x+2x+3x+4x + .... 125x +126x = ?
x+2x+4x+...+125x+126x bang cach thu gon cac bieu thuc sau:
Đề không đúng
Tự dưng 1x+2x+4x+......+125x + 126x là sao
thu gon bieu thuc sau:
x+2x+3x+4x+...+125x+126
a, (2x^3 - 5x^2 - x + 1):( 2x + 1 )
b, ( 4x^3 - 2x^4 + x^5 - 3x^2 + 1 ):( x^2 - 2x + 3 )
c, ( -3x^3 + 7x^2 - 17x + 10 ):( 3x - 1 )
d, ( x^2 - 2x + 1 ):( x - 1 )
e, [ x^2 - 4 + ( x - 2 )^2 ]:( x - 2 )
f, ( 125x^3 + 1 ):( 5x + 1 )
Bài 8: Phân tích đa thức thành nhân tử.
a, x^4 - y^4
b, x^2 - 3y^2
c, (3x - 2y)^2 - (2x - 3y)^2
d, 9(x -y)^2 - 4(x + y)^2
e, (4x^2 - 4x + 1) - (x+1)^2
f, x^3 + 27
g, 27x^3 - 0,001
h, 125x^3 - 1
a) \(x^4-y^4\)
\(=\left(x^2\right)^2-\left(y^2\right)^2\)
\(=\left(x^2-y^2\right)\left(x^2+y^2\right)\)
\(=\left(x+y\right)\left(x-y\right)\left(x^2+y^2\right)\)
b) \(x^2-3y^2\)
\(=x^2-\left(y\sqrt{3}\right)^2\)
\(=\left(x-y\sqrt{3}\right)\left(x+y\sqrt{3}\right)\)
c) \(\left(3x-2y\right)^2-\left(2x-3y\right)^2\)
\(=\left(3x-2y+2x-3y\right)\left(3x-2y-2x+3y\right)\)
\(=\left(5x-5y\right)\left(x+y\right)\)
\(=5\left(x-y\right)\left(x+y\right)\)
d) \(9\left(x-y\right)^2-4\left(x+y\right)^2\)
\(=\left[3\left(x-y\right)+2\left(x+y\right)\right]\left[3\left(x-y\right)-2\left(x+y\right)\right]\)
\(=\left(3x-3y+2x+2y\right)\left(3x-3y-2x-2y\right)\)
\(=\left(5x-y\right)\left(x-5y\right)\)
e) \(\left(4x^2-4x+1\right)-\left(x+1\right)^2\)
\(=\left(2x-1\right)^2-\left(x+1\right)\)
\(=\left(2x-1+x+1\right)\left(2x-1-x-1\right)\)
\(=3x\left(x-2\right)\)
f) \(x^3+27\)
\(=x^3+3^3\)
\(=\left(x+3\right)\left(x^2-3x+9\right)\)
g) \(27x^3-0,001\)
\(=\left(3x\right)^3-\left(0,1\right)^3\)
\(=\left(3x-0,1\right)\left(9x^2+0,3x+0,01\right)\)
h) \(125x^3-1\)
\(=\left(5x\right)^3-1^3\)
\(=\left(5x-1\right)\left(25x^2+5x+1\right)\)
Phân tích các đa thức sau thành nhân tử
a)x^2.(y-z)+y^2.(z-x)+z^2.(x-y)
b) (2x+5)^2-(x-9)^2
c)(2x+3)^2- 25.(x-1)^2
d)(4x^2-3x-18)^2-(4x^2+3x)^2
e)-4x^2+12xy-9y^2+25
f) 8x^2-2
g)8x^3-64
h)125x^3+1
Giải phương trình
a) 125x3-(2x+1)3-(3x-1)3=0
b) 8x3=(4x+1)3-(2x+1)3=0
c) (x-3)3+(x+1)3=8(x-1)3
a: Đặt 2x+1=a; 3x-1=b
Phương trình trở thành \(\left(a+b\right)^3-a^3-b^3=0\)
\(\Leftrightarrow3ab\left(a+b\right)=0\)
=>5x(2x+1)(3x-1)=0
hay \(x\in\left\{0;-\dfrac{1}{2};\dfrac{1}{3}\right\}\)
c: Đặt x-3=a; x+1=b
Theo đề, ta có phương trình \(a^3+b^3=\left(a+b\right)^3\)
