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a) x^3 +5x^2+6x
= x^3+2x^2+3x^2+6x
=x*(x+3)*(x+2)
b) x^2-6x+8
= x^2-2x-4x+8
=(x-2)*(x-4)
c)2x^2+98+28x-8y^2
=2(x^2+14x+49-4y^2)
=2*[(x+7)^2-4y^2]
2*(x-7-2y)*(x-7+2y)
a) x2 + 2x2 + 3x2 + 6x
= x( x+2 ) + 3( x+2 )
=(x+3)(x+2)
b) x2 - 2x - 4x + 8
=x(x-2)-4(x-2)
=(x-4)(x-2)
c)
\(6x^4-11x^2+3=6x^4-9x^2-2x^2+3\)
\(=3x^2\left(2x^2-3\right)-\left(2x^2-3\right)=\left(2x^2-3\right)\left(3x^2-1\right)\)
\(B=x^4+3x^3-x^2+3x^3+9x^2-3x-x^2-3x+1\)
\(=\left(x^2+3x-1\right)^2\)
$ 2x^3 - x^2 + 5x + 3 \\ = 2x^3 + x^2 - 2x^2 - x + 6x + 3 \\ = x^2(2x + 1) - x(2x + 1) + 3(2x + 1) \\ = (2x + 1)(x^2 - x + 3) $
\(2x^3-x^2+5x+3\)
= \(2x^3+x^2-2x^2-x+6x+3\)
\(=x^2\left(2x+1\right)-x\left(2x+1\right)+3\left(2x+1\right)\)
\(=\left(2x+1\right)\left(x^2-x+3\right)\)
Vì \(x^2-x+3=\left(x-\dfrac{1}{2}\right)^2-\dfrac{1}{4}+3>0\)
Nên
\(2x^3-x^2+5x+3=\left(2x+1\right)\left(x^2-x+3\right)\)
a, (x-2)(3x-2)
b, (x+5)(x+6)
c, (x+1)(x+4)
d (x-5y)(x-2y)
e, (x-6)(x-3)
f, (x-2)(x-1)
g (x-2)(3x+1)
h (2x-3)(x+2)
i
a)x2-2xy+y2-2x+2y
=(x2-2xy+y2-2x+2y+1)-1
=(x-y-1)2-1
=(x-y-2)(x-y)
b)x3-49x
=x(x2-72)
=x(x+7)(x-7)
c)x2-y2+6x+9
=(x2+6x+9)-y2
=(x+3)2-y2
=(x-y+3)(x+y+3)
d)x2-6x+5
=x2-5x-x+5
=x(x-5)-(x-5)
=(x-1)(x-5)
1) \(7x-6x^2-2\)
\(=-6x^2+3x+4x-2\)
\(=-3x\left(2x-1\right)+2\left(2x-1\right)\)
\(=\left(2x-1\right)\left(2-3x\right)\)
2) \(x^2-4x+3\)
\(=x^2-3x-x+3\)
\(=x\left(x-3\right)-\left(x-3\right)\)
\(=\left(x-3\right)\left(x-1\right)\)
3) \(2x^2+3x-5\)
\(=2x^2-2x+5x-5\)
\(=2x\left(x-1\right)+5\left(x-1\right)\)
\(=\left(2x+5\right)\left(x-1\right)\)
4) \(16x-5x^2-3\)
\(=-5x^2+15x+x-3\)
\(=-5x\left(x-3\right)+\left(x-3\right)\)
\(=\left(1-5x\right)\left(x-3\right)\)
a, \(x^4-6x^3+11x^2-6x+1=0\)
\(\Rightarrow\left(x^2-3x+1\right)^2=0\)
\(\Rightarrow x^2-3x+1=0\)
\(\Rightarrow x=\frac{\pm\sqrt{5}+3}{2}\)
Chúc bạn học tốt
\(x^4-\left(6x^2-2x^2\right)+\left(9x^2-6x+1\right)=0\)
\(x^4-2x^2\left(3x-1\right)+\left(3x-1\right)^2=0\)
\(\left(x^2-3x+1\right)^2=0\)
tự làm
B) \(\left(6x^4-18x^3\right)+\left(13x^{^3}-39x^2\right)+\left(x-3x\right)-\left(2x-6\right)=0\)
\(6x^3\left(x-3\right)+13x^2\left(x-3\right)+x\left(x-3\right)-2\left(x-3\right)=0\)
\(\left(x-3\right)\left(6x^3+13x^2-2\right)=0\)
\(\left(x-3\right)\left(6x^3+12x^2+x^2+2x-x-2\right)\)
\(\left(x-3\right)\left\{6x^2\left(x+2\right)+x\left(x+2\right)-\left(x+2\right)\right\}\)
\(\left(x-3\right)\left(x+2\right)\left(6x^2-x-1\right)\)
\(\left(x-3\right)\left(x+2\right)\left(6x^2-3x+2x-1\right)\)
\(\left(x-3\right)\left(x+2\right)\left(3x\left(2x-1\right)+\left(2x-1\right)\right)\)
\(\left(x-3\right)\left(x+2\right)\left(2x-1\right)\left(3x+1\right)=0\)
câu C nghĩ đã
a, \(x^4-6x^3+11x^2-6x+1=0\)
=> \(x^4-6x^3+9x^2+2x^2-6x+1=0\)
=> \(x^2+3x+1=0\)
=> \(\Delta\) =\(b^2-4c\)
=\(3^2.4=5\)
Nên \(\sqrt{\Delta}=5\)
x= \(\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{-3+\sqrt{5}}{2}\)
hoặc x= \(\dfrac{b+\sqrt{\Delta}}{2a}=\dfrac{3+\sqrt{5}}{2}\)
a) x =0 không là nghiệm
với x khác 0 ; chia cả 2 vế của pt cho x2
\(6x^2+5x-38+5.\frac{1}{x}+6.\frac{1}{x^2}=0\Leftrightarrow6\left(x^2+2+\frac{1}{x^2}\right)+5\left(x+\frac{1}{x}\right)-50=0\)
\(\Leftrightarrow6\left(x+\frac{1}{x}\right)^2+5\left(x+\frac{1}{x}\right)-50=0\) đặt \(t=\left(x+\frac{1}{x}\right)\)=> 6t2 +5t -50 =0 => t= -10/3 hoặc t =5/2
+x +1/x = -10/3 => 3x2 +10x+3 =0 => x =-3 ; x =-1/3
+x+1/x = 5/2 => 2x2 -5x +2 =0 => x=2; x =1/2
b) a4 +2a2 + 1 - a2 = ( a2 +1)2 -a2 = (a2 -a+1)(a2+a+1)
vao cau hoi tuong tu