a) Ta có: \(-4x^2-7x=0\)
\(\Leftrightarrow x\left(-4x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\-4x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\-4x=7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-7}{4}\end{matrix}\right.\)
Vậy: \(S=\left\{0;-\dfrac{7}{4}\right\}\)
A(2,3x-6,5)(0,1x+2)=0
B(2x+7)(x-5)(5x+1)=0
C(x-1)(2x+7)(x2+2)=0
D(4x-10)(24+5x)=0
E(3,5-7x)(0,1x+2,3)=0
F (5x+2)(x+7)=0
G15 (x+9)(x-3)(x+21)=0
H (x2+1)(×2-4x+4)=0
I(3x-2)(2 (x+3)/9-4x-3/5)=0
Thu gọn biểu thức a)x^2-4x(3x-4)+7x-5 b)7x(x^2-5)-3x^2y(xy-6y^2) c)(5x+4)(2x-7) d)4x(2x-3)-5x(x-2)+x^2(3-x)
1) 2x^4-7x^2-4=0
2)(x62+5x^2)-2(x^2+5x)-24=0
3)x^2-2x-3(x-1)+3=0
4)(x+1/x)^2+2(x+1/x)-8=0
5)x(x+1)(x+2)(2x+3)-18=0
7)(x^2+4x+7)=(x+4)nhân vs căn bậc hai cua x^2 +2
a,4x-10=0 b, 7-3x=9-x c, 2x-(3-5x) = 4(x+3)
d, 5-(6-x)=4(3-2x) e, 4(x+3)=-7x+17 f, 5(x-3) - 4=2(x-1)+7
g, 5(x-3)-4=2(x-1)+7 h,4(3x-2)-3(x-4)=7x+20
Giải phương trình bậc 3:
a)2x^3+5x^2-3x-10=0
b)x^3-2x^2+7x+66=0
c)x^3+3x-4=0
d)x^3+7x^2-48=0
e)4x^3+4x^2-x+14=0
f)3x^3-4x^2+5x+500=0
Giải phương trình
1) 16-8x=0
2) 7x+14=0
3) 5-2x=0
4) 3x-5=7
5) 8-3x=6
6) 8=11x+6
7)-9+2x=0
8) 7x+2=0
9) 5x-6=6+2x
10) 10+2x=3x-7
11) 5x-3=16-8x
12)-7-5x=8+9x
13) 18-5x=7+3x
14) 9-7x=-4x+3
15) 11-11x=21-5x
16) 2(-7+3x)=5-(x+2)
17) 5(8+3x)+2(3x-8)=0
18) 3(2x-1)-3x+1=0
19)-4(x-3)=6x+(x-3)
20)-5-(x+3)=2-5x
Giải các phương trình sau:
g/ x(x + 3)(x – 3) – (x + 2)(x2 – 2x + 4) = 0
h/ (3x – 1)(x2 + 2) = (3x – 1)(7x – 10)
i/ (x + 2)(3 – 4x) = x2 + 4x + 4
k/ x(2x – 7) – 4x + 14 = 0
m/ x2 + 6x – 16 = 0
n/ 2x2 + 5x – 3 = 0
1 phân tích đa thức thành nhân tử a. 7x^2-5x-2
b. x^3-7x^2-4x+10
2 tìm x biết 5.(2x-1)^2-3.(2x-1)=0
3 chứng minh x^2-4x+7>0
