Ta có: \(\left(2x+1\right)\left(x^2+2\right)=0\)
mà \(x^2+2\ge2>0\forall x\)
nên 2x+1=0
\(\Leftrightarrow2x=-1\)
hay \(x=-\frac{1}{2}\)
Vậy: \(x=-\frac{1}{2}\)
Ta có: \(\left(2x+1\right)\left(x^2+2\right)=0\)
mà \(x^2+2\ge2>0\forall x\)
nên 2x+1=0
\(\Leftrightarrow2x=-1\)
hay \(x=-\frac{1}{2}\)
Vậy: \(x=-\frac{1}{2}\)
Tìm x)
x(1-2x) +(x-2).(2x-3) = 0
x(2x -4) -2x (x+3) =0
(2x+3)2+(x-3).(2x+3) =0
10y2 -2xy +25 +x2 +30y
(2x-1)2 +(2x+1)2 -2(2x-2)(2x+1) +x = 12
\(\text{x^2 -2x = 24 }\)
\(\left(x+2\right)^2-9=0\)
tìm x help me :)))
Câu 1. Giải các phườn trình sau:
a, 3x+6=0
b, 2x-10=0
c, 3x-7=11
d, 3x-9=0
e, 3x(2-x) =15(x-2)
f, (x+5)(x+4)=0
g, x(x+4)=0
h, (2x -4)(x-2)=0
i, (x+1/5)(2x-3)=0
k, x²-4x=0
m, 4x²-1=0
n, x²-6x+9=0
l, (3x-5)²-(x+4)²=0
o, 7x(x+2)-5(x+2)=0
p, 3x(2x-5)-4x+10=0
q, (2-2x)-x²+1=0
r, x(1-3x)=5(1-3x)
s, 2x-3/4+x+1/6=3
t, x-3/4-2x+1/3=x/6
u, x+1/13+x+2/12=x+3/11+x+4/10
v, 2x+1/15+2x+2/14=2x+3/13+2x+4/12
Giúp e nha mn. E cảm ơn trc ạ!
Tìm x:
a) 2x (x-5) -(x2-10x +25)=0
b) x2 - 9 +3x(x+3) = 0
c) x3 - 16x = 0
d) (2x+3)(x-2) - (x2 -4x+4) = 0
e) 9x2 -(x2 -2x +1)=0
f)x3-4x2 -9x +36 = 0
g) 3x - 6 = (x-1).(x-2)
i) (x-2).(x+2) +(2x+1)2 =-5x.(x-3) =5
k) x2 -1 = (x-1).(2x+3)
l) (2x-1)2 +(x+3).(x-3) -5x(x-2)=6
cho x,y là các số hữu tỉ thoả mãn \(\dfrac{\text{1-2x}}{\text{1-x}}+\dfrac{\text{\text{1-2y}}}{\text{1-y}}\) cmr x^2+y^2 -xy là bình phương một số hữu tỉ
\(P=\left(\dfrac{x^3+2x}{x^3+1}+\dfrac{2x}{x^2-x+1}\right)\text{/}\dfrac{x^2+2x+1}{x-1}\)
a)rút gọn P
b)tại P=-5
a) (x-5).(x-1) > 0
b) (2x-3).(x+1) < 0
c) \(2x^2-3x+1>0\)
d) \(\frac{3x-2}{x-2}>0\)
e) \(\frac{3x-1}{2x-3}< \frac{3}{2}\)
f) \(\frac{x-5}{x^2+1}< 0\)
g) \(\frac{2x-1}{5x-1}< \frac{2}{5}\)