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a)4x3y-6xy2
=2xy(2x2-3y)
b)4x2-4x+1
=(2x)2-2*2x*1+12
=(2x-1)2
c)x2-2xy-3x+6y
=x(x-2y)-3(x-2y)
=(x-3)(x-2y)
d)x3-2x2+x-xy2
=x(x2-2x+1-y2)
=x[(x-1)2-y2]
=x(x-y-1)(x+y-1)
e)x2-x+y2-y-x2y2+xy
=xy2-x+y2-y-x2y2+x2-xy2+xy
=(xy2-x+y2-y)-x(xy2-x+y2-y)
=(1-x)(xy2-x+y2-y)
=(1-x)[xy2+xy+y2-(xy+y+x)]
=(1-x)[y(xy+y+x)-(xy+y+x)]
=(1-x)(y-1)(xy+y+x)
Bài 2:
a)x(x-y)+y(y-x)
=x2-xy+y2-xy
=(x-y)2.Tại x=53 và y=3 ta có:
N=(53-3)2=502=2500
b) x2013-53x2012+103x2011-51x2010
=x2010(x3-53x2+103x-51)
=x2010[x3-2x2+x-51x2+102x-51]
=x2010[x(x2-2x+1)-51(x2-2x+1)]
=x2010(x-51)(x2-2x+1).Tại x=51 ta có:
M=512010(51-51)(512-2*51+1)=0
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\(b.x^4+4x^2-5=x^4-x^2+5x^2-5\)
\(=x^2\left(x^2-1\right)+5\left(x^2-1\right)\)
\(=\left(x^2+5\right)\left(x^2-1\right)\)
\(=\left(x^2+5\right)\left(x-1\right)\left(x+1\right)\)
\(c.x^3-19x-30=x^3-25x+6x-30\)
\(=x\left(x-5\right)\left(x+5\right)+6\left(x-5\right)\)
\(=\left(x-5\right)\left(x^2+5x+6\right)\)
\(=\left(x-5\right)\left(x^2+2x+3x+6\right)\)
\(=\left(x-5\right)\left[x\left(x+2\right)+3\left(x+2\right)\right]\)
\(=\left(x-5\right)\left(x+2\right)\left(x+3\right)\)
\(x^2+4x+3\)
\(=\left(x+1\right)\left(x+3\right)\)
\(2x^2+3x-5\)
\(\left(x-1\right)\left(x+\frac{5}{2}\right)\)
\(x^2\left(x+1\right)-\left(x+1\right)\left(3x+1\right)+7x-x^2\)
\(=x^3+x^2-3x^2-4x-1+7x-x^2\)
\(=x^3-3x^2+3x-1\)
\(=\left(x-1\right)^3\)
a)\(8+\left(4x+3\right)^3\)
\(\Leftrightarrow2^3+\left(4x+3\right)^3\)
\(\Leftrightarrow\left(2+4x+3\right)\left[2^2-2.\left(4x+3\right)+\left(4x+3\right)^2\right]\)
\(\Leftrightarrow\left(5+4x\right)\left[4-8x-6+16x^2+24x+9\right]\)
\(\Leftrightarrow\left(5+4x\right)\left(16x^2+16x+7\right)\)
b)\(81-\left(9-x\right)^2\)
\(\Leftrightarrow9^2-\left(9-x\right)^2\)
\(\Leftrightarrow\left(9-9+x\right)\left(9+9-x\right)\)
\(\Leftrightarrow x\left(18-x\right)\)
Bài 1 :
\(A=\left(x-1\right)\left(x-2\right)\left(x+7\right)\left(x+8\right)+8\)
\(A=\left[\left(x-1\right)\left(x+7\right)\right]\left[\left(x-2\right)\left(x+8\right)\right]+8\)
\(A=\left(x^2+6x-7\right)\left(x^2+6x-16\right)+8\)
Đặt \(a=x^2+6x-7\)
\(A=a\left(a-9\right)+8\)
\(A=a^2-9a+8\)
\(A=a^2-8a-a+8\)
\(A=a\left(a-8\right)-\left(a-8\right)\)
\(A=\left(a-8\right)\left(a-1\right)\)
Thay a vào là xong bạn :)