![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
x⁸ + x⁴ + 1
= x⁸ + 2x⁴ + 1 - x⁴
= (x⁴ + 1)² - x⁴
= (x⁴ + 1)² - (x²)²
= (x⁴ + 1 + x²)(x⁴ + 1 - x²)
= (x⁴ + x² + 1)(x⁴ - x² + 1)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(5x^2+5xy-x-y\)
\(=5x.\left(x+y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(5x-1\right)\)
b) \(5x^2-10y+5y^2-20z^2\)
\(=5.\left(x^2-2y+y^2-4z^2\right)\)
Đề sai ở đâu đó.
c) \(4x^2-y^2+4x+1\)
\(=\left(4x+4x^2+1\right)-y^2\)
\(=\left(2x+1\right)^2-y^2\)
\(=\left(2x+y+1\right)\left(2x-y+1\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
(x2 - x + 1)2 - 5x(x2 - x + 1) + 4x2
Đặt x2 - x + 1 = a
<=> a2 - 5xa + 4x2 = x2 - 4xa - xa + 4x2
= a(a - 4x) - x(a - 4x) = (a - x)(a - 4x)
= (x2 - x + 1 - x)(x2 - x + 1 - 4x)
= (x2 - 2x + 1)(x2 - 5x + 1) = (x - 1)2(x2 - 5x + 1)
Đặt x2 - x + 1 = y
đthức <=> y2 - 5xy + 4x2
= y2 - xy - 4xy + 4x2
= y( y - x ) - 4x( y - x )
= ( y - x )( y - 4x )
= ( x2 - x + 1 - x )( x2 - x + 1 - 4x )
= ( x2 - 2x + 1 )( x2 - 5x + 1 )
= ( x - 1 )2( x2 - 5x + 1 )
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
(2x – 3) . (x2 – 5x + 1)
= 2x. (x2 – 5x + 1) + (-3). (x2 – 5x + 1)
= 2x . x2 + 2x . (-5x) + 2x . 1 + (-3).x2 + (-3).(-5x) + (-3). 1
= 2x3 + (-10x2 ) + 2x + (-3x2) + 15x + (-3)
= 2x3 + (-10x2 + -3x2) + (2x + 15x) + (-3)
![](https://rs.olm.vn/images/avt/0.png?1311)
casio fx 570 thì ấn mode => 5 => 3 sau điền hệ số a;b;c
casio fx 580 thì ấn mode => 9 => 2 => 2 => điền hệ số a;b;c
có cả cách này à =)))
menu setup -> 9 -> 2 - > 2 (pt cần phân tích) -> nhập hệ số của pt vào từng biến thích hợp -> ''=''
VD : \(A=x^2+4x-5\)có nghiệm \(x_1=1;x_2=-5\)
vậy đa thức cần phân tích là : \(\left(x-1\right)\left(x+5\right)=x^2+5x-x-5\)
Vậy \(A=x^2+4x-5=x^2+5x-x-5=\left(x-1\right)\left(x+5\right)\)
tương tự nhé
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(a-b\right)^2-\left(b-a\right)\\ =\left(a-b\right)\left(a-b\right)+\left(a-b\right)\\ =\left(a-b\right)\left(a-b+1\right)\)
`HaNa☘D`
\(\left(a-b\right)^2-\left(b-a\right)\)
\(=\left(a-b\right)^2+\left(a-b\right)\)
\(=\left(a-b\right)\left(a-b+1\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
19) Ta có: \(-x^2-4x-4\)
\(=-\left(x^2+4x+4\right)\)
\(=-\left(x+2\right)^2\)
20) Ta có: \(-4x^2-12x-9\)
\(=-\left(4x^2+12x+9\right)\)
\(=-\left(2x+3\right)^2\)
21) Ta có: \(-4x^2-4x-1\)
\(=-\left(4x^2+4x+1\right)\)
\(=-\left(2x+1\right)^2\)
22) Ta có: \(-x^2+6x-9\)
\(=-\left(x^2-6x+9\right)\)
\(=-\left(x-3\right)^2\)
23) Ta có: \(-x^2+10x-25\)
\(=-\left(x^2-10x+25\right)\)
\(=-\left(x-5\right)^2\)
24) Ta có: \(-x^2+8x-16\)
\(=-\left(x^2-8x+16\right)\)
\(=-\left(x-4\right)^2\)
25) Ta có: \(-4x^2+12x-9\)
\(=-\left(4x^2-12x+9\right)\)
\(=-\left(2x-3\right)^2\)
26) Ta có: \(a^2-a+b-b^2\)
\(=\left(a-b\right)\left(a+b\right)-\left(a-b\right)\)
\(=\left(a-b\right)\left(a+b-1\right)\)
13) Ta có: \(y^2-2xy+2x-y\)
\(=y\left(y-2x\right)-\left(y-2x\right)\)
\(=\left(y-2x\right)\left(y-1\right)\)
14) Ta có: \(x-2xy+4y-2\)
\(=x\left(1-2y\right)-2\left(1-2y\right)\)
\(=\left(1-2y\right)\left(x-2\right)\)
15) Ta có: \(x^2-2xy+x-2y\)
\(=x\left(x-2y\right)+\left(x-2y\right)\)
\(=\left(x-2y\right)\left(x+1\right)\)
16) Ta có: \(xy-z-y+xz\)
\(=x\left(y+z\right)-\left(y+z\right)\)
\(=\left(y+z\right)\left(x-1\right)\)
17) Ta có: \(2xy+3z-6y-xz\)
\(=\left(2xy-xz\right)+\left(3z-6y\right)\)
\(=x\left(2y-z\right)-3\left(2y-z\right)\)
\(=\left(2y-z\right)\left(x-3\right)\)
18) Ta có: \(2xy-2z+4y-xz\)
\(=\left(2xy+4y\right)+\left(xz+2z\right)\)
\(=2y\left(x+2\right)+z\left(x+2\right)\)
\(=\left(x+2\right)\left(2y+z\right)\)
đặt a=x^2-5x
(x^2-5x)^2+10(x^2-5x+24)
=a^2+10(a+24)
=a^2+10a+24
=a^2+6a+4a+24
=a(a+6)+4(a+6)
=(a+6)(a+4)
=(x^2-5x+6)(x^2-5x+4)
đây nha bạn![undefined](https://i.imgur.com/129RNT7h.jpg)