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\(a,3x^2+2x-1\)
\(\Leftrightarrow3x^2+3x-x-1\)
\(\Leftrightarrow3x\left(x+1\right)-\left(x+1\right)\)
\(\Leftrightarrow\left(x+1\right)\left(3x-1\right)\)
\(b,x^3+6x^2+11x+6\)
\(\Leftrightarrow x^3+3x^2+3x^2+9x+2x+6\)
\(\Leftrightarrow x^2\left(x+3\right)+3x\left(x+3\right)+2\left(x+6\right)\)
\(\Leftrightarrow\left(x+3\right)\left(x^2+3x+2\right)\)
\(\Leftrightarrow\left(x+3\right)\left(x+1\right)\left(x+2\right)\)
\(c,x^4+2x^2-3\)
\(\Leftrightarrow x^4-x^3+x^3-x^2+3x^2-3\)
\(\Leftrightarrow x^3\left(x-1\right)+x^2\left(x-1\right)+3\left(x-1\right)\left(x+1\right)\)
\(\Leftrightarrow\left(x-1\right)\left(x^3+x^2+3x+3\right)\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x+3\right)\)
\(d,ab+ac+b^2+2bc+c^2\)
\(\Leftrightarrow a\left(b+c\right)+\left(b+c\right)^2\)
\(\Leftrightarrow\left(b+c\right)\left(a+b+c\right)\)
3x^2+2x-1=3x^2+3x-x-1=3x(x+1)-(x+1)=(x+1)(3x-1)
x^4+2x^2-3=x^4+3x^2-x^2 -3=x^2(x^2+3)-(x^2+3)=(x^2+3)(x^2-1)
\(x^3+2x-3\)
\(=x^3-x+3x-3\)
\(=x\left(x^2-1\right)+3\left(x-1\right)\)
\(=x\left(x-1\right)\left(x+1\right)+3\left(x-1\right)\)
\(=\left(x-1\right)\left(x\left(x+1\right)+3\right)\)
\(x^3+5x^2+8x+4\)
\(=\left(x+1\right)\left(x+2\right)\)
Cây cuối tương tự câu đầu thôi:
\(x^3-7x+6\)
\(=x^3-x-6x+6\)
.......
a^10-a^7+a^7-a^4+a^4-a+a+a^5-a^2+a^2+1
=(a^10-a^7)+(a^7-a^4)+(a^4-a) + (a^5-a^2) + (a^2+a+1)
=a^7(a^3-1)+a^4(a^3-1) +a(a^3-1)+a^2(a^3-1) + (a^2+a+1)
=(a^7+a^4+a+a^2)(x-1)(a^2+a+1)+(a^2+a+1)
Bạn làm tiếp đặt a^2+a+1 làm nhân tử chung ..các câu sau cũng như thế nhé ^.^
Làm được câu a thôi nhé
Cách 1:
a10 + a5 + 1
= a10 - a9 + a7 - a6 + a5 - a3 + a2 + a9 - a8 + a6 - a5 + a4 - a3 + a + a8 - a7 + a5 - a4 + a2 - a + 1
nhóm 7 hạng tử ta đc :
= a2(a8 - a7 + a5 - a4 + a3 - a + 1) + a(a8 - a7 + a5 - a4 + a3 - a + 1) + (a8 - a7 + a5 - a4 + a3 - a + 1)
= (a2 + a + 1)(a8 - a7 + a5 - a4 + a3 - a + 1)
Cách 2:
x10 + x5 + 1 = (x10 - x) + (x5 - x2) + (x2 + x + 1)
= x.[(x3)3 - 1] + x2.(x3 - 1) + (x2 + x + 1)
= x.(x3 - 1).(x6 + x3 + 1) + x2.(x3 - 1) + (x2 + x + 1)
= (x2 + x + 1). [x.(x -1).(x6 + x3 + 1) + x2 + 1 ]
P.s:Ko chắc ^^!
\(a^2+b^2+2a-2b-2ab=a^2-2ab+b^2+2\left(a-b\right)\)
\(=\left(a-b\right)^2+2\left(a-b\right)\)
\(=\left(a-b\right)\left(a-b+2\right)\)
\(4a^2-4b^2-4a+1=4a^2-4a+1-\left(2b\right)^2\)
\(=\left(2a-1\right)^2-\left(2b\right)^2\)
\(=\left(2a-1-2b\right)\left(2a-1+2b\right)\)
- x3-5x2 +8x-4=x3-x2-4x2+4x+4x-4=x2(x-1)-4x(x-1)+4(x-1)=(x-1)(x2-4x+4)+(x-1)(x-2)2 +> x=1:2
a, \(a^5+a+1=a^5+a^4+a^3-a^4-a^3-a^2+a^2+a+1\)
\(=a^3\left(a^2+a+1\right)-a^2\left(a^2+a+1\right)+\left(a^2+a+1\right)\)
\(=\left(a^2+a+1\right)\left(a^3-a^2+1\right)\)