=>3ab(a+b)=0
=>(x-3)(x+1)(2x-2)=0
hay \(x\in\left\{3;1;-1\right\}\)
Phân tích thành nhân tử : 8x⁴y+12x²y²+20x³y⁴;4x²-1; 125x³+1; x⁴+2x²+1; 4x²-12xy+y². Cho mình lời giải chi tiết nha.Thanks
`4x^2-1`
`=(2x)^2-1`
`=(2x-1)(2x+1)`
`125x^3+1`
`=(5x)^3+1`
`=(5x+1)(25x^2-5x+1)`
`x^4+2x^2+1`
`=(x^2+1)^2`
`4x^2-12xy+y^2`
`=4x^2-12xy+9y^2-8y^2`
`=(2x-3y)^2-(2sqrt2y)^2`
`=(2x-2sqrt2y-3y)(2x+2sqrt2y-3y)`
Phân tich các đa thức sau thành nhân tử:
a) x3 + x2 - 2x - 8
b) 125x3 - 10x2 + 2x - 1
c) x3 - 4x2 + 12x - 27
d) x4 + 2x3 - 4x - 4
e) x3 - 3x2 - 3x + 1
Cố giúp 1 ít cx đc, giành thời gian giúp mk giải bt là ok r .^.
a) \(x^3+x^2-2x-8\)
\(=\left(x^3-2x^2\right)+\left(3x^2-6x\right)+\left(4x-8\right)\)
\(=x^2\left(x-2\right)+3x\left(x-2\right)+4\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+3x+4\right)\)
b) \(125x^3-10x^2+2x-1\)
\(=\left(125x^3-25x^2\right)+\left(15x^2-3x\right)+\left(5x-1\right)\)
\(=25x^2\left(5x-1\right)+3x\left(5x-1\right)+\left(5x-1\right)\)
\(=\left(5x-1\right)\left(25x^2+3x+1\right)\)
c) \(x^3-4x^2+12x-27\)
\(=\left(x^3-3x^2\right)-\left(x^2-3x\right)+\left(9x-27\right)\)
\(=x^2\left(x-3\right)-x\left(x-3\right)+9\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2-x+9\right)\)
d) Đề sai sai, nghiệm ra khá xấu nên bạn xem lại nhé
e) \(x^3-3x^2-3x+1\)
\(=\left(x^3+x^2\right)-\left(4x^2+4x\right)+\left(x+1\right)\)
\(=x^2\left(x+1\right)-4x\left(x+1\right)+\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-4x+1\right)\)
Đề k sai nên chắc phải hỏi lại cô thôi. Dù sao cx camon bn nhiều
bài 2: phân tích đa thức thành nân tử
a, x4-y4
e, (4x2-4x+1)-(x+1)2
g, 27x3-0.001
c, (3x-2y)2(2x-3y)2
k, 125x3-1
a) x4 - y4
= (x2)2 - (y2)2
= (x2 - y2)(x2 + y2)
= (x - y)(x + y)(x2 + y2)
a.\(x^4-y^4=\left(x^2\right)^2-\left(y^2\right)^2=\left(x^2-y^2\right)\left(x^2+y^2\right)=\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)\)g,\(27x^3-0.001=27x^3-0=27x^3\)
a: \(=\left(x^2-y^2\right)\left(x^2+y^2\right)\)
\(=\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)\)
e: \(=\left(2x-1\right)^2-\left(x+1\right)^2\)
\(=\left(2x-1-x-1\right)\left(2x-1+x+1\right)\)
\(=3x\left(x-2\right)\)
g: \(=\left(3x-\dfrac{1}{10}\right)\left(9x^2+\dfrac{3}{10}x+\dfrac{1}{100}\right)\)
k: \(=\left(5x-1\right)\left(25x^2+5x+1\right)\)