`(5-2x)^3-27=0`
`<=>(5-2x)^3=27`
`<=>(5-2x)^3=3^3`
`<=>5-2x=3`
`<=>2x=2<=>x=1`
`(5-2x)^3 -27=0`
`<=> (5-2x)^3=0+27`
`<=> (5-2x)^3=27`
`<=> (5-2x)^3= 3^3`
`<=> 5-2x=3`
`=> 2x=5-3`
`=> 2x= 2`
`=>x=2:2`
`=>x=1`
vậy `x=1`
`(5-2x)^3-27=0`
`<=>(5-2x)^3=27`
`<=>(5-2x)^3=3^3`
`<=>5-2x=3`
`<=>2x=2<=>x=1`
`(5-2x)^3 -27=0`
`<=> (5-2x)^3=0+27`
`<=> (5-2x)^3=27`
`<=> (5-2x)^3= 3^3`
`<=> 5-2x=3`
`=> 2x=5-3`
`=> 2x= 2`
`=>x=2:2`
`=>x=1`
vậy `x=1`
Tìm x
a, \(9x^2-49=0\) b, \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-1\right)-27=0\)
c, (x-1)(x+2)-x-2=0 d, \(x\left(3x+2\right)+\left(x+1\right)^2-\left(2x-5\right)\left(2x+5\right)=0\)
e, (4x+1)(x-2)-(2x-3)(2x+1)=7
Tìm x biết:
\(a.\left(x+1\right)\left(x^2-x+1\right)-x\left(x^2-5\right)=71\)
\(b.\left(2x-3\right)^3-8x\left(x-1\right)^2+4x\left(4x+1\right)+27=0\)
BÀI 6 tìm x
1,\(2x\left(x-5\right)-\left(3x+2x^2\right)=0\) 2,\(x\left(5-2x\right)+2x\left(x-1\right)=13\)
3,\(2x^3\left(2x-3\right)-x^2\left(4x^2-6x+2\right)=0\) 4,\(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)
5,\(6x^2-\left(2x-3\right)\left(3x+2\right)=1\) 6,\(2x\left(1-x\right)+5=9-2x^2\)
Tìm x bt
1) \(3.\left(2x-1\right).\left(3x-1\right)-\left(2x-3\right).\left(9x-1\right)-3=-3\)\(-3\)
2) \(3x-1.\left(2x+7\right)-x+1.\left(6x-5\right)=x+2-\left(x-5\right)\)
3) \(3xy.\left(x+y\right)-\left(x+y\right).\left(x^2+y^2+2xy\right)+y^3=27\)
4) \(5.x+1.\left(1-x\right).\left(2x^2+3\right)=0\)
giúp mk vs mai mk đi hk rùi
tìm x
a/ \(4x^2-12x=-9\)
b/ \(\left(5-2x\right)\left(2x+7\right)=4x^2-25\)
c/ \(x^3+27+\left(x+3\right)\left(x-9\right)=0\)
d/ \(4\left(2x+7\right)^2-9\left(x+3\right)^2=0\)
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/ \(4x^2+9x+5=0\)
Tìm \(x\):
\(8\)) \(1-\left(x-6\right)=4\left(2-2x\right)\)
\(9\))\(\left(3x-2\right)\left(x+5\right)=0\)
\(10\))\(\left(x+3\right)\left(x^2+2\right)=0\)
\(11\))\(\left(5x-1\right)\left(x^2-9\right)=0\)
\(12\))\(x\left(x-3\right)+3\left(x-3\right)=0\)
\(13\))\(x\left(x-5\right)-4x+20=0\)
\(14\))\(x^2+4x-5=0\)
Giải phương trình
a. \(\frac{1}{27}\cdot\left(x-3\right)^3-\frac{1}{125}\cdot\left(x-5\right)^3=0\)
b.\(125x^3-\left(2x+1\right)^3-\left(3x-1\right)^3=0\)
c.\(\left(x-3\right)^3+\left(x+1\right)^3=8\cdot\left(x-1\right)^3\)
d.\(\left(x^2-3x+2\right)\cdot\left(x^2+15x+56\right)+8=0\)
e.\(\left(2x^2-3x+1\right)\cdot\left(2x^2+5x+1\right)-9x^2=0\)
f.\(\left(x+6\right)^4+\left(x+8\right)^4=272\)
bài 7 tìm x
1,x(x+3)-5(x+3)=0 2,5x(x-1)=x-1
3,(x+1)=(x+1)\(^2\) 4,x(2x-3)-2(3-2x)=0
5,\(\left(x-2\right)^2-4=0\) 6,\(36x^2=49\)
7,\(2x\left(x-6\right)-x+6=0\) 8,\(3x\left(2x-1\right)-24x+12=0\)
9,\(x^2-6x+8=0\) 10,\(x^2+2x-15=0\)
bài 1: phân tích đa thức thành nhân tử
a,2x+10y
b,x\(^2+4x+4\)
c,\(x^2-y^2+10y-25\)
bài 2 tìm x, biết
a,\(x^2-3x+x-3=0\)
b,\(2x\left(x-3\right)-\dfrac{1}{2}\left(4x^2-3\right)=0\)
c,\(x^2-\left(x-3\right)\left(2x-5\right)=9\)