x - 6 = 0
3) (x2 + 1) (x2 + 4x + 4) = 0
4) x3 + x2 + x + 1 = 0
5) x2 - 7x + 6 = 0
6) 2x2 - 3x - 5 = 0
7) x2 + x - 12 = 0
8) 2x3 + 6x2 = x2 + 3x
9) (3x - 1) (x2 + 2) = (3x - 1)(7x - 10)
Bài 2: Cho biểu thức A = (5x - 3y + 1) (7x + 2y -2) a) Tìm x sao cho với y = 2 thì A = 0 b) Tìm y sao cho với x = -2 thì A = 0
Bài 1: Giải các pt sau: 1) x2 + 5x + 6 = 0
2) x2 - x - 6 = 0
3) (x2 + 1) (x2 + 4x + 4) = 0
4) x3 + x2 + x + 1 = 0
5) x2 - 7x + 6 = 0
6) 2x2 - 3x - 5 = 0
7) x2 + x - 12 = 0
8) 2x3 + 6x2 = x2 + 3x
9) (3x - 1) (x2 + 2) = (3x - 1)(7x - 10)
Bài 2: Cho biểu thức A = (5x - 3y + 1) (7x + 2y -2) a) Tìm x sao cho với y = 2 thì A = 0 b) Tìm y sao cho với x = -2 thì A = 0
Bài 1)1)\(x^2+5x+6=x^2+3x+2x+6\)=0
=x(x+3)+2(x+3)=(x+2)(x+3)=0
Dễ rồi
2)\(x^2-x-6=0=x^2-3x+2x-6=0\)
=x(x-3)+2(x-3)=0
=(x+2)(x-3)=0
Dễ rồi
3)Phương trình tương đương:\(\left(x^2+1\right)\left(x+2\right)^2=0\)
Vì \(x^2+1>0\)
=>\(\left(x+2\right)^2=0\)
Dễ rồi
4)Phương trình tương đương\(x^2\left(x+1\right)+\left(x+1\right)\)=0
=> \(\left(x^2+1\right)\left(x+1\right)=0Vì\) \(x^2+1>0\)
=>x+1=0
=>..................
5)\(x^2-7x+6=x^2-6x-x+6\) =0
=x(x-6)-(x-6)=0
=(x-1)(x-6)=0
=>.....
6)\(2x^2-3x-5=2x^2+2x-5x-5\)=0
=2x(x+1)-5(x+1)=0
=(2x-5)(x+1)=0
7)\(x^2-3x+4x-12\)=x(x-3)+4(x-3)=(x+4)(x-3)=0
Dễ rồi
Nghỉ đã hôm sau làm mệt
