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a) \(x^3-x^2-5x+125\)
\(=\left(x+5\right)\left(x^2-5x+25\right)-x\left(x+5\right)\)
\(=\left(x+5\right)\left(x^2-6x+25\right)\)
b) \(5x^2-5xy-3x+3y\)
\(=5x\left(x-y\right)-3\left(x-y\right)\)
\(=\left(x-y\right)\left(5x-3\right)\)
c) \(x^2-2x-4y^2+1\)
\(=\left(x-1\right)^2-4y^2\)
\(=\left(x-2y-1\right)\left(x+2y-1\right)\)
\(x^2+4x+3\)
\(=\left(x+1\right)\left(x+3\right)\)
\(2x^2+3x-5\)
\(\left(x-1\right)\left(x+\frac{5}{2}\right)\)
Bài 1:
a) 2x^2 -3x + 1 = 2x^2 -2x -x +1 = 2x.(x-1) - (x-1) = (x-1).(2x-1)
b) 2x^3y - 2xy^3 - 4xy^2 - 2xy = 2xy.(x^2 - y^2 - 2y -1) = 2xy.[ x^2 - (y^2 + 2y+1)] = 2xy.[x^2 - (y+1)^2]
= 2xy.(x-y-1).(x+y+1)
c) (x^2 + x+3).(x^2 + x +5) - 8 = (x^2+x+4-1).(x^2+x+4+1) - 8 = (x^2+x+4)^2 - 1 - 8 = (x^2+x+4)^2 - 3^2
= (x^2+x+4-3).(x^2+x+4+3) = (x^2+x+1).(x^2+x+7)
Bài 2:
a) (x+2).(x^2-2x+4) - (x^3+2x) = 0
x^3 + 8 - x^3 - 2x = 0
8 - 2x = 0
x = 4
b) x^2 - 2x - 8 = 0
x^2 +2x - 4x - 8 = 0
x.(x+2) - 4.(x+2) = 0
(x+2).(x-4) = 0
...
bn tự làm tiếp nha
a) \(=\left(x-2y\right)\left(x^2+5x\right)\)
b) \(=\left(x-1\right)\left(x^2+2x+1\right)=\left(x-1\right)\left(x+1\right)^2\)
c) \(=\left(x^2+1-2x\right)\left(x^2+1+2x\right)\)
\(=\left(x^2-2x+1\right)\left(x^2+2x+1\right)\)
\(=\left(x-1\right)^2\left(x+1\right)^2\)
d) \(=3\left(x+3\right)-\left(x-3\right)\left(x+3\right)\)
\(=\left(x+3\right)\left(3-x+3\right)\)
\(=\left(x+3\right)\left(6-x\right)\)
e) \(=\left(x^2-\frac{1}{3}x\right)\left(x^2+\frac{1}{3}x\right)\)
f) \(=2x\left(x-y\right)-16\left(x-y\right)\)
\(=2\left(x-y\right)\left(x-8\right)\)
1.
a) khong bit bạn xem đề sai k
b) x2 -4x +4 - (x+3)(x-3) = 0
<=> x2 -4x + 4 - x2 +9 = 0
<=> -4x = -13 <=> x= 13/4
c) x2 -3x +2 = 0
<=> x2 -x -2x +2 = 0
<=> x(x-1) - 2( x-1) = 0
<=> (x-1)(x-2) = 0 <=> x = 1 hoặc 2
2.
a) x4 -8 = (x2 -4)(x2 +4) = (x-2)(x+2)(x2 +4)
b)x2 -y2 -2x+2y = (x-y)(x+y) - 2(x-y) = (x-y)(x+y-2)
c)x2 - 5x +6 = x2 -3x -2x +6 = x(x-3) - 2(x-3) = (x-2)(x-3)
a) \(\left(3x-1\right)^2+ \left(x+3\right)^2-5\left(2x-3\right)\left(x+3\right)=0\)
\(\Leftrightarrow-15x+55=0\)
\(\Leftrightarrow-15x=0-55\)
\(\Leftrightarrow-15x=-55\)
\(\Leftrightarrow x=\frac{-55}{-15}\)
\(\Rightarrow x=\frac{11}{3}\)
b) \(x^2-4x+4-\left(x+3\right)\left(x-3\right)=0\)
\(\Leftrightarrow x^2-4x-\left(x+3\right)\left(x-3\right)=4-0\)
\(\Leftrightarrow x^2-4x-\left(x+3\right)\left(x-3\right)=4\)
\(\Leftrightarrow4x+9=4\)
\(\Leftrightarrow4x=4+9\)
\(\Leftrightarrow4x=13\)
\(\Rightarrow x=\frac{13}{4}\)
a) x3 + 2x2 + x
= x3 + x2 + x2 + x
= x2 ( x + 1 ) + x ( x + 1 )
= ( x2 + x ) ( x + 1 )
1) (x + 2)(x - 2) - (x + 3)(x + 1)
= x^2 - 4 - (x - 3)(x + 1)
= x^2 - 4 - x^2 + 2x + 3
= 2x - 1
2) a) 5(x - y) - 3x(y - x)
= 5x - 5y - 3x(y - x)
= 5x - 5y - 3xy + 3x2
b) 5x^2 - 16 + 3
= (5x^2 - x) + (-15x + 3)
= x(5x - 1) - 3(5x - 1)
= (5x - 1)(x - 3)
3) a) 2x(x + 3) + 12 - 2x^2 = 0
<=> 2x(x + 3) + 12 - 2x^2 = 0 - 12
<=> 2x(x + 3) - 2x^2 = -12
<=> x = -2
b) x^3 - 16x = 0
<=> x(x + 4)(x - 4) = 0
<=> x = 0
<=> x = 0; x = +- 4
c) (2x - 1)^2 = (x + 3)^2
<=> 4x^2 - 4x + 1 = x^2 + 6x + 9
<=> 4x^2 - 4x + 1 = x^2 + 6x + 9 - 9
<=> 4x^2 - 4x - 8 = x^2 + 6x
<=> 4x^2 - 4x - 8 = x^2 + 6x - 6x
<=> 4x^2 -10x - 8 = x^2
<=> 3x^2 - 10x - 8 = 0
<=> x = 4, x = -2/3
d) x^2 - x - 6 = 0
<=> x = -2; x = 3
\(\left(x+2\right)\left(x-2\right)-\left(x+3\right)\left(x+1\right)\)
\(=x^2-4-\left(x^2+4x+3\right)\)
\(=x^2-4-x^2-4x-3\)
\(=-4x-7\)