`[x-5]/[x-3] < 0`
`<=>[({(x-5 > 0),(x-3 < 0):}),({(x-5 < 0),(x-3 > 0):}):}`
`<=>[({(x > 5),(x < 3):}\text{ (Vô lí)}),({(x < 5),(x > 3):}<=>3 < x < 5):}`
Vậy `3 < x < 5`
a,x+5/x-1+8/x^2-4x+3=x+1/x-3 b,x-4/x-1-x^2+3/1-x^2+5/x+1=0 c,3x/4-5=3-x/2+5x-1/6 d,(x-2)(x+2)-(x-3)(x+4)-2x+3=0 e,(x-1)^2+2(x+1)=5x+5 g,(x-3)(x+4)x=0
Tìm x,biết
1) 3x^2 - 4x = 0
2) (x^2 - 5x) + x - 5 = 0
3) x^2 - 5x + 6 = 0
4) 5x(x-3) - x+3 = 0
5) x^2 - 2x + 5 = 0
6) x^2 + x -6 = 0
1) x(x-5)-4x+20=0
2) 3(x+1)+x(x+1)
3) 2x^3+x=0
4) x^3-16x=0
5) x^2+6x=-9
6) x^4-2x^3+10x^2-20x=0
7) (2x-3)^2=(x+5)^2
a, (x-1)^3=x-1
b, (x-2)^3=4.(x-2)
c. (x+3)^5=9.(x+5)^3
d,(x-5)^10=(5-x)^7
e, (2.x-1)^2020-(2x-1)^2018=0
f, x4-8x=0
g, x^3-16x=0
h, 4.x^2-x^4=0
i, x^4+9x^2=0
k, x^5+8x^2=0
l, 81.x^2-9x^4=0
m, 32x^5+x^10=0
n, (11-15)^2-5.(3^2-4.x^2)= -9
HELP ME ! MK CẦN GẤP ...
CẢM ƠN CÁC BN NHÌU NHA...
Tìm x
1) (2x-1)(x+3)(2-x)=0
2)x^3 + x^2 + x + 1 = 0
3) 2x(x-3)+5(x-3) =0
4)x(2x-7)-(4x-14)=0
5) 2x^3 + 3x^2 + 2x + 3 = 0
1) Tim x
a) 3x(x-1)+x-1=0
b)2(x+3)-x^2-3x=0
làm tương tự như bài này nè!
x(x-2)+x-2=0
=>x*((x-2)-x-3)=0
=>(x-2)(x+10=0
hoac x-3=0 =>x=3
hoac 5x-1=0 =>x=1 phan 5
vậy x = 1 phần 5 ;x=3
hoac 5x-1=0 => x=1 phan 5
Tìm x, biết :
a/ \(\dfrac{1}{3}x\left(x^2-4\right)=0\)
b/ \(x\left(x+5\right)=x+5\)
c/ \(x^3-\dfrac{1}{9}x=0\)
3)\(^2-\left(x+5\right)^2=0\)
e/ \(\left(x+2\right)^2-\left(x-2\right)\left(x+2\right)=0\)
f/ \(x\left(2x-3\right)-6+4x=0\)
g/ \(2\left(3x-2\right)^2-9x^2+4=0\)
h/ \(x^2\left(x+1\right)+2x\left(x+1\right)=0\)
i/ \(4x^2+9x+5=0\)
giải phương trình sau
1/ ( x-5)^2 +3(x-5) =0
2/ ( x^2-9) +2 (x-3) =0
3/ ( 2x+1)^2+(x-1)(2x+1)=0
4/ (x-1) (x+3) +( x+3)^2=0
5/ ( x+5)^2 -16x^2 =0
6/ x^3-2x^2-x+2=0
1) (x+2)(x-3) = 0 2) (2x + 3)(-x + 7) = 0
3) (x-1)(x+5)(-3x+8) = 0 4) x(x2-1) = 0
