a/ (3x-4) (2x-1) - (x-1) (6x-5)
b/ (x-2)^2 + (2x+3)^2 -8x
a/ (3x-4) (2x-1) - (x-1) (6x-5)
b/ (x-2)^2 + (2x+3)^2 -8x
\(a,\left(3x-4\right)\left(2x-1\right)-\left(x-1\right)\left(6x-5\right)\)
\(=6x^2-3x-8x+4-\left(6x^2-5x-6x+5\right)\)
\(=6x^2-3x-8x+4-6x^2+5x+6x-5\)
\(=-1\)
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\(b,\left(x-2\right)^2+\left(2x+3\right)^2-8x\)
\(=x^2-4x+4+4x^2+12x+9-8x\)
\(=5x^2+13\)
4x^2 (5x-3y) + x^2 (4x+y)+11x^2y
\(4x^2\left(5x-3y\right)+x^2\left(4x+y\right)+11x^2y\)
\(=20x^3-12yx^2+x^2\left(4x+y\right)+11x^2y\)
\(=24x^3-12yx^2+yx^2+11x^2y\)
\(=24x^3+11x^2y-11yx^2\)
3x-2x=7-x
`3x-2x=7-x`
`3x-2x+x =7`
`2x=7`
`=> x=7/2`
Vậy `x=7/2`
\(3x-2x-7+x=0< =>2x=7< =>x=\dfrac{7}{2}\)
2x²y×(x-1)
tính giá trị của biểu thức :
C=(2x+1)^2+(3x-1)^2+(4x+2)(3x-1) tại x=1/5
mình đang cần gấp
`C=(2x+1)^2+(3x-1)^2+(4x+2)(3x-1)`
`C=4x^2+4x+1+9x^2-6x+1+12x^2-4x+6x-2`
`C=25x^2`
Thay `x=1/5` vào `C` có: `25.(1/5)^2=5^2 . 1/[5^2]=1`
Tìm x biết (4x-1)²=4
`(4x-1)^2 = 4`
`<=> (4x-1)^2 = 2^2`
`@ TH1: 4x - 1 = - 2`
`<=> 4x = - 2 + 1`
`<=> 4x = - 1`
`<=> x = - 1/4`
`@ TH2: 4x - 1 = 2`
`<=> 4x = 2 + 1`
`<=> 4x = 3`
`<=> x = 3/4`
(4x - 1)² = 4
(4x - 1)² = 2² hoặc (4x - 1)² = (-2)²
4x - 1 = 2 hoặc 4x - 1 = -2
*) 4x - 1 = 2
4x = 2 + 1
4x = 3
x = 3/4
*) 4x - 1 = -2
4x = -2 + 1
4x = -1
x = -1/4
Vậy x = -1/4; x = 3/4
m.n ơi giúp mik với
Giúp vs ạ tính mãi ko ra
= (x - x2 )(4-3x) - (3x2 +5x - 12x - 20)
= 4x - 4x2 - 3x2 + 3x3 -3x2 - 5x + 12x + 20
= 3x3 - 10x + 11x +20
e)
= x.x-x.1 .(4-3x) - (x-4)(3x+5)
= x2 - x(4 - 3x) - ( x-4 )(3x+5)
= x2(4 -3x) - x ( 3x- 5) -4( 3x- 5)
= 4x2- 3x3 - 3x2 + 5x - 12x + 20
= - 3x3 + 4x2 - 3x2 + 5x - 12x + 20
= - 3x3 + x2 - 7x + 20