Phân tích thành nhân tử \(x^3-19x-30\)
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NT
0
B1
23 tháng 8 2017
x^3-19x-30
=x^3-25x+6x-30
=x(x^2-25)+6(x-5)
=x(x+5)(x-5)+6(x-5)
=(x-5)(x^2+5x+6)
=(x-5)(x^2+2x+3x+6)
=(x-5)[x(x+2)+3(x+2)]
=(x-5)(x+2)(x+3)
PN
0
NN
0
TD
2
H
2 tháng 12 2018
\(a,2x+2y-x^2-xy=2x+2y-\left(x^2+xy\right).\)
\(=2\left(x+y\right)-x\left(x+y\right)\)
\(=\left(2-x\right)\left(x+y\right)\)
10 tháng 9 2020
Ta có