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a, \(5x-xy+y^2-5y\)
\(=x\left(5-y\right)-y\left(5-y\right)\)
\(=\left(5-y\right)\left(x-y\right)\)
b, Có: \(x^2+2x+1-y^2\)
<=> \(\left(x+1\right)^2-y^2\)
<=> \(\left(x+1-y\right)\left(x+1+y\right)\)
Với x = 84; y = 15 ta có:
\(\left(x+1-y\right)\left(x+1+y\right)=\left(84+1-15\right)\left(84+1+15\right)\)
<=> \(70.100=7000\)
Bài 1. Tính:
a) \(x^2\left(x-2x^3\right)\)
\(=x^3-2x^5\)
b) \(\left(x^2+1\right)\left(5-x\right)\)
\(=5x^2-x^3+5-x\)
c. \(\left(x-2\right)\left(x^2+3x-4\right)\)
\(=x^3+3x^2-4x-2x^2-6x+8\)
\(=x^3+x^2-10x+8\)
d) \(\left(x-2\right)\left(x-x^2+4\right)\)
\(=x^2-x^3+4x-2x+2x^2-8\)
\(=3x^2-x^3+2x-8\)
e) \(\left(x^2-1\right)\left(x^2+2x\right)\)
\(=x^4+2x^3-x^2-2x\)
f) \(\left(2x-1\right)\left(3x+2\right)\left(3-x\right)\)
\(=\left(6x^2+4x-3x-2\right)\left(3-x\right)\)
\(=\left(6x^2+x-2\right)\left(3-x\right)\)
\(=18x^2+3x-6-6x^3-x^2+2x\)
\(=17x^2+5x-6-6x^3\)
g) \(\left(x+3\right)\left(x^2+3x-5\right)\)
\(=x^3+3x^2-5x+3x^2+9x-15\)
\(=x^3+6x^2+4x-15\)
h) \(\left(xy-2\right)\left(x^3-2x-6\right)\)
\(=x^4y-2x^2y-6xy-2x^3+4x+12\)
i) \(\left(5x^3-x^2+2x-3\right)\left(4x^2-x+2\right)\)
\(=20x^3-5x^4+10x^3-4x^4+x^3-2x^2+8x^3-2x^2+4x-12x^2+3x-6\)
\(=39x^3-9x^4-16x^2+7x-6\)
Bài 5: Tìm x, biết
1) \(\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=6\)
\(\Leftrightarrow\left(x^2-4x+4\right)-\left(x^2-9\right)-6=0\)
\(\Leftrightarrow x^2-4x+4-x^2+9-6=0\)
\(\Leftrightarrow-4x+7=0\)
\(\Leftrightarrow-4x=-7\)
\(\Leftrightarrow x=\dfrac{-7}{-4}=\dfrac{7}{4}\)
Vậy \(x=\dfrac{7}{4}\)
2) \(4\left(x-3\right)^2-\left(2x-1\right)\left(2x+1\right)=10\)
\(\Leftrightarrow4\left(x^2-6x+9\right)-\left(4x^2-1\right)-10=0\)
\(\Leftrightarrow4x^2-24x+36-4x^2+1-10=0\)
\(\Leftrightarrow-24x+27=0\)
\(\Leftrightarrow-24x=-27\)
\(\Leftrightarrow x=\dfrac{-27}{-24}=\dfrac{9}{8}\)
Vậy \(x=\dfrac{9}{8}\)
4) \(\left(x-4\right)^2-\left(x-2\right)\left(x+2\right)=6\)
\(\Leftrightarrow\left(x^2-8x+16\right)-\left(x^2-4\right)-6=0\)
\(\Leftrightarrow x^2-8x+16-x^2+4-6=0\)
\(\Leftrightarrow-8x+14=0\)
\(\Leftrightarrow-8x=-14\)
\(\Leftrightarrow x=\dfrac{-14}{-8}=\dfrac{7}{4}\)
Vậy \(x=\dfrac{7}{4}\)
5) \(9\left(x+1\right)^2-\left(3x-2\right)\left(3x+2\right)=10\)
\(\Leftrightarrow9\left(x^2+2x+1\right)-\left(9x^2-4\right)-10=0\)
\(\Leftrightarrow9x^2+18x+9-9x^2+4-10=0\)
\(\Leftrightarrow18x+3=0\)
\(\Leftrightarrow18x=-3\)
\(\Leftrightarrow x=\dfrac{-3}{18}=\dfrac{-1}{6}\)
Vậy \(x=\dfrac{-1}{6}\)
a) x2 – y 2 – 5x + 5y = (x2 – y 2 ) – (5x – 5y)
= (x + y) (x – y) – 5(x – y) = (x - y) (x + y – 5)
b) 2x2 – 5x – 7 = 2x2 + 2x – 7x – 7
= (2x2 + 2x) – (7x + 7) = 2x(x +1) – 7(x + 1)
= (x + 1)(2x – 7)
Chúc bn học tốt =)
b1 :
a. =(x-1)(5x-3x)=2x(x-1)
b.=(3x)2 +2.3x.y +y2= (3x + y)2
c. =(x+y+x-y)(x+y-x+y)=2x.2y=4xy
d.=(x3)2-(y3)2= (x3+y3)(x3-y3)
bài 1
a, 5x(x-1) - 3x(x-1)
= (5x-3x)(x-1)
= 2x(x-1)
b, 9x2 + 6xy + y2
= (3x)2 + 2.3x.y + y2
= (3x+y)2
c, (x+y)2 - (x-y)2
= [(x+y)-(x-y)][(x+y)+(x-y)]
= (x+y-x+y)(x+y+x-y)
= 2y.2x
= 4xy
d, x6-y6
= (x3)2 - (y3)2
= (x3-y3)(x3+y3)
Câu 1:
a) x2(x - 2x3)
= x3 - 2x5
