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\(x^8+3x^4+4\)
\(=\left(x^8-x^6+2x^4\right)+\left(x^6-x^4+2x^2\right)+\left(2x^4-2x^2+4\right)\)
\(=x^4\left(x^4-x^2+2\right)+x^2\left(x^4-x^2+2\right)+2\left(x^4-x^2+2\right)\)
\(=\left(x^4+x^2+2\right)\left(x^4-x^2+2\right)\)
\(4x^4+4x^3+5x^2+2x+1\)
\(=\left(4x^4+2x^3+2x^2\right)+\left(2x^3+x^2+x\right)+\left(2x^2+x+1\right)\)
\(=2x^2\left(2x^2+x+1\right)+x\left(2x^2+x+1\right)+\left(2x^2+x+1\right)\)
\(=\left(2x^2+x+1\right)^2\)
\(1,\)\(3x-3y-x^2+2xy-y^2\)
\(=-\left(x^2-2xy+y^2\right)+\left(3x-3y\right)\)
\(=-\left(x-y\right)^2+3\left(x-y\right)\)
\(=\left(x-y\right)\left[-\left(x-y\right)+3\right]\)
\(=\left(x-y\right)\left(3-x+y\right)\)
\(2,\)\(49\left(x-4\right)^2-9\left(x+2\right)^2\)
\(=\left[7\left(x-4\right)\right]^2-\left[3\left(x-2\right)\right]^2\)
\(=\left(7x-28-3x+6\right)\left(7x-28+3x-6\right)\)
\(=\left(4x-22\right)\left(10x+34\right)\)
\(3,\)\(x^4+4x^2-5\)
\(=x^4-x^2+5x^2-5\)
\(=x^2\left(x^2-1\right)+5\left(x^2-1\right)\)
\(=\left(x^2+5\right)\left(x-1\right)\left(x+1\right)\)
a, \(\left(x+1\right)^2-25=\left(x+1-5\right)\left(x+1+5\right)=\left(x-4\right)\left(x+6\right)\)
b, \(\left(xy+4\right)^2-4\left(x+y\right)^2=\left(xy+4\right)^2-\left(2x+2y\right)^2=\left(xy+4-2x-2y\right)\left(xy+4+2x+2y\right)\)
c, xem lại đề nhé
bài 1, bạn tự làm nhé đặt chia đi bạn
bài 2
a,\(\left(x^2-2xy+y^2\right)+2\left(x-y\right)=\left(x-y\right)^2+2\left(x-y\right)=\left(x-y\right)\left(x-y+2\right)\)
\(b,=a^2-2a-5a+10=a\left(a-2\right)-5\left(a-2\right)=\left(a-2\right)\left(a-5\right)\)
x^4 + 3x^2 - 4x - 12
= x^3 (x -2) + 3(x - 2)(x +2) +2x(x -2)
=(x -2)(x^3 + 3x + 6 + 2x)
= (x -2)(x^3 + 5x + 6 )
= (x - 2)(x^3 + x^2 -x^2 - x + 6x + 6)
= (x -2)[x^2(x+1) -x(x+1)+6(x+1)]
=(x-2)(x+1)(x^2-x+6)
\(\left(x+3\right)^2-16\)
\(=\left(x+3-4\right)\left(x+3+4\right)\)
\(=\left(x-1\right)\left(x+7\right)\)
\(49\left(x-4\right)^2-9\left(x+2\right)^2\)
\(=\left(7x-28\right)^2-\left(3x+6\right)^2\)
\(=\left(7x-28-3x-6\right)\left(7x-28+3x+6\right)\)
\(=\left(4x-34\right)\left(10x-22\right)\)
\(=4\left(2x-17\right)\left(5x-11\right)\)
cảm ơn nha