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Phân tích đa thức thành nhân tử:
a) \(xy+y^2-x-y=y\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(y-1\right)\)
b) \(25-x^2+4xy-4y^2=25-\left(x^2-4xy+4y^2\right)=25-\left(x-2y\right)^2\)
\(=\left(5-x+2y\right)\left(5+x-2y\right)\)
Rút gọn biểu thức;
\(A=\left(6x+1\right)^2+\left(3x-1\right)^2-2\left(3x-1\right)\left(6x+1\right)\)
\(=\left[\left(6x+1\right)-\left(3x-1\right)\right]^2=\left(6x+1-3x+1\right)=\left(3x+2\right)^2\)
Tìm a để đa thức.. Bạn chia cột dọ thì da
\(xy+y^2-x-y=\left(xy+y^2\right)-\left(x+y\right)=y\left(x+y\right)-\left(x+y\right)=\left(y-1\right)\left(x+y\right)\)b)\(25-\left(x^2-4xy+4y^2\right)=5^2-\left(x-2y\right)^2=\left(x-2y+5\right)\left(5-x+2y\right)\)
2x2-3xy-2y2=2x2+xy-4xy-2y2=(2x2+xy)-(4xy+2y2)=x(2x+y)-2y(2x+y)=(2x+y)(x-2y)
a) x2 - 2xy - 4 + y2
= (x - y)2 - 22
= (x - y - 2)(x - y + 2)
b) x2 + y2 - 1 - 2xy
= (x - y)2 - 12
= (x - y - 1)(x - y + 1)
c) 25 - x2 + 4xy - 4y2
= 52 - (x - 2y)2
= (5 - x + 2y)(5 + x - 2y)
B1 :
a, B = (x+1)^2+(y-2)^2 = (99+1)^2+(102-2)^2 = 100^2+100^2 = 20000
b, = (2x^2+16x+32)-2y^2
= 2.(x+4)^2-2y^2
= 2.[(x+4)^2-y^2] = 2.(x+4-y).(x+4+y)
c, <=> (x^2-3x)+(2x-6) = 0
<=> (x-3).(x+2) = 0
<=> x-3=0 hoặc x+2=0
<=> x=3 hoặc x=-2
B2 :
P = (3-x).(x+3)/x.(x-3) = -(x+3)/x = -x-3/x
k mk nha
Bai 1
a)B=(x+1)2+(y-2)2
Voi x=99,y=102
=>B= 1002+1002
=20000
b)\(2x^2-2y^2+16x+32\)
=\(2\left[\left(x^2+8x+16\right)-y^2\right]\)
=\(2\left[\left(x+4\right)^2-y^2\right]\)
=2(x-y+4)(x+y+4)
c)\(x^2-3x+2x-6=0\)
=>x(x-3)+2(x-3)=0
=>(x-3)(x+2)=0
=>x=-2;3
Bai 2
\(P=\frac{9-x^2}{x^2-3x}\)
=\(-\frac{x^2-9}{x\left(x-3\right)}\)
=\(-\frac{\left(x-3\right)\left(x+3\right)}{x\left(x-3\right)}\)
=\(\frac{-x-3}{x}\)
Rút gọn
\(\left(2x+1\right)\left(4x^2-3x+1\right)+\left(2x-1\right)\left(4x^2+3x+1\right)\)
\(=8x^3-12x^2+2x+4x^2-3x+1+8x^3+12x^2+2x-4x^2-3x-1\)
\(=16x^3-2x\)
Phân tích đa thức thnahf nhân tử
\(4y^2+16y-x^2-8x\)
\(=\left(4y^2-x^2\right)+\left(16y-8x\right)\)
\(=\left(2y-x\right)\left(2y+x\right)+8\left(2y-x\right)\)
\(=\left(2y-x\right)\left(2y+x+8\right)\)
Chứng minh .............
Có: \(x^2+x+1=\left(x^2+2\cdot x\cdot\frac{1}{2}+\frac{1}{4}\right)+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\)
Vì: \(\left(x+\frac{1}{2}\right)^2\ge0\)
=> \(\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\)
Kết luận......
a) \(x^2+5x+6=x^2+2x+3x+6=x\left(x+2\right)+3\left(x+2\right)=\left(x+3\right)\left(x+2\right)\)
b) \(x^2-4x+3=x^2-x-3x+3=x\left(x-1\right)-3\left(x-1\right)=\left(x-3\right)\left(x-1\right)\)
c) \(x^2+5x+4=x^2+x+4x+4=x\left(x+1\right)+4\left(x+1\right)=\left(x+4\right)\left(x+1\right)\)
d) \(x^2-x-6=x^2+2x-3x-6=x\left(x+2\right)-3\left(x+2\right)=\left(x-3\right)\left(x+2\right)\)
\(x^4+x^3+2x^2+x+1\)
\(=\left(x^4+2x^2+1\right)+\left(x^3+x\right)\)
\(=\left(x^2+1\right)^2+x\left(x^2+1\right)\)
\(=\left(x^2+1\right)\left(x^2+1+x\right)\)
x^4+x^3+2x^2+x+1
=(x^4+2x^2+1)+(x^3+x)
=(x^2+1)^2+x(x^2+1)
=(x^2+1)(x^2+x+1)
a)\(5x^2-4\left(x^2-2x+1\right)-5=5\left(x^2-1\right)-4\left(x-1\right)^2=5\left(x-1\right)\left(x+1\right)-4\left(x-1\right)^2=\left(x-1\right)\left(5x+5-4x+4\right)=\left(x-1\right)\left(x+9\right)\)
b) \(9x^2+6x-4y^2+4y=\left(9x^2+6x+1\right)-\left(4y^2-4y+1\right)=\left(3x+1\right)^2-\left(2y-1\right)^2=\left(3x+1-2y+1\right)\left(3x+1+2y-1\right)=\left(3x-2y+2\right)\left(3x+2y\right)\)
a: \(5x^2-4\left(x^2-2x+1\right)-5\)
\(=5x^2-4x^2+8x-4-5\)
\(=x^2+8x-9\)
\(=\left(x+9\right)\left(x-1\right)\)
b: \(9x^2+6x-4y^2+4y\)
\(=\left(3x+2y\right)\left(3x-2y\right)+2\left(3x+2y\right)\)
\(=\left(3x+2y\right)\left(3x-2y+2\right)\)