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a) 8x^2 - 2x - 1
=8x2+2x-4x-1
=2x(4x+1)-(4x+1)
=(2x-1)(4x+1)
b) 6x^2 + 7xy + 2y^2
=4xy+6x2+4y2+3xy
=2x(2y+3x)+y(2y+3x)
=(2y+3x)(y+2x)
c) chịu
d)x^3 + x + 2
Ta thấy :x=-1 là nghiệm của đa thức (đây là dùng pp nhẩm nghiệm nhé)
=>đa thức có 1 hạng tử là x+1
=>(x+1)(x2-x+2) (nếu bn cần cách khác thì nhắn vs mk)
e) x^3 - 2x - 1
lí luận tương tự phần d
=>(x+1)(x2-x-1)
f) x^3 + 3x^2 - 4
lí luận tương tự phần d
=(x-1)(x2+4x+4)
=(x-1)(x+2)2
g) x^2 - 15x + 14
=x2-x-14x+14
=x(x-1)-14(x-1)
=(x-14)(x-1)
a) \(8x^2-2x-1=\left(4x^2-2x\right)+\left(4x^2-1\right)=2x\left(2x-1\right)+\left(2x-1\right)\left(2x+1\right)=\left(2x-1\right)\left(4x+1\right)\)
b) \(6x^2+7xy+2y^2=\left(6x^2+3xy\right)+\left(4xy+2y^2\right)=3x\left(2x+y\right)+2y\left(2x+y\right)=\left(2x+y\right)\left(3x+2y\right)\)
c) \(9x^2-9xy-4y^2=\left(9x^2-y^2\right)-\left(9xy+3y^2\right)=\left(3x-y\right)\left(3x+y\right)-3y\left(3x+y\right)=\left(3x+y\right)\left(3x-4y\right)\)
d) \(x^3+x+2=\left(x^3+1\right)+\left(x+1\right)=\left(x+1\right)\left(x^2-x+1\right)+\left(x+1\right)=\left(x+1\right)\left(x^2-x+2\right)\)
e) \(x^3-2x-1=\left(x^3-x\right)-\left(x+1\right)=x\left(x-1\right)\left(x+1\right)-\left(x+1\right)=\left(x+1\right)\left(x^2-x-1\right)\)
f) \(x^3+3x^2-4=\left(x^3-1\right)+\left(3x^2-3\right)=\left(x-1\right)\left(x^2+x+1\right)+3\left(x-1\right)\left(x+1\right)=\left(x-1\right)\left(x^2+x+1+3x+3\right)=\left(x-1\right)\left(x^2+4x+4\right)=\left(x-1\right)\left(x+2\right)^2\)
g) \(x^2-15x+14=x^2-x+14-14x=x\left(x-1\right)-14\left(x-1\right)=\left(x-1\right)\left(x-14\right)\)
TÌM X
a) (3x+2)(2x+9)-(6x+1)(x+2)=7
=> 6x2 + 31x +18 - 6x2 - 13x - 2 - 7 = 0
=> 18x + 9 = 0 => 9(2x + 1) = 0 => 2x + 1 = 0 => x = -1/2
b) (x-2)(x+5)-(x+3)(x+2)=-6
=> x2 + 3x - 10 - x2 - 5x -6 + 6 = 0 => -2x -10 = 0 => -2(x + 5) = 0
=> x + 5 = 0 => x = -5
c) 3(2x-1)(3x-1)-(2x-3)(9x-1)=0
=> 18x2 - 15x +3 - 18x2 + 29x -3 = 0 => 14x = 0 => x = 0
a) \(\left(3x+2\right)\left(2x+9\right)-\left(6x+1\right)\left(x+2\right)=7\\\Rightarrow 6x^2+31x+18-6x^2-16x-2-7=0\\ \Rightarrow18x+9=0\Rightarrow9\left(2x+1\right)=0\Rightarrow2x+1=0\Rightarrow x=-\frac{1}{2}\)
b) \(\left(x-2\right)\left(x+5\right)-\left(x+3\right)\left(x+2\right)=-6\\ \Rightarrow x^2+3x-10-x^2-5x-6+6=0\\ \Rightarrow-2x-10=0\\ \Rightarrow-2\left(x+5\right)=0\\ \Rightarrow x+5=0\\ \Rightarrow x=-5\)
c) \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)=0\\ \Rightarrow18x^2-15x+3-18x^2+29x-3=0\\ \Rightarrow14x=0\\ \Rightarrow x=0\)
+) <=> \(x^3-3x^2+3x-1+3x^2+6x+8-x^3=17\)
<=>9x=10
<=> x=\(\frac{10}{9}\)
+) \(x\left(x^2-25\right)-x^3-8=3\)<=> \(x^3-x^3-25x=3+8\)
<=> x=\(-\frac{11}{25}\)
x2-7x+12
=x2-3x-4x+12
=x(x-3)-4(x-3)
=(x-3)(x-4)
x4-4x2+4x-1
=x4-1-4x2+4x
=(x2-1)(x2+1)-4x(x-1)
=(x-1)(x+1)(x2+1)-4x(x-1)
=(x-1)[(x+1)(x2+1)-4x]
=(x-1)(x3+x2+x+1-4x)
=(x-1)(x3+x2-3x+1)
6x4-11x2+3
=6x4-2x2-9x2+3
=2x2(3x2-1)-3(3x2-1)
=(3x2-1)(2x2-3)
1.\(x^2-2x-4y^2-4y=\left(x+2y\right)\left(x-2y\right)-2\left(x+2y\right)=\left(x+2y\right)\left(x-2y-2\right)\)
2.\(x^4+2x^3-4x-4=\left(x^2+2\right)\left(x^2-2\right)+2x\left(x^2-2\right)=\left(x^2-2\right)\left(x^2+2x-2\right)\)
3.\(3x^2-3y^2-2\left(x-y\right)^2=3\left(x-y\right)\left(x+y\right)-2\left(x-y\right)\left(x-y\right)=\left(x-y\right)\left(3x+3y-2x+2y\right)\)\(=\left(x-y\right)\left(x+5y\right)\)
4.\(x^3-4x^2-9x+36=x^2\left(x-4\right)-9\left(x-4\right)=\left(x-3\right)\left(x+3\right)\left(x-4\right)\)
5.\(\left(x-1\right)\left(2x+1\right)+3\left(x-1\right)\left(x+2\right)\left(2x+1\right)=\left(x-1\right)\left(2x+1\right)\left(1+3x+6\right)\)\(=\left(x-1\right)\left(2x+1\right)\left(3x+7\right)\)
6.\(\left(6x+3\right)-\left(2x-5\right)\left(2x+1\right)=3\left(2x+1\right)-\left(2x-5\right)\left(2x+1\right)\)\(=\left(2x+1\right)\left(3-2x-5\right)=\left(2x+1\right)\left(-2-2x\right)=-2\left(2x+1\right)\left(x+1\right)\)
7.\(\left(x-5\right)^2+\left(x+5\right)\left(x-5\right)+\left(x-5\right)\left(2x+1\right)=\left(x-5\right)\left(x-5+x+5+2x+1\right)\)\(=\left(x-5\right)\left(4x+1\right)\)
8.\(\left(3x-2\right)\left(4x-3\right)+\left(3x-2\right)\left(x-1\right)-2\left(3x-2\right)\left(x+1\right)\)\(=\left(3x-2\right)\left(4x-3+x-1-2x-2\right)=\left(3x-2\right)\left(3x-6\right)=3\left(3x-2\right)\left(x-2\right)\)