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x^4+2008x^2+2007x+2008
=x^4+2008x^2+2008x-x+2008
=(x^4-x)+(2008x^2+2008x+2008)
=x(x^3-1)+2008(x^2+x+1)
=x(x-1)(x^2+x+1)+2008(x^2+x+1)
=(x^2+x+1)(x^2-x+2008)
x4+2008x2+2007x+2008
<=> x4-x+2008x2+2008x+2008
<=> x(x3-1)+2008(x2+x+1)
<=> x(x-1)(x2+x+1)+2008(x2+x+1)
<=> (x2+x+1)(x2-x+2008)
\(\left(x^4+x^2+1\right)+\left(2007x^2+2007x+2007\right)\)
=\(\left(x^2+x+1\right)\left(x^2-x+1\right)+2007\left(x^2+x+1\right)\)
=\(\left(x^2+x+1\right)\left(x^2-x+2008\right)\)
=x4+2008x2+2008x-x+2008
=(x4-x)+(2008x2+2008x+2008)
=x(x3-1)+2008(x2+x+1)
=x(x-1)(x2+x+1)+2008(x2+x+1)
=(x2+x++1)(x2-x+2008)
a) \(x^8+x+1\)
\(=x^8+x^7+x^6-x^7-x^6-x^5+x^5+x^4+x^3-x^4-x^3-x^2+x^2+x+1\)
\(=x^6\left(x^2+x+1\right)-x^5\left(x^2+x+1\right)+x^3\left(x^2+x+1\right)-x^2\left(x^2+x+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^6-x^5+x^3-x^2+1\right)\)
b) \(x^4+2008x^2+2007x+2008\)
\(=x^4+x^3+x^2-x^3-x^2-x+2008x^2+2008x+2008\)
\(=x^2\left(x^2+x+1\right)-x\left(x^2+x+1\right)+2008\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+2008\right)\)
1) tách 2008x^2 thành 2007x^2 + x rồi nhóm mấy cái có hệ số là 2007 thành 1 nhóm
b) áp dụng hằng đẳng thức
2) từ a+b+c => a+c = -b
mà a-b/c +1 = a+c-b/c
làm tương tự với mấy phân số còn lại là ra dpcm
\(A=x^5-x^4+3x^3+3x^2-x+1\)
\(A=\left(x^5+x^4\right)+\left(-2x^4-2x^3\right)+\left(5x^3+5x^2\right)+\left(-2x^2-2x\right)+\left(x+1\right)\)
\(A=x^4\left(x+1\right)-2x^3\left(x+1\right)+5x^2\left(x+1\right)-2x\left(x+1\right)+\left(x+1\right)\)
\(A=\left(x+1\right)\left(x^4+2x^3+5x^2-2x+1\right)\)
a) \(x^2+4\)
\(=x^2+4+4x-4x\)
\(=\left(x^2+2.x.2+2^2\right)-4x\)
\(=\left(x+2\right)^2-\left(2\sqrt{x}\right)^2\)
\(=\left(x+2-2\sqrt{x}\right)\left(x+2+2\sqrt{x}\right)\)
c) \(x^2+7x+6\)
\(=x^2+x+6x+6\)
\(=\left(x^2+x\right)+\left(6x+6\right)\)
\(=x\left(x+1\right)+6\left(x+1\right)\)
\(=\left(x+1\right)\left(x+6\right)\)
d) \(x^4+2008x^2+2007x+2008\)
\(=x^4+2008x^2+2008x-x+2008\)
\(=\left(x^4-x\right)+\left(2008x^2+2008x+2008\right)\)
\(=x\left(x^3-1\right)+2008\left(x^2+x+1\right)\)
\(=x\left(x-1\right)\left(x^2+x+1\right)+2008\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left[x\left(x-1\right)+2008\right]\)
\(=\left(x^2+x+1\right)\left(x^2-x+2008\right)\)
a)x4+4
=x4+4x2+4-4x2
=(x4+4x2+4)-4x2
=(x2+2)2-4x2
=(x2+2-2x)(x2+2+2x)
b)(x+2)(x+3)(x+4)(x+5)-24
=[(x+2)(x+5)][x+3)(x+4)]-24
=(x2+5x+2x+10)(x2+4x+3x+12)-24
=(x2+7x+10)(x2+7x+12)-24
=Đặt x2+7x+10=a ta có
a(a+2)-24
=a2+2a-24
=a2+6a-4a-24
=(a2+6a)-(4a+24)
=a(a+6)-4(a+6)
=(a+6)(a-4)
thay a=x2+7x+10
(x2+7x+10+6)(x2+7x+10-4)
=(x2+7x+16)(x2+7x+6)
=(x2+7x+16)(x2+x+6x+6)
=(x2+7x+16)[(x2+x)+(6x+6)]
=(x2+7x+16)[x(x+1)+6(x+1)]
=(x2+7x+16)(x+1)(x+6)
c)x2+7x+6
=x2+x+6x+6
=(x2+x)+(6x+6)
=x(x+1)+6(x+1)
=(x+1)(x+6)
taị sao bn cứ thick sửa đề của mk vậy
mk thích thì mk sửa thôi Bao Than Đen
ko xửa thì làm mãi ko ra