b) (x2 + 1)(5 - x)
= - x3 + 5x2 - x + 5
c) (x - 2)(x2 + 3x - 4)
= x3 + 3x2 - 4x - 2x2 - 6x + 8
= x3 + x2 - 10x + 8
d) (x - 2)(x - x2 + 4)
= x2 - x3 + 4x - 2x + 2x2 - 8
= -x3 + 3x2 + 2x - 8
e) (x2 - 1)(x2 + 2x)
= x4 + 2x3 - x2 - 2x
f) (2x - 1)(3x + 2)(3 - x)
= (6x2 + 4x - 3x - 2)(3 - x)
= (6x2 + x - 2)(3 - x)
= 18x2 - 6x3 + 3x - x2 - 6 + 2x
= -6x3 + 17x2 + 5x - 6
g) (x + 3)(x2 + 3x - 5)
= x3 + 3x2 - 5x + 3x2 + 9x - 15
= x3 + 6x2 + 4x - 15
h) (xy - 2)(x3 - 2x - 6)
= x4y - 2x2y - 6xy - 2x3 + 4x + 12
i) (5x3 - x2 + 2x - 3)(4x2 - x + 2)
= 20x5 - 5x4 + 10x3 - 4x4 + x3 - 2x2 + 8x3 - 2x2 + 4x - 12x2 + 3x - 6
= 20x5 - 9x4 + 19x3 - 16x2 + 7x - 6
Bạn ơi mk làm 3 phần a,b,c rồi đấy, bạn vào xem đi
Chúc bạn học tốt!
Phân tích đa thức thành nhân tử
a,\(xy+y^2-x-y\)
=\(\left(xy+y^2\right)-\left(x+y\right)\)
=\(y\left(x+y\right)-\left(x+y\right)\)
=\(\left(x+y\right)\left(y-1\right)\)
b, \(25-x^2+4xy-4y^2\)
=\(-\left(x^2-4xy+4y^2-25\right)\)
=\(-\left[\left(x-2y\right)^2-5^2\right]\)
=\(-\left(x-2y-5\right)\left(x-2y+5\right)\)
a) xy + y2 - x - y
= x(x+y) - (x+y)
= (x+y)(x - 1)
b) 25 - x2 + 4xy - 4y2
= 52 -( x2 - 4xy + 4y2)
Tự làm nha 
Bài 2:
a, \(x^{16}-1=\left(x^8\right)^2-1^2\)
\(=\left(x^8-1\right)\left(x^8+1\right)\)
\(=\left(x^4-1\right)\left(x^4+1\right)\left(x^8+1\right)\)
\(=\left(x^2-1\right)\left(x^2+1\right)\left(x^4+1\right)\left(x^8+1\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x^2+1\right)\left(x^4+1\right)\left(x^8+1\right)\)
b, \(x^6-y^6=\left(x^3\right)^2-\left(y^3\right)^2\)
\(=\left(x^3-y^3\right)\left(x^3+y^3\right)\)
\(=\left(x-y\right)\left(x^2+xy+y^2\right)\left(x+y\right)\left(x^2-xy+y^2\right)\)
Chúc bạn học tốt!!!
Bài 1:
a, \(4x^2-25-\left(2x-5\right)\left(2x+7\right)=0\)
\(\Rightarrow\left(2x-5\right)\left(2x+5\right)-\left(2x-5\right)\left(2x+7\right)=0\)
\(\Rightarrow\left(2x-5\right)\left(2x+5-2x-7\right)=0\)
\(\Rightarrow2x-5=0\Rightarrow x=\dfrac{5}{2}\)
b, \(2x^3+3x^2-2x-3=0\)
\(\Rightarrow2x^3-2x^2+5x^2-5x+3x-3=0\)
\(\Rightarrow2x^2\left(x-1\right)+5x\left(x-1\right)+3\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right)\left(2x^2+5x+3\right)=0\)
\(\Rightarrow\left(x-1\right)\left(2x^2+2x+3x+3\right)=0\)
\(\Rightarrow\left(x-1\right)\left[2x\left(x+1\right)+3\left(x+1\right)\right]=0\)
\(\Rightarrow\left(x-1\right)\left(x+1\right)\left(2x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\x+1=0\\2x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=-\dfrac{3}{2}\end{matrix}\right.\)
c, \(x^3+27+\left(x+3\right)\left(x-9\right)=0\)
\(\Rightarrow\left(x+3\right)\left(x^2+3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)
\(\Rightarrow\left(x+3\right)\left(x^2+3x+9+x-9\right)=0\)
\(\Rightarrow\left(x+3\right)\left(x^2+4x\right)=0\)
\(\Rightarrow x\left(x+3\right)\left(x+4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x+3=0\\x+4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=-4\end{matrix}\right.\)
Chúc bạn học tốt!!!
TK
a) x2 + xy –x – y = x(x + y) – (x + y) = (x + y)(x -1 ).
b) a2 – b2 + 8a + 16 = (a2 + 8a + 16) – b2 = (a + 4)2 – b2
= (a + 4 – b)(a + 4 + b).