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Câu a nhé: 2x . x^2 - 2x . 7x - 2x . 3 = 2x^3 - 14x^2 - 6x
a) \(3x^2-3y^2-x-y\)
\(\Leftrightarrow3\left(x^2-y^2\right)-x-y\)
\(\Leftrightarrow3\left(x-y\right)\left(x+y\right)-\left(x+y\right)\)
\(\Leftrightarrow3\left(x-y\right)\)
d) \(3x^2-7x+4\)
\(\Leftrightarrow3x^2-7x+7-3\)
\(\Leftrightarrow\left(3x^2-3\right)-\left(7x-7\right)\)
\(\Leftrightarrow3\left(x^2-1\right)-7\left(x-1\right)\)
\(\Leftrightarrow3\left(x-1\right)\left(x+1\right)-7\left(x-1\right)\)
\(\Leftrightarrow\left(x-1\right)\left(3\left(x+1\right)-7\right)\)
\(\Leftrightarrow\left(x+1\right)\left(3x-6\right)\)
e) \(-2x^2+3x-1\)
\(\Leftrightarrow\left(-2x^2-1^2\right)+3x\)
\(\Leftrightarrow\left(-2x-1\right)\left(-2x+1\right)+3x\)
f) \(x^2+2xy+y^2-2x-2y\)
\(\Leftrightarrow\left(x+y\right)^2-2\left(x+y\right)\)
\(\Leftrightarrow\left(x+y\right)^2-2\left(x+y\right)\)
k) \(2x^2+5x+3\)
\(\Leftrightarrow2x^2+2x+3x+3\)
\(\Leftrightarrow2x\left(x+1\right)+3\left(x+1\right)\)
\(\Leftrightarrow\left(2x+3\right)\left(x+1\right)\)
l) \(x^2-2x-y^2+1\)
\(\Leftrightarrow\left(x^2-2x+1\right)-y^2\)
\(\Leftrightarrow\left(x-1\right)^2-y^2\)
\(\Leftrightarrow\left(x-1-y\right)\left(x-1+y\right)\)
a) \(3x^2-3y^2-x-y\)
\(\Leftrightarrow3\left(x^2-y^2\right)-x-y\)
\(\Leftrightarrow3\left(x-y\right)\left(x+y\right)-\left(x+y\right)\)
\(\Leftrightarrow3\left(x-y\right)\)
d) \(3x^2-7x+4\)
\(\Leftrightarrow3x^2-7x+7-3\)
\(\Leftrightarrow\left(3x^2-3\right)-\left(7x-7\right)\)
\(\Leftrightarrow3\left(x^2-1\right)-7\left(x-1\right)\)
\(\Leftrightarrow3\left(x-1\right)\left(x+1\right)-7\left(x-1\right)\)
\(\Leftrightarrow\left(x-1\right)\left(3\left(x+1\right)-7\right)\)
\(\Leftrightarrow\left(x+1\right)\left(3x-6\right)\)
e) \(-2x^2+3x-1\)
\(\Leftrightarrow\left(-2x^2-1^2\right)+3x\)
\(\Leftrightarrow\left(-2x-1\right)\left(-2x+1\right)+3x\)
f) \(x^2+2xy+y^2-2x-2y\)
\(\Leftrightarrow\left(x+y\right)^2-2\left(x+y\right)\)
\(\Leftrightarrow\left(x+y\right)^2-2\left(x+y\right)\)
k) \(2x^2+5x+3\)
\(\Leftrightarrow2x^2+2x+3x+3\)
\(\Leftrightarrow2x\left(x+1\right)+3\left(x+1\right)\)
\(\Leftrightarrow\left(2x+3\right)\left(x+1\right)\)
l) \(x^2-2x-y^2+1\)
\(\Leftrightarrow\left(x^2-2x+1\right)-y^2\)
\(\Leftrightarrow\left(x-1\right)^2-y^2\)
\(\Leftrightarrow\left(x-1-y\right)\left(x-1+y\right)\)
Bài 2:
\(A=x^2+4y^2-2x+10-4xy-4y\)
\(=\left(x^2+4xy+4y^2\right)-2\left(x+2y\right)+10\)
\(=\left(x+2y\right)^2-2\left(x+2y\right)+10\)
Thay x + 2y = 5 vào biểu thức A ta được: \(A=5^2-2.5+10=25\)
\(B=\left(x^2+4xy+4y^2\right)-2\left(x+2y\right)\left(y-1\right)+y^2-2y+1\)
\(=x^2+4xy+4y^2-2xy+2x-4y^2+4y+y^2-2y+1\)
\(=x^2+2xy+y^2+2x+2y+1\)
\(=\left(x+y\right)^2+2\left(x+y\right)+1\)
Thay x + y = 5 vào biểu thức B ta được: \(B=5^2+2.5+1=25+10+1=36\)
\(C=x^2-y^2-4x=\left(x^2-4x+4\right)-y^2-4\)
\(=\left(x-2\right)^2-y^2-4\) \(=\left(x-y-2\right)\left(x-2+y\right)-4\)
Thay x + y = 2 vào C ta được: \(C=\left(x-2-y\right)\left(2-2\right)-4=0-4=-4\)
\(D=x^2+y^2+2xy-4x-4y-3\)
\(=\left(x+y\right)^2-4\left(x+y\right)-3\) Thay x + y = 4 vào D ta được:
\(D=4^2-4.4-3=16-16-3=-3\)
Bài 3:
a) \(N=-9x^2+12x-5=-\left(9x^2-12x+4\right)-1\)
\(=-\left(3x-2\right)^2-1\)
Do \(\left(3x-2\right)^2\ge0\) nên \(-\left(3x-2\right)^2-1< 0\)
Vậy N < 0
b) ghi đề cẩn thận lại đi, mk k hiểu
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)
Lời giải:
a. $x^2-y^2+8x-8y=(x^2-y^2)+(8x-8y)$
$=(x-y)(x+y)+8(x-y)=(x-y)(x+y+8)$
b. $4x^2+4xy+y^2-4x-2y=(4x^2+4xy+y^2)-(4x+2y)$
$=(2x+y)^2-2(2x+y)=(2x+y)(2x+y-2)$
c.
$x^3+y^3+4x+4y=(x^3+y^3)+(4x+4y)$
$=(x+y)(x^2-xy+y^2)+4(x+y)$
$=(x+y)(x^2-xy+y^2+4)$
d.
$=(x^3-3x^2y+3xy^2-y^3)+(x^2-y^2)=(x-y)^3+(x-y)(x+y)$
$=(x-y)[(x-y)^2+(x+y)]=(x-y)(x^2-2xy+y^2+x+y)$
e.
$x^2+3x+2=(x^2+x)+(2x+2)=x(x+1)+2(x+1)$
$=(x+1)(x+2)$
f.
$x^2-7x+6=(x^2-6x)-(x-6)=x(x-6)-(x-6)=(x-1)(x-6)$
h.
Đa thức không phân tích được thành nhân tử với hệ số nguyên.