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Bài 1 :
x2-2x+2>0 với mọi x
=x2-2.x.1/4+1/16+31/16
=(x-1/4)2 + 31/16
Vì (x-1/4)2 \(\ge\) 0 nên (x-1/4)2 + 31/16 \(\ge\) 0 với mọi x (đfcm)
Bài 3
a) 2x(x - 3) - x + 3 = 0
2x(x - 3) - (x - 3) = 0
(x - 3)(2x - 1) = 0
x - 3 = 0 hoặc 2x - 1 = 0
*) x - 3 = 0
x = 3
*) 2x - 1 = 0
2x = 1
x = 1/2
Vậy x = 1/2; x = 3
b) (3x - 1)(2x + 1) - (x + 1)² = 5x²
6x² + 3x - 2x - 1 - x² - 2x - 1 - 5x² = 0
(6x² - x² - 5x²) + (3x - 2x - 2x) = 0 + 1 + 1
-x = 2
x = -2
Bài 2
a) 5x² + 30y
= 5(x² + 6y)
b) x³ - 2x² - 4xy² + x
= x(x² - 2x - 4y² + 1)
= x[(x² - 2x + 1) - 4y²]
= x[(x - 1)² - (2y)²]
= x(x - 1 - 2y)(x - 1 + 2y)
a)x^2-(a+b)x+ab
= x^2 - ax - bx + ab
= (x^2 - ax) - (bx - ab)
= x(x-a) - b(x-a)
= (x-b)(x-a)
b)7x^3-3xyz-21x^2+9z
=
c)4x+4y-x^2(x+y)
= 4(x + y) - x^2(x+y)
= (4-x^2) (x+y)
= (2-x)(2+x)(x+y)
d) y^2+y-x^2+x
= (y^2 - x^2) + (x+y)
= (y-x)(y+x)+ (x+y)
= (y-x+1) (x+y)
e)4x^2-2x-y^2-y
= [(2x)^2 - y^2] - (2x +y)
= (2x-y)(2x+y) - (2x+y)
= (2x -y -1)(2x+y)
f)9x^2-25y^2-6x+10y
=
1) \(x^3+2x-3\)
\(=\left(x^3-x^2\right)+\left(x^2-x\right)+\left(3x-3\right)\)
\(=x^2\left(x-1\right)+x\left(x-1\right)+3\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2+x+3\right)\)
2) \(x^3-6x+4\)
\(=\left(x^3-2x^2\right)+\left(2x^2-4x\right)-\left(2x-4\right)\)
\(=x^2\left(x-2\right)+2x\left(x-2\right)-2\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+2x-2\right)\)
3) \(x^3-2x^2+1\)
\(=\left(x^3-x^2\right)-\left(x^2-x\right)-\left(x-1\right)\)
\(=x^2\left(x-1\right)-x\left(x-1\right)-\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2-x-1\right)\)
4) \(x^3+5x^2-12\)
\(=\left(x^3+2x^2\right)+\left(3x^2+6x\right)-\left(6x+12\right)\)
\(=x^2\left(x+2\right)+3x\left(x+2\right)-6\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2+3x-6\right)\)
\(\left(x-3\right)\left(x-1\right)-3\left(x-3\right)\)
\(=\left(x-3\right)\left(x-1-3\right)\)
\(=\left(x-3\right)\left(x-4\right)\)
\(\left(x-1\right)\left(2x+1\right)+3\left(x-1\right)\left(x+2\right)\left(2x+1\right)\)
\(=\left(x-1\right)\left(2x+1\right)\left(1+3x+6\right)\)
\(=\left(x-1\right)\left(2x+1\right)\left(3x+7\right)\)
a) \(\left(x+3\right)\left(5x-1\right)=\left(5x+1\right)\left(x-2\right)\)
\(\left(x+3\right)\left(5x-1\right)-\left(5x+1\right)\left(x-2\right)=0\)
\(\left(x+3\right)\left(5x-1\right)+\left(5x-1\right)\left(x-2\right)=0\)
\(\left(x+3+x-2\right)\left(5x-1\right)=0\)
\(\left(2x-1\right)\left(5x-1\right)=0\)
Xảy ra 2 trường hợp:
TH1:2x-1=0⇒x=\(\dfrac{1}{2}\)
TH2:5x-1=0⇒x=\(\dfrac{1}{5}\)
b: Ta có: \(\left(2x-3\right)\left(x+1\right)=2x\left(x-1\right)\)
\(\Leftrightarrow2x^2+2x-3x-3-2x^2+2x=0\)
\(\Leftrightarrow x=3\)