Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(3x\left(x-5\right)-x\left(4+3x\right)=43\)
\(\Leftrightarrow3x^2-15x-4x-3x^2=43\)
\(\Leftrightarrow-19x=43\)
\(\Leftrightarrow x=\frac{-43}{19}\)
Bài 2
a) 4x(x-3)-3x+9
=4x(x-3)-3(x-3)
= (x-3)(4x-3)
b) x3+2x2-2x-4
=(x3+2x2)-(2x+4)
=x2(x+2)-2(x+2)
=(x+2)(x2-2)
c) 4x2-4y+4y-1
=4x2-1
=(2x-1)(2x+1)
d) x5-x
=x(x4-1)
=x(x2-1)(x2+1)
a) 4x(x-3)-3x+9
= 4x(x-3) - 3(x-3)
= (x-3)(4x-3)
b)x3 + 2x2 - 2x - 4
= x2(x + 2) - 2(x + 2)
= (x+2)(x2-2)
c) 4x2 - 4y +4y -1
= [(2x)2-12] + (-4y+4y)
= (2x+1)(2x-1)
d) x5-x
= x(x4 - 1)
b. x4 - x2 - 2x - 1
=x4-(x2+2x+1)
=x4-(x+1)2
=(x2-x-1)(x2+x+1)
d. ( x2 + 3x + 1 ) ( x2 + 3x - 3 ) - 5
Đặt x2+3x=y
=> (y+1)(y-3)-5=y2-2y-8=(y-1)2-9
=(y-4)(y+2)
=(x2+3x-4)(x2+3x+2)=(x-1)(x+4)(x+1)(x+2)
Bài 1 :
a) \(x^4-4x^2-4x-1\)
\(=x^4-\left(4x^2+4x+1\right)\)
\(=x^4-\left(2x+1\right)^2\)
\(=\left(x^2-2x-1\right)\left(x^2+2x+1\right)\)
b) \(x^2+2x-15\)
\(=x^2+2x+1-16\)
\(=\left(x+1\right)^2-4^2\)
\(=\left(x+1+4\right)\left(x+1-4\right)=\left(x+5\right)\left(x-3\right)\)
c) \(x^3y-2x^2y^2+5xy\)
\(=xy\left(x^2-2xy+5\right)\)
B2:
a) \(2\left(x-1\right)^2-\left(2x+3\right)\left(2x-3\right)\)
\(=2\left(x^2-2x+1\right)-\left(4x^2-9\right)\)
\(=2x^2-4x+2-4x^2+9\)
\(=-2x^2-4x+11\)
b) \(\left(x+3\right)^2-2\left(x+3\right)\left(x-3\right)+\left(x-3\right)^2\)
\(=\left(x+3-x+3\right)^2=6^2=36\)
c) \(4\left(x-1\right)\left(x+3\right)+5\left(2x+1\right)^2-2\left(5-3x\right)^2\)
\(=4\left(x^2+2x-3\right)+5\left(4x^2+4x+1\right)-2\left(9x^2-30x+25\right)\)
\(=4x^2+8x-12+20x^2+20x+5-18x^2+60x-50\)
\(=6x^2+88x-57\)
\(a,\left(4x^3-5xy+2x\right)\left(-\dfrac{1}{2}xy\right)\\ =4x^3\cdot\left(-\dfrac{1}{2}xy\right)-5xy\cdot\left(-\dfrac{1}{2}xy\right)+2x\cdot\left(-\dfrac{1}{2}xy\right)\\ =-2x^4y+\dfrac{5}{2}x^2y^2-x^2y\)
\(b,2x^2\left(x^2+3x+\dfrac{1}{2}\right)\\ =2x^2\cdot x^2+2x^2\cdot3x+2x^2\cdot\dfrac{1}{2}\\ =2x^4+6x^3+x^2\)
\(c,\dfrac{2}{3}x^3\left(x+x^2-\dfrac{3}{4}x^5\right)\\ =\dfrac{2}{3}x^3\cdot x+\dfrac{2}{3}x^3\cdot x^2-\dfrac{2}{3}x^3\cdot\dfrac{3}{4}x^5\\ =\dfrac{2}{3}x^4+\dfrac{2}{3}x^5-\dfrac{1}{2}x^8\)
\(d,3x\left(2x^3-\dfrac{1}{3}x^2-4x\right)\\ =3x\cdot2x^3-3x\cdot\dfrac{1}{3}x^2-3x\cdot4x\\ =6x^4-x^3-12x^2\)
a; (4\(x^3\) - 5\(xy\) + 2\(x\))(- \(\dfrac{1}{2}\)\(xy\))
= - 4\(x^3\).\(\dfrac{1}{2}xy\)+ 5\(xy\).\(\dfrac{1}{2}\)\(xy\) - 2\(x\).\(\dfrac{1}{2}\)\(xy\)
= - (4. \(\dfrac{1}{2}\)).(\(x^3\).\(x\)),y + (5.\(\dfrac{1}{2}\)).(\(x.x\)).(y.y)2 - (2.\(\dfrac{1}{2}\))(\(x.x\)).y
= - 2\(x^4\)y + \(\dfrac{5}{2}\)\(x^2\)y2 - \(x^2\)y