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a) \(36-12x+x^2\) \(=6^2-2.6.x+x^2\)
\(=\left(6-x\right)^2\)
b) \(4x^2+12x+9=\left(2x\right)^2+2.2x.3+3^2\)
\(=\left(2x+3\right)^2\)
c) \(-25x^6-y^8+10x^3y^4=-\left[25x^6-10x^3y^4+y^8\right]\)
\(=-\left[\left(5x^3\right)^2-2.5x^3.y^4+\left(y^4\right)^2\right]\)
\(=-\left(5x^3-y^4\right)^2\)
d) \(\dfrac{1}{4}x^2-5xy+25y^2=\left(\dfrac{1}{2}x\right)^2-2.\dfrac{1}{2}x.5y+\left(5y\right)^2\)
\(=\left(\dfrac{1}{2}x-5y\right)^2\)
Học tốt~~~
a. \(36-12x+x^2=6^2-2.6.x+x^2=\left(6-x\right)^2\)
b. \(4x^2+12x+9=\left(2x\right)^2+2.2x.3+3^2=\left(2x+3\right)^2\)
c: \(=-\left(25x^6-10x^3y^4+y^8\right)\)
\(=-\left(5x^3-y^4\right)^2\)
d: \(=\left(\dfrac{1}{2}x\right)^2-2\cdot\dfrac{1}{2}x\cdot5y+\left(5y\right)^2=\left(\dfrac{1}{2}x-5y\right)^2\)
7, 4x mũ 2 - 12x + 9 - y mũ 2 = -(y-2x+3) (y+2x-3)
8, 16x mũ 2 - 4y mũ 2 + 4y - 1 = -(2y - 4x - 1) (2y+4x-1)
9, 25 - x mũ 2 - 12x - 36 = -(x+1) (x+11)
10, x mũ 2 - 9 - 5 ( x + 3 ) = (x-8) (x+3)
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Câu hỏi của Phạm Đỗ Bảo Ngọc - Toán lớp 8 - Học trực tuyến OLM
6, \(x^2-1+2xy+y^2=\left(x+y\right)^2-1=\left(x+y-1\right)\left(x+y+1\right)\)
7, \(4x^2-12x+9-y^2=\left(2x-3\right)^2-y^2=\left(2x-3-y\right)\left(2x-3+y\right)\)
8, \(16x^2-4y^2+4y-1=16x^2-\left(2y-1\right)^2=\left(4x-2y+1\right)\left(4x+2y-1\right)\)
9, \(25-x^2-12x-36=25-\left(x+6\right)^2=\left(5-x-6\right)\left(5+x+5\right)=-\left(x+1\right)\left(x+10\right)\)
10, \(x^2-9-5\left(x+3\right)=\left(x-3\right)\left(x+3\right)-5\left(x+3\right)=\left(x+3\right)\left(x-8\right)\)
1, \(x^2\left(x-3\right)-4x+12=x^2\left(x-3\right)-4\left(x-3\right)\)
\(=\left(x-2\right)\left(x+2\right)\left(x-3\right)\)
2, \(2a\left(x+y\right)-x-y=2a\left(x+y\right)-\left(x+y\right)=\left(2a-1\right)\left(x+y\right)\)
3, \(2x-4+5x^2-10x=2\left(x-2\right)+5x\left(x-2\right)=\left(2+5x\right)\left(x-2\right)\)
4, sửa đề :
\(6x^2-12x-7x+14=6x\left(x-2\right)-7\left(x-2\right)=\left(6x-7\right)\left(x-2\right)\)
5, \(xy-y^2-3x+3y=y\left(x-y\right)-3\left(x-y\right)=\left(y-3\right)\left(x-y\right)\)
a) x2(x-3)-4x+12
=x2(x-3)-4(x-3)
=(x-3)(x2-4)
=(x-3)(x-2)(x+2)
b) 2a(x+y)-x-y
=2a(x+y)-(x+y)
=(x+y)(2a-1)
c) 2x-4+5x2-10x
=2(x-2)+5x(x-2)
=(x-2)(2+5x)
d) 5x2-12x-7x+14
=5x2-19x+14
e) xy-y2-3x+3y
=y(x-y)-3(x-y)
=(x-y)(y-3)
#H
1) 3x2y+6xy+3y= 3y.(x2+2x+1) = 3y.(x+1)2
2) 12x-4x2-9+a2 = a2-(4x2-12x+9)= a2-(2x-3)2= (a+2x-3).(a-2x+3)
3) x3-7x-6 = x3-2x2+2x2-4x-3x+6 = x2.(x-2)+2x.(x-2)-3.(x-2)= (x-2).(x2+2x-3) = (x-2).(x2+x-3x-3)= (x-2).(x+1).(x-3)
a: \(x^2+12x+36=0\)
=>\(x^2+2\cdot x\cdot6+6^2=0\)
=>\(\left(x+6\right)^2=0\)
=>x+6=0
=>x=-6
b: \(4x^2-4x+1=0\)
=>\(\left(2x\right)^2-2\cdot2x\cdot1+1^2=0\)
=>\(\left(2x-1\right)^2=0\)
=>2x-1=0
=>2x=1
=>x=1/2
c: \(x^3+6x^2+12x+8=0\)
=>\(x^3+3\cdot x^2\cdot2+3\cdot x\cdot2^2+2^3=0\)
=>\(\left(x+2\right)^3=0\)
=>x+2=0
=>x=-2
Giải:
a) \(36-12x+x^2\)
\(=6^2-2.6.x+x^2\)
\(=\left(6-x\right)^2\)
b) \(4x^2+12x+9\)
\(=\left(2x\right)^2+2.3.2x+3^2\)
\(=\left(2x+3\right)^2\)
c) \(-25x^6-y^8+10x^3y^4\)
\(=-\left(25x^6+y^8-10x^3y^4\right)\)
\(=-\left(5x^3-y^4\right)^2\)
d) \(\dfrac{1}{4}x^2-5xy+25y^2\)
\(=\left(\dfrac{1}{2}x\right)^2-2.\dfrac{1}{2}x.5y+\left(5y\right)^2\)
\(=\left(\dfrac{1}{2}x-5y\right)^2\)
Vậy ...
a, 36-12x+x2=62-2.6.x+x2=(6-x)2
b, 4x2+12x+9=(2x)2+2.2x.3+32=(2x+3)2
c, -256-y8+10x3y4=-(256+y8-10x3y4=-(53-y4)2
d, 1/4x2-5xy+25y2=(1/2x)2-2.1/2x.5y+(5y)2=(1/2-5y)2