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a) Áp dụng HĐT 1 thu được ( 2 x + y ) 2 .
b) Áp dụng HĐT 3 với A = 2x + l; B = x - l thu được
[(2x +1) + (x -1)] [(2x +1) - (x -1)] rút gọn thành 3x(x + 2).
c) Ta có: 9 - 6x + x 2 - y 2 = ( 3 - x ) 2 - y 2 = (3 - x - y)(3 -x + y).
d) Ta có: -(x + 2) + 3( x 2 - 4) = -{x + 2) + 3(x + 2)(x - 2)
= (x + 2) [-1 + 3(x - 2)] = (x + 2)(3x - 7).
a) x3-2x2-x+2
=x(x2-1)+2(-x2+1)
=x(x2-1)-2(x2-1)
=(x2-1)(x-2)
b)
x2+6x-y2+9
=x2+6x+9-y2
=(x+3)2-y2
=(x+3-y)(x+3+y)
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)
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: \(A=x^3y-12xy-x^2y\)
\(=xy\cdot x^2-xy\cdot12-xy\cdot x\)
\(=xy\left(x^2-x-12\right)\)
\(=xy\left(x^2-4x+3x-12\right)\)
\(=xy\left[x\left(x-4\right)+3\left(x-4\right)\right]\)
\(=xy\left(x-4\right)\left(x+3\right)\)
c: \(C=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-120\)
=(x+1)(x+4)(x+2)(x+3)-120
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-120\)
\(=\left(x^2+5x\right)^2+10\left(x^2+5x\right)+24-120\)
\(=\left(x^2+5x\right)^2+10\left(x^2+5x\right)-96\)
\(=\left(x^2+5x+16\right)\left(x^2+5x-6\right)\)
\(=\left(x^2+5x+16\right)\left(x+6\right)\left(x-1\right)\)
d: \(D=x^5-x^4+x^2-1\)
\(=\left(x^5-x^4\right)+\left(x^2-1\right)\)
\(=x^4\left(x-1\right)+\left(x-1\right)\left(x+1\right)\)
\(=\left(x-1\right)\left(x^4+x+1\right)\)
a) Xem lại đề
b) x³ - 4x²y + 4xy² - 9x
= x(x² - 4xy + 4y² - 9)
= x[(x² - 4xy + 4y² - 3²]
= x[(x - 2y)² - 3²]
= x(x - 2y - 3)(x - 2y + 3)
c) x³ - y³ + x - y
= (x³ - y³) + (x - y)
= (x - y)(x² + xy + y²) + (x - y)
= (x - y)(x² + xy + y² + 1)
d) 4x² - 4xy + 2x - y + y²
= (4x² - 4xy + y²) + (2x - y)
= (2x - y)² + (2x - y)
= (2x - y)(2x - y + 1)
e) 9x² - 3x + 2y - 4y²
= (9x² - 4y²) - (3x - 2y)
= (3x - 2y)(3x + 2y) - (3x - 2y)
= (3x - 2y)(3x + 2y - 1)
f) 3x² - 6xy + 3y² - 5x + 5y
= (3x² - 6xy + 3y²) - (5x - 5y)
= 3(x² - 2xy + y²) - 5(x - y)
= 3(x - y)² - 5(x - y)
= (x - y)[(3(x - y) - 5]
= (x - y)(3x - 3y - 5)
a) xy+3x-7y-21
=x(y+3)-7(x+3)
=(x-7)(y+3)
b)2xy-15-6x-5y
=2x(y-3)-5(-3+y)
=(2x-5)(y-3)
c)2x^2y+2xy^2-2x-2y
=2x(xy-1)+2y(xy-1)
=(2x+2y)(xy-1)
x(x+3)-5x(x-5)-5(x+3)
=(x-5)(x+3)-5x(x-5)
=(x-5)(x+3-5x)
Câu cuối mình bị nhầm dòng cuối phải là (x-5)(x+3+x-5)=(x-5)(2x-2)nha bạn
\(b,x^3-2x^2-4xy^2+x\)
\(=x\left(x^2-2x-4y^2+1\right)\)
\(=x\left[\left(x^2-2x+1\right)-4y^2\right]\)
\(=x\left[\left(x-1\right)^2-\left(2y\right)^2\right]\)
\(=x\left(x-1-2y\right)\left(x-1+2y\right)\)
\(=x\left(x-2y-1\right)\left(x+2y-1\right)\)
\(---\)
\(c,\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-8\)
\(=\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]-8\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-8\) (1)
Đặt \(y=x^2+7x+10\), thay vào (1) ta được:
\(y\left(y+2\right)-8\)
\(=y^2+2y+1-9\)
\(=\left(y+1\right)^2-3^2\)
\(=\left(y+1-3\right)\left(y+1+3\right)\)
\(=\left(y-2\right)\left(y+4\right)\)
\(=\left(x^2+7x+10-2\right)\left(x^2+7x+10+4\right)\)
\(=\left(x^2+7x+8\right)\left(x^2+7x+14\right)\)
#Ayumu