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a) \(^{x^4-y^4}\)
\(=\left(x^2-y^2\right)\left(x^2+y^2\right)\)
\(=\left[\left(x-y\right).\left(x+y\right)\right].\left(x^2-y^2\right)\)
\(=\left(x-y\right).\left(x+y\right).\left(x^2-y^2\right)\)
c) \(\left(3x-2y\right)^2-\left(2x-3y\right)^2\)
\(=\left[\left(3x-2y\right)+\left(2x-3y\right)\right].\left[\left(3x-2y\right)-\left(2x-3y\right)\right]\)
\(=\left(3x-2y+2x-3y\right)\left(3x-2y-2x+3y\right)\)
b) \(x^2-3y^2\)
\(=\left(x-3y\right)\left(x+3y\right)\)
d) \(9\left(x-y\right)^2-4\left(x+y\right)^2\)
\(=9\left(x-y\right)^2+4\left(x-y\right)^2\)
\(=\left(x-y\right).\left(9+4\right)\)
\(=\left(x-y\right).13\)
\(=13\left(x-y\right)\)
f) \(x^3+27\)
\(=x^3+3^3\)
\(=\left(x+3\right)\left(x^2-x.3+3^2\right)\)
h) \(125x^3-1\)
\(=\left(5x\right)^3-1^3\)
\(=\left(5x-1\right)\left(5x^2+5x.1+1^2\right)\)
\(=\left(5x-1\right)\left(5x^2+5x+1\right)\)
\(a,x^4-y^4=\left(x^2+y^2\right)\left(x^2-y^2\right)=\left(x^2+y^2\right)\left(x+y\right)\left(x-y\right)\)
\(b,x^2-3y^2=\left(x+\sqrt{3}y\right)\left(x-\sqrt{3}y\right)\)
cn lại tg tự nha bn
Bài làm:
a, 1-4x2
=1-(2x)2
=(1-2x).(1+2x)
b, 8-27x3
=23-(3x)3
=(2-3x).(4+6x+9x2)
Các câu còn lại bạn dùng hằng đẳng thức là phân tích được ra thôi
1 - 4x^2
= 1^2 - ( 2x )^2
= ( 1 - 2x ) ( 1 + 2x )
8 - 27x^ 3
= 2^3 - ( 3x )^3
= ( 2 - 3x ) [ 2^2 + 2 * 3x + ( 3x )^2 ]
= ( 2 - 3x ) ( 4 + 6x + 9x^2 )
= ( 2 - 3x ) ( 9x^2 + 6x + 4 )
27 + 27x + 9x^2 + x^3
= x^3 + 9x^2 + 27x + 27
= x^3 + 3x^2 + 6x^2 + 18x + 9x + 27
= x^2 ( x + 3 ) + 6x ( x + 3 ) + 9 ( x + 3 )
= ( x + 3 ) ( x^2 + 6x + 9 )
= ( x + 3 ) ( x + 3 )^2
= ( x + 3 )^3
x^2 + 4x - 5
= x^2 - x + 5x - 5
= x ( x - 1 ) + 5 ( x - 1 )
= ( x + 1 ) ( x - 5 )
a) \(x^2-12x+36=\left(x-6\right)^2\)
b) \(x^2-7=\left(x-\sqrt{7}\right)\left(x+\sqrt{7}\right)\)
c) \(1-27x^3=\left(1-3x\right)\left(1+3x+9x^2\right)\)
d) mk chỉnh lại đề chút xíu nhé. đúng thì bn tham khảo:
\(\left(x-y\right)^2-4y^2=\left(x-y-2y\right)\left(x-y+2y\right)=\left(x-3y\right)\left(x+y\right)\)
e) \(y^3+3y^2+3y+1=\left(y+1\right)^3\)
f) \(x^2+14x+49=\left(x+7\right)^2\)
\(\left(x-1\right)^2-25\)
\(=x^2-2x+1-25\)
\(=x^2-2x-24\)
\(=x^2-6x+4x-24\)
\(=x.\left(x-6\right)+4.\left(x-6\right)\)
\(=\left(x+4\right).\left(x-6\right)\)
a, \(1-2y+y^2=\left(y+1\right)^2=\left(y+1\right)\left(y+1\right)\)
b, \(\left(x+1\right)^2-25=\left(x+1\right)^2-5^2=\left(x+1-5\right)\left(x+1+5\right)=\left(x-4\right)\left(x+6\right)\)
c, \(1-4x^2=1^2-\left(2x\right)^2=\left(1-2x\right)\left(1+2x\right)\)
d, \(8-27x^3=2^3-\left(3x\right)^3=\left(2-3x\right)\left(4+6x+9x^2\right)\)
a/ x^3-6x^2+12x-8
=(x-2)^3
b/x^2+5x+4
=x^2+x+4x+4
=x(x+1)+4(x+1)
=(x+1)(x+4)
c/ 16^2-9(x+1)^2=0
<=> (4x-3x-3)(4x+3x+3)=0
<=>x-3=0 hay 7x+3=0
<=> x=3 hay x=-3/7
d/ x^3-2x^2-x+2
=x^2(x-2)-(x-2)
=(x-2)(x^2-1)
=(x-2)(x-1)(x+1)
e/x^2+y^2-2xy-x+y
=(x-y)^2-(x-y)
f/x^3+y^3+3y^2+3y+1
=x^3+(y+1)^3
=(x+y+1)[x^2-xy-x+(y+1)^2]
=(x+y+1)(x^2-xy-x+y^2+2y+1)
b. x2+2.5/2x+(5/2)2-(5/2)2+4
= (x+5/2)2-25/4+4
=(x+5/2)2-(3/2)2
= x+ 5/2 -3/2 ) . (x+5/2-3/2)
= (x+1 ) (x+2)
c.
(4x)2- [3(x+1)]2 =0
[4x-3(x+1)] [4x+3(x+1)] =0
(x-3) (7x+3) =0
<=> x-3 =0 => x = 3
7x+3=0 => x= -3/7
d. x3-2x2-x+2
= (x3-2x2) - (x+2)
= x2 (x-2) - (x-2)
= (x-2) (x2-1)
CHÚC BẠN HỌC TỐT
* Tớ còn a, e, và f sorry nó k dễ để suy nghĩ trong thơi gian ngắn được nên tớ bỏ !! Ahihihih
\(27x^3-27x^2+18x-4\)
\(=27x^3-9x^2-18x^2+6x+12x-4\)
\(=9x\left(3x-1\right)-6x\left(3x-1\right)+4\left(3x-1\right)\)
\(=\left(3x-1\right)\left(9x^2-6x+4\right)\)
\(x^2+2xy+y^2-x-y-12\)
\(=\left(x+y\right)^2-\left(x+y\right)-12\)
\(=\left(x+y\right)^2-4\left(x+y\right)+3\left(x+y\right)-12\)
\(=\left(x+y\right)\left(x+y-4\right)+3\left(x+y-4\right)=\left(x+y+3\right)\left(x+y-4\right)\) \(P=\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\) (nhóm 2 cái đầu với cuối lại với nhau, 2 cái giữa vào 1 nhóm)
Đặt \(x^2+7x+11=a\)
Ta có: \(P=\left(a-1\right)\left(a+1\right)-24\)
\(=a^2-25=\left(a-5\right)\left(a+5\right)\)
\(=\left(x^2+7x+6\right)\left(x^2+7x+16\right)=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)
d, \(4x^4-32x^2+1\)
\(=4x^4+4x^2+1-36x^2\)
\(=\left(2x+1\right)^2-\left(6x\right)^2=\left(2x^2-6x+1\right)\left(2x^2+6x+1\right)\)
a) (x + y)2 - 2(x + y) + 1
= (x + y)2 - 2.1.(x + y) + 1
= (x + y - 1)2
b) x3 + 1 - x2 - x
= (x3 - x2) - (x - 1)
= x2(x - 1) - (x - 1)
= (x2 - 1)(x - 1) = (x - 1)(x + 1)(x - 1) = (x - 1)2(x + 1)
c) 27x3 - 0,001
= \(\left(3x\right)^3-\frac{1}{1000}=\left(3x\right)^3-\left(\frac{1}{10}\right)^3=\left(3x-\frac{1}{10}\right)\left(9x^2+\frac{3}{10}x+\frac{1}{100}\right)\)
d) 125x3 - 1 =(5x)3 - 1 = (5x - 1)(25x2 + 5x + 1)
e) (x2 + 4)2 - 16x2
= (x2 + 4)2 - (4x)2
= (x2 - 4x + 4)(x2 + 4x + 4)
= (x - 2)2(x + 2)2
= [(x - 2)(x + 2)]2
a.\(\left(x+y\right)^2-2\left(x+y\right)+1\)
\(=\left(x+y\right)^2-2.\left(x+y\right).1+1^2\)
\(=\left[\left(x+y\right)-1\right]^2\)
\(=\left(x+y-1\right)^2\)
b.\(x^3+1-x^2-x\)
\(=\left(x^3-x^2\right)+\left(1-x\right)\)
\(=x^2\left(x-1\right)-\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2-1\right)\)
\(=\left(x-1\right)^2\left(x+1\right)\)