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\(x^3+1-x^2-x\)
\(=\left(x+1\right)\left(x^2+x+1\right)-x\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2+x+1-x\right)\)
\(=\left(x+1\right)\left(x^2+1\right)\)
đáp án
=( x + 1 ) . ( x2 + 1 )
hok tốt
okazki
Mik nghĩ đề câu sau là thek này:
\(x^3+6x^2+3x+1\)
\(=\left(x+1\right)\left(x^2+x+1\right)+6x\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2+7x+1\right)\)
a)\(x^2y-x^3-9y+9y\)
\(=x^2\left(y-x\right)-9\left(y-x\right)\)
\(=\left(x^2-9\right)\left(y-x\right)\)
\(=\left(x+3\right)\left(x-3\right)\left(y-x\right)\)
\(b,9x^2-1=\left(3x+1\right)\left(3x-1\right)\)
\(c,\left(x-y\right)4-4=\left(x-y-1\right)4\)
\(1,x^2y-x^3-9y+9x=x^2\left(y-x\right)-9\left(y-x\right).\)
\(\left(y-x\right)\left(x^2-9\right)=\left(y-x\right)\left(x-3\right)\left(x+3\right)\)
\(2,9x^2-1=\left(3x\right)^2-1^2=\left(3x+1\right)\left(3x-1\right)\)
\(3,\left(x-y\right)4-4=4\left(x-y-1\right)\)
\(4,\)\(9\left(x-y\right)^2=3^2\left(x-y\right)^2=\left(3x-3y\right)^2\)
\(5,3x^2-6ab+3b^2-12c^2???\)
\(6,x^2-25+y^2+2xy=\left(x+y\right)^2-25\)
\(=\left(x+y-5\right)\left(x+y+5\right)\)
\(4x^2-28=0\)
\(\Leftrightarrow4\left(x^2-7\right)=0\)
\(\Leftrightarrow x^2-7=0\)
\(\Leftrightarrow x^2=7\)
\(\Leftrightarrow x=\pm\sqrt{7}\)
x2 - 6x + 9
= (x -3)2 (hàng đẳng thức đáng nhớ số 2)
x2 + x + 1/4
= x2 + 2.x.1/2 + 1/4
= (x +1/2)2 (hàng đẳng thức 1)
x2-6x+9=(x+3)2
x2+x+\(\frac{1}{4}\)=\(\left(x+\frac{1}{2}\right)^2\)
Học tốt!
Ít thôi -..-
a) ( 3x + 2 )( 2x + 9 ) - ( x + 3 )( 6x + 1 ) = ( x + 1 )2 - ( x + 2 )( x - 2 )
<=> 6x2 + 31x + 18 - ( 6x2 + 19x + 3 ) = x2 + 2x + 1 - ( x2 - 4 )
<=> 6x2 + 31x + 18 - 6x2 - 19x - 3 = x2 + 2x + 1 - x2 + 4
<=> 12x + 15 = 2x + 5
<=> 12x - 2x = 5 - 15
<=> 10x = -10
<=> x = -1
b) ( 2x + 3 )( x - 4 ) + ( x - 5 )( x - 2 ) = ( 3x - 5 )( x - 4 )
<=> 2x2 - 5x - 12 + x2 - 7x + 10 = 3x2 - 17x + 20
<=> 3x2 - 12x - 2 = 3x2 - 17x + 20
<=> 3x2 - 12x - 3x2 + 17x = 20 + 2
<=> 5x = 22
<=> x = 22/5
c) ( x + 2 )3 - ( x - 2 )3 - 12x( x - 1 ) = -8
<=> x3 + 6x2 + 12x + 8 - ( x3 - 6x2 + 12x - 8 ) - 12x2 + 12x = -8
<=> x3 + 6x2 + 12x + 8 - x3 + 6x2 - 12x + 8 - 12x2 + 12x = -8
<=> 12x + 16 = -8
<=> 12x = -24
<=> x = -2
d) ( 3x - 1 )2 - 5( x + 1 ) + 6x - 3.2x + 1 - ( x - 1 )2 = 16
<=> 9x2 - 6x + 1 - 5x - 5 + 6x - 6x + 1 - ( x2 - 2x + 1 ) = 16
<=> 9x2 - 11x - 3 - x2 + 2x - 1 = 16
<=> 8x2 - 9x - 4 = 16
<=> 8x2 - 9x - 4 - 16 = 0
<=> 8x2 - 9x - 20 = 0
( Đến đây bạn có hai sự lựa chọn : 1 là vô nghiệm
2 là nghiệm vô tỉ =) )
a) (3x + 2)(2x + 9) - (x + 3)(6x + 1) = (x + 1)2 - (x + 2)(x - 2)
=> 3x(2x + 9) + 2(2x + 9) - x(6x + 1) - 3(6x + 1) = x2 + 2x + 1 - x(x - 2) - 2(x - 2)
=> 6x2 + 27x + 4x + 18 - 6x2 - x - 18x - 3 = x2 + 2x + 1 - x2 + 2x - 2x + 4
=> (6x2 - 6x2) + (27x + 4x - x - 18x) + (18 - 3) = (x2 - x2) + (2x + 2x - 2x) + (1 + 4)
=> 12x + 15 = 2x + 5
=> 12x + 15 - 2x - 5 = 0
=> 10x + 10 = 0
=> 10x = -10 => x = -1
b) (2x + 3)(x - 4) + (x - 5)(x - 2) = (3x - 5)(x - 4)
=> 2x(x - 4) + 3(x - 4) + x(x - 2) - 5(x - 2) = 3x(x - 4) - 5(x - 4)
=> 2x2 - 8x + 3x - 12 + x2 - 2x - 5x + 10 = 3x2 - 12x - 5x + 20
=> (2x2 + x2) + (-8x + 3x - 2x - 5x) + (-12 + 10) = 3x2 - 17x + 20
=> 3x2 - 12x - 2 = 3x2 - 17x + 20
=> 3x2 - 12x - 2 - 3x2 + 17x - 20 = 0
=> (3x2 - 3x2) + (-12x + 17x) + (-2 - 20) = 0
=> 5x - 22 = 0
=> 5x = 22 => x = 22/5
c) (x + 2)3 - (x - 2)3 - 12x(x - 1) = -8
=> x3 + 6x2 + 12x + 8 - (x3 - 6x2 + 12x - 8) - 12x2 + 12x = -8
=> x3 + 6x2 + 12x + 8 -x3 + 6x2 - 12x + 8 - 12x2 + 12x = -8
=> (x3 - x3) + (6x2 + 6x2 - 12x2) + (12x - 12x + 12x) + (8 + 8) = -8
=> 12x + 16 = -8
=> 12x = -24
=> x = -2
Còn bài cuối làm nốt
a/ 4x2+x-4x-1
x(4x+1)-(4x+1)
(4x+1)(x-1)
b/(6-11)x2+3
-5x2+3
c/x2-3xy-4xy+12y2
x(x-3y)-4y(x-3y)
(x-3y)(x-4y)
d/(x-y)2+3(x-y)
(x-y+3)(x-y)
e/(2-12)x2+17x-2
-10x2+17x-2
g/x3+x2+2x2+2x+4x+4
x2(x+1)+2x(x+1)+4(x+1)
(x+1)(x2+2x+4)
h/x3+2x2+7x2+14x+12x+24
x2(x+2)+7x(x+2)+12(x+2)
(x+2)(x2+7x+12)
(x+2)(x2+4x+3x+12)
(x+2)(x+4)(x+3)
Giải:
a) 4x2 - 3x - 1 = 4x2 - 4x + x - 1 = 4x(x - 1) + (x -1) = (x - 1)(4x +1)
b) 6x2 - 11x + 3 = 6x2 - 2x - 9x + 3 = 2x(3x - 1) - 3(3x - 1) = (3x - 1)(2x - 3)
c) x2 - 7xy + 12y2 = x2 - 6xy + 9y2 - xy +3y2 = (x - 3y)2 - y(x - 3y) = (x - 3y)( x - 3y - y) = (x - 3y)(x - 4y)
d) x2 - 2xy + y2 + 3x - 3y = (x - y)2 + 3(x - y) = (x - y)(x - y + 3)
e)Sửa đề: x2 → x3
2x3 - 12x2 + 17x - 2 = 2x3 - 4x2 - 8x2 + 16x + x - 2 = (2x2- 8x + 1)(x -2)
f) x3 - 3x + 2 = x3 - x - 2x + 2 = x(x + 1)(x - 1) - 2(x - 1) = (x - 1)(x2 + x - 2) = (x - 1)2(x + 2)
g) x3 + 3x2 + 6x + 4 = x3 + 3x2 + 3x + 1 + 3x + 3 = (x +1)3 + (x + 1) = (x + 1)(x2 + 2x + 4 )
h) x3 + 9x2 + 26x + 24 = x3 + 4x2 + 5x2 + 20x + 6x + 24 = (x + 4)(x2 + 5x + 6) = (x + 4)(x + 3)(x + 2)ư
Chúc bạn học tốt@@
a. \(\left(x+y\right)^3+\left(x-y\right)^3\)
\(=x^3+3x^2y+3xy^2+y^3+x^3-3x^2y+3xy^2-y^3\)
\(=2x^3+6xy^2\)
\(=2x\left(x^2+6y^2\right)\)
b. \(x^3-y^3+2x^2-2y^2\)
\(=\left(x-y\right)\left(x^2+xy+y^2\right)+2\left(x^2-y^2\right)\)
\(=\left(x-y\right)\left(x^2+xy+y^2\right)+2\left(x+y\right)\left(x-y\right)\)
\(=\left(x-y\right)\left(x^2+xy+y^2+2x+2y\right)\)
c. \(x^3-y^3-3x^2+3x-1\)
\(=\left(x^3-3x^2+3x-1\right)-y^3\)
\(=\left(x-1\right)^3-y^3\)
\(=\left(x-1-y\right)\left[\left(x-1\right)^2+\left(x-1\right)y+y^2\right]\)
\(=\left(x-y-1\right)\left(x^2+y^2+xy-2x-y+1\right)\)
\(x^3+1-x^2-x\)
\(=\left(x+1\right)\left(x^2-x+1\right)-x\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-2x+1\right)\)
\(=\left(x+1\right)\left(x-1\right)^2\)
\(x+y-x^3-y^3\)
\(=\left(x+y\right)-\left(x+y\right)\left(x^2-xy+y^2\right)\)
\(=\left(x+y\right)\left(1-x^2+xy-y^2\right)\)