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a.100x2-(x2+25)2=(10x)2-(x2+25)2=(10x-x2-25)(10x+x2+25)=(10x-x2-25)(x+5)2
b.1+(x-y+5)2-2(x-y+5)=(x-y+4)2
\(a,100x^2-\left(x^2+25\right)\)
\(=\left(10x-x^2+25\right)\left(10x+x^2+25\right)\)
\(b,1+\left(x-y+5\right)^2-2\left(x-y+5\right)^2\)
\(=\left(1-x+y-5\right)^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 3
a) x² + 10x + 25
= x² + 2.x.5 + 5²
= (x + 5)²
b) 8x - 16 - x²
= -(x² - 8x + 16)
= -(x² - 2.x.4 + 4²)
= -(x - 4)²
c) x³ + 3x² + 3x + 1
= x³ + 3.x².1 + 3.x.1² + 1³
= (x + 1)³
d) (x + y)² - 9x²
= (x + y)² - (3x)²
= (x + y - 3x)(x + y + 3x)
= (y - 2x)(4x + y)
e) (x + 5)² - (2x - 1)²
= (x + 5 - 2x + 1)(x + 5 + 2x - 1)
= (6 - x)(3x + 4)
Bài 4
a) x² - 9 = 0
x² = 9
x = 3 hoặc x = -3
b) (x - 4)² - 36 = 0
(x - 4 - 6)(x - 4 + 6) = 0
(x - 10)(x + 2) = 0
x - 10 = 0 hoặc x + 2 = 0
*) x - 10 = 0
x = 10
*) x + 2 = 0
x = -2
Vậy x = -2; x = 10
c) x² - 10x = -25
x² - 10x + 25 = 0
(x - 5)² = 0
x - 5 = 0
x = 5
d) x² + 5x + 6 = 0
x² + 2x + 3x + 6 = 0
(x² + 2x) + (3x + 6) = 0
x(x + 2) + 3(x + 2) = 0
(x + 2)(x + 3) = 0
x + 2 = 0 hoặc x + 3 = 0
*) x + 2 = 0
x = -2
*) x + 3 = 0
x = -3
Vậy x = -3; x = -2
![](https://rs.olm.vn/images/avt/0.png?1311)
c: \(=\dfrac{x^3+2x^2+x^2+2x-10x-20}{x+2}\)
\(=x^2+x-10\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Tải trên điện thoaaij về phần mềm PhotoMath thì bạn sẽ có đáp án và bài giải bài thực hiện phép tính này. Nếu thắc mắc về cánh sử dụng thì seach mạng.
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 3:
b. $B=(x+y)(2x-y)+(xy^4-x^2y^2):(xy^2)$
$=(2x^2-xy+2xy-y^2)+(y^2-x)$
$=2x^2+xy-y^2+y^2-x=2x^2+xy-x$
Bài 4:
a. $25x^3-10x^2+x=x(25x^2-10x+1)=x(5x-1)^2$
b. $x^2-9x+9y-y^2=(x^2-y^2)-(9x-9y)=(x-y)(x+y)-9(x-y)=(x-y)(x+y-9)$
c. $16-x^2-4y^2-4xy=16-(x^2+4y^2+4xy)$
$=4^2-(x+2y)^2=(4-x-2y)(4+x+2y)$
![](https://rs.olm.vn/images/avt/0.png?1311)
a.
$12x^3y-24x^2y^2+12xy^3=12xy(x^2-2xy+y^2)=12xy(x-y)^2$
b.
$x^2-6x+xy-6y=(x^2+xy)-(6x+6y)=x(x+y)-6(x+y)=(x-6)(x+y)$
c.
$2x^2+2xy-x-y=2x(x+y)-(x+y)=(x+y)(2x-1)$
d.
$x^3-3x^2+3x-1=(x-1)^3$
e.
$3x^2-3y^2-12x-12y=(3x^2-3y^2)-(12x+12y)$
$=3(x-y)(x+y)-12(x+y)=(x+y)[3(x-y)-12]=3(x-y)(x-y-4)$
f.
$x^2-2xy-x^2+4y^2=4y^2-2xy=2y(2y-x)$
a) \(100x^2-\left(x^2+25\right)^2\)
\(=\left(10x\right)^2-\left(x^2+25\right)^2\)
\(=\left(10x+x^2+25\right)\left(10x-x^2-25\right)\)
\(=\left(x+5\right)^2\left(10x-x^2-25\right)\)
b) \(1+\left(x-y+5\right)^2-2\left(x-y+5\right)\)
\(=\left[\left(x-y+5\right)-1\right]^2\)
\(=\left(x-y+4\right)^2\)
c) \(\left(x^2+4y^2-5\right)^2-16\left(x^2y^2+2xy+1\right)\)
\(=\left(x^2+4y^2-5\right)^2-4^2\left(xy+1\right)^2\)
\(=\left[\left(x^2+4y^2-5\right)-4\left(xy+1\right)\right]\left[\left(x^2+4y^2-5\right)+4\left(xy+1\right)\right]\)
\(=\left(x^2+4y^2-5-4xy-4\right)\left(x^2+4y^2-5+4xy+4\right)\)
\(=\left(x^2-4xy+4y^2-9\right)\left(x^2+4xy+4y^2-1\right)\)
\(=\left[\left(x-2y\right)^2-3^2\right]\left[\left(x+2y\right)^2-1^2\right]\)
\(=\left(x-2y-3\right)\left(x-2y+3\right)\left(x+2y-1\right)\left(x+2y+1\right)\)
d) \(\left(x^2+8x-34\right)^2-\left(3x^2-8x-2\right)^2\)
\(=\left[\left(x^2+8x-34\right)-\left(3x^2-8x-2\right)\right]\left[\left(x^2+8x-34\right)+\left(3x^2-8x-2\right)\right]\)
\(=\left(x^2+8x-34-3x^2+8x+2\right)\left(x^2+8x-34+3x^2-8x-2\right)\)
\(=\left(-2x^2+16x-32\right)\left(4x^2-36\right)\)
\(=-2\left(x^2-8x+16\right)\left[\left(2x\right)^2-6^2\right]\)
\(=-2\left(x-4\right)^2\left(2x-6\right)\left(2x+6\right)\)
\(=-2\left(x-4\right)^24\left(x-3\right)\left(x+3\right)\)
\(=8\left(x-4\right)^2\left(x-3\right)\left(x+3\right)\)