\(3x^2-4x-1=0\)
\(\text{Δ}=\left(-4\right)^2-4\cdot3\cdot\left(-1\right)=16+12=28>0\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{4-\sqrt{28}}{6}=\dfrac{4-2\sqrt{7}}{6}=\dfrac{2-\sqrt{7}}{3}\\x_2=\dfrac{4+\sqrt{28}}{6}=\dfrac{2+\sqrt{7}}{3}\end{matrix}\right.\)
/2x+3x/-3x+2=0
/2+3x/=4x-3
/7x+1/-/5x+6/=0
a, |x-1\2|+2x=6
b, 3x+|2\3-x|+2=0
c, 7x+6+|3x-18|=x-1
d,4x-|8-4x|-6=0
e,12-|3x-4|=8-2x
a , | 4x + 2020 | = 0
b , | 2x + 1/4 | + | -5 | = | -14 |
c , | 2020 - 5x | - | 3 | = - | -8 |
d , | x mũ 2 + 4x | = 0
e , | x-1 | + 3x = 1
g , | 2-3x | + 3x = 2
h , | 5x-4 | + 5x = 4
i , | x - 1/4 | - | 2x + 5 | = 0
k , | 5x - 7 | - | 8-5x | = 0
n , | x mũ 3 - 4x | =0
|2 + 3x| = |4x - 3|
|3/2x + 1/2| = |4x - 1|
|5/4x - 7/2| - |5/8x + 3/5| = 0
|7/5x + 3/2| - |4/3 - 1/4| = 0
Tìm x:
1)3x^2-17x+10=0
2)x^4-4x^3+5x^2-4x+4+0
|2/3x-1|.|3/4x+1/2|=0
(2/3x -1 ) x (3/4x + 1/2 ) =0
1) (5-3x) . (1+x) = 0
2) X\(^2\)-3x = 0
4x + 2x\(^2\)= 0
x mũ 2-x=0
x mũ 2 +2x=0
x mũ 2-3x=0
3x mũ 2 +4=0
[1/4x-1]+[5/6x-2]-[3/8x+5]-3,5=0
TÌM NGHIỆM CỦA ĐA THỨC SAU
3x2-6x=0
4x2-3x-1=0
