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á,(x+7)2-y2
= (x+7+y)(x+7-y)
b, (x+1)(x-25)
chúc bạn luôn thành công trong cuộc sống nha
a)
\(x^2-y^2+14x+49\)
\(=\left(x^2+14x+49\right)-y^2\)
\(=\left(x^2+2.7.x+7^2\right)-y^2\)
\(=\left(x+7\right)^2-y^2\)
\(=\left(x+y+7\right)\left(x-y+7\right)\)
Vậy ...
Dễ thì tự mà làm đi ??? ( Đừng có ném gạch )
==
#Thiên_Hy
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a, \(16x^3+54y^3=2\left(8x^3+27y^3\right)=2\left(2x+3y\right)\left(4x^2-12xy+9y^2\right)\)
b, \(5x^2\left(x-1\right)+10xy\left(x-1\right)-5y^2\left(1-x\right)\)
\(=\left(5x^2+10xy+5y^2\right)\left(x-1\right)=5\left(x^2+2xy+y^2\right)\left(x-1\right)=5\left(x+1\right)^2\left(x-1\right)\)
bổ sung phần a hộ mình
\(=2\left(2x+3y\right)\left(4x^2-12xy+9y^2\right)=2\left(2x+3y\right)\left(2x-3y\right)^2\)
- 9y3x – 36y2x= 9xy(y2–4) =9xy(y–4)(y+4)
- 64-y^2 – x^2 – 2xy = 64– (x^2 + 2xy + y^2) = 82 – (x+y)2 = (8 – x –y)(8+x+y)
- x^2 + x – 30= x^2 + x – 25 – 5 = (x2 – 25)(x – 5)= (x-5)(x+5)(x-5)= (x-5)^2 (x+5)
a) 4x2 - 20x + 25 - 36y2
= (2x - 5)2 - 36y2
= (2x - 5 - 6y)(2x - 5 + 6y)
b) x3 + x2 - 2x - 8
= (x3 - 8) + (x2 - 2x)
= (x - 2)(x2 + 2x + 4) + x(x - 2)
= (x - 2)(x2 + 2x + 4 + x)
= (x - 2)(x2 + 3x + 4)
d) x4 + 6x3 + 9x2 - 16
= x2(x2 + 6x + 9) - 16
= x2(x + 3)2 - 16
= (x2 + 3x)2 - 16
= (x2 + 3x - 4)(x2 + 3x + 4)
= (x2 + 4x - x - 4)(x2 + 3x + 4)
= [x(x + 4) - (x + 4)](x2 + 3x + 4)
= (x - 1)(x + 4)(x2 + 3x + 4)
\(\left(x-3\right)\left(x-5\right)\left(x-6\right)\left(x-10\right)-24x^2\)
\(=\left(x^2-13x+30\right)\left(x^2-11x+30\right)-24x^2\)
Đặt \(t=x^2-11x+30\)
\(\Rightarrow\left(x-3\right)\left(x-5\right)\left(x-6\right)\left(x-10\right)-24x^2\)
\(=t.\left(t-2x\right)-24x^2\)
\(=t^2-2xt-24x^2\)
\(=\left(t^2-2xt+x^2\right)-25x^2\)
\(=\left(t-x\right)-\left(5x\right)^2\)
\(=\left(t-6x\right)\left(t+4x\right)\)
\(=\left(x^2-17x+30\right)\left(x^2-7x+30\right)\)
Tham khảo nhé~
\(=6\left(4x^2-9y^2+6y-1\right)\)
\(=6\left[\left(2x\right)^2-\left(3y-1\right)^2\right]\)
\(=6\left(2x-3y+1\right)\left(2x+3y-1\right)\)
\(24x^2+36y-54y^2-6\)
\(=6\left(4x^2+6y-9y^2-1\right)\)
\(=6\left[4x^2-\left(9y^2-6y+1\right)\right]\)
\(=6\cdot\left[\left(2x\right)^2-\left(3y-1\right)^2\right]\)
\(=6\left(2x-3y+1\right)\left(2x+3y-1\right)\)