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9 tháng 8 2019
\(a,x^4+4x^2-5\)
\(=x^4+4x^2+4-9\)
\(=\left(x^2+2\right)^2-3^2\)
\(=\left(x^2+5\right)\left(x^2-1\right)\)
28 tháng 9 2016
1
x2-y2-x+y
=(x-y)(x+y)-(x-y)
=(x-y)(x+y-1)
2
b(1-a)-x(1-a3)
=(1-a)-x(1-a)(1+a+a2)
=(1-a)(1-x-ax-xa2)
3
x4+4x2-5
=x4+4x2+4-9
=(x2+2)2_32
=(x2+2-3)(x2+2+3)
=(x2-1)(x2+5)
29 tháng 9 2016
c6
4a2b2-(a2+b2-c2)2
=(2ab)2-(a2+b2-c2)2
=(2ab-a2-b2+c2)(2ab+a2+b2-c2)
HV
2
BA
12 tháng 10 2019
\(a,x^2-x-6\)
\(=x^2+2x-3x-6\)
\(=x\cdot\left(x+2\right)-3\cdot\left(x+2\right)\)
\(=\left(x+2\right)\cdot\left(x-3\right)\)
12 tháng 10 2019
b) x4 + 4x2 - 5
= [ ( x2 )2 + 2.2.x2 + 4 ] - 9
= ( x2 + 2 )2 - 32
= ( x2 + 2 - 3 )( x2 + 2 + 3 )
= ( x2 - 1 )( x2 + 5 )
c) x3 - 19x - 30
= x3 - 9x - 10x - 30
= x.( x2 - 9 ) - 10.( x - 3 )
= x.( x - 3 ).( x + 3 ) - 10.( x - 3 )
= ( x - 3 ).[ x.( x - 3 ) - 10 ]
= ( x - 3 ).( x2 - 3x - 10 )
= ( x - 3 ).( x2 + 2x - 5x - 10 )
= ( x - 3 ). [ x.( x + 2 ) - 5.( x + 2 ) ]
= ( x - 3 )( x + 2 )( x - 5 )
4 tháng 8 2020
Bài 4:
a) Ta có: \(a^4+a^2+1\)
\(=a^4+2a^2+1-a^2\)
\(=\left(a^2+1\right)^2-a^2\)
\(=\left(a^2-a+1\right)\left(a^2+a+1\right)\)
b) Ta có: \(a^4+a^2-2\)
\(=a^4+2a^2-a^2-2\)
\(=a^2\left(a^2+2\right)-\left(a^2+2\right)\)
\(=\left(a^2+2\right)\left(a^2-1\right)\)
\(=\left(a^2+2\right)\left(a-1\right)\left(a+1\right)\)
c) Ta có: \(x^4+4x^2-5\)
\(=x^4+5x^2-x^2-5\)
\(=x^2\left(x^2+5\right)-\left(x^2+5\right)\)
\(=\left(x^2+5\right)\left(x^2-1\right)\)
\(=\left(x^2+5\right)\left(x-1\right)\left(x+1\right)\)
d) Ta có: \(x^3-19x-30\)
\(=x^3-25x+6x-30\)
\(=x\left(x^2-25\right)+6\left(x-5\right)\)
\(=x\left(x-5\right)\left(x+5\right)+6\left(x-5\right)\)
\(=\left(x-5\right)\left(x^2+5x\right)+6\left(x-5\right)\)
\(=\left(x-5\right)\left(x^2+5x+6\right)\)
\(=\left(x-5\right)\left(x^2+2x+3x+6\right)\)
\(=\left(x-5\right)\left[x\left(x+2\right)+3\left(x+2\right)\right]\)
\(=\left(x-5\right)\left(x+2\right)\left(x+3\right)\)
e) Ta có: \(x^3-7x-6\)
\(=x^3-4x-3x-6\)
\(=x\left(x^2-4\right)-3\left(x+2\right)\)
\(=x\left(x-2\right)\left(x+2\right)-3\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2-2x\right)-3\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2-2x-3\right)\)
\(=\left(x+2\right)\left(x^2-3x+x-3\right)\)
\(=\left(x+2\right)\left[x\left(x-3\right)+\left(x-3\right)\right]\)
\(=\left(x+2\right)\left(x-3\right)\left(x+1\right)\)
f) Ta có: \(x^3-5x^2-14x\)
\(=x\left(x^2-5x-14\right)\)
\(=x\left(x^2-7x+2x-14\right)\)
\(=x\left[x\left(x-7\right)+2\left(x-7\right)\right]\)
\(=x\left(x-7\right)\left(x+2\right)\)
MT
25 tháng 7 2015
\(a,x^2-5=x^2-\left(\sqrt{5}\right)^2=\left(x-\sqrt{5}\right)\left(x+\sqrt{5}\right)\)
\(b,x^4+x^3+x+1=x^3.\left(x+1\right)+\left(x+1\right)\)
\(=\left(x+1\right).\left(x^3+1\right)=\left(x+1\right)\left(x+1\right)\left(x^2-x+1\right)\)
\(=\left(x+1\right)^2\left(x^2-x+1\right)\)
\(c,x^3-19x-30=x^3-25x+6x-30\)
\(=x.\left(x^2-25\right)+6.\left(x-5\right)\)
\(=x.\left(x-5\right)\left(x+5\right)+6.\left(x-5\right)\)
\(=\left(x-5\right).\left[x\left(x+5\right)+6\right]\)
\(=\left(x-5\right).\left(x^2+5x+6\right)\)
\(=\left(x-5\right).\left(x^2+2x+3x+6\right)\)
\(=\left(x-5\right)\left[x.\left(x+2\right)+3.\left(x+2\right)\right]\)
\(=\left(x-5\right)\left(x+2\right)\left(x+3\right)\)
Đầu bài hỏi j thế nhỉ
tìm nhân tử nha