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4x4-32x2+1
=4x4+12x3+2x2-12x3-36x2-6x+2x2+6x+1
=2x2.(2x2+6x+1)-6x.(2x2+6x+1)+(2x2+6x+1)
=(2x2+6x+1)(2x2- 6x+1)
= 4x4 - 4x2 + 1 - 28x2 = [(2x2)2 - 2.2x2 .1 + 12 ] - 28x2 = (2x2 - 1)2 - (\(\sqrt{28}\).x)2
= (2x2 - 1 - \(\sqrt{28}\)x) .(2x2 -1 + \(\sqrt{28}\)x) = (2x2 - 2\(\sqrt{7}\)x - 1). (2x2 + 2\(\sqrt{7}\)x -1)
\(8x+12y-32x^2y^2\)
\(=4\left(2x+3y-8x^2y^2\right)\)
thế này thì đã đúng chưa?
Đề sai nhé .Sửu lại
\(x^2-4x^2y^2+4+4x\)
\(=\left(x^2+4x+4\right)-4x^2y^2\)
\(=\left(x+2\right)^2-\left(2xy\right)^2\)
\(=\left(x+2+2xy\right)\left(x+2-2xy\right)\)
12x2 - 32x2 +25x -6
= x ( 12x - 32x + 25 - 6 )
= x ( -20x + 19 )
bị sai đề đúng không?????????
k cho mk mấy cái đi a...hi..hi...
\(\left(1+x^2\right)^2-4x\left(1-x^2\right)\)
\(\Leftrightarrow\left(1+x^2\right)^2+4x\left(1+x^2\right)\)
\(\Leftrightarrow\left(1+x^2\right)\times\left[\left(1+x^2\right)+4\right]\)
( 1+x2 )2 -4x( 1- x2 )
=x4+2x2+1-4x+4x3
=x3+2x2-x+2x3+4x2-2x-x2-2x+1
=x(x2+2x-1)+2x(x2+2x-1)-(x2+2x-1)
=(x2+2x-1)(x2+2x-1)
=(x2+2x-1)2
\(\left(1+x\right)^2-4x\left(1-x^2\right)\)
\(=\left(1+x\right)^2-4x\left(1-x\right)\left(1+x\right)\)
\(=\left(1+x\right)\left(1+x-4\left(1-x\right)\right)\)
\(=\left(1+x\right)\left(1+x-4+4x\right)\)
\(=\left(1+x\right)\left(5x-3\right)\)
\(x^8+3x^4+4\)
\(=\left(x^8-x^6+2x^4\right)+\left(x^6-x^4+2x^2\right)+\left(2x^4-2x^2+4\right)\)
\(=x^4\left(x^4-x^2+2\right)+x^2\left(x^4-x^2+2\right)+2\left(x^4-x^2+2\right)\)
\(=\left(x^4+x^2+2\right)\left(x^4-x^2+2\right)\)
\(4x^4+4x^3+5x^2+2x+1\)
\(=\left(4x^4+2x^3+2x^2\right)+\left(2x^3+x^2+x\right)+\left(2x^2+x+1\right)\)
\(=2x^2\left(2x^2+x+1\right)+x\left(2x^2+x+1\right)+\left(2x^2+x+1\right)\)
\(=\left(2x^2+x+1\right)^2\)
\(\left(2x^2+6x+1\right)^2-26x^2\left(2x^2+6x+1\right)+26x^2\left(2x^2+6x+1\right)-26x^2\)
\(-26x^2\left\{\left(2x^2+6x+1\right)^2-\left(2x^2+6x+1\right)+\left(2x^2+6x+1\right)\right\}\)
\(-26x^2\left\{\left(2x^2+6x+1\right)\left(2x^2+6x+1\right)-1+1\right\}\)
\(-26x^2\left(2x^2+6x+1\right)\left(2x^2+6x+1\right)\)
\(-26x^2\left(2x^2+6x+1\right)^2\)
nhầm là 4x4 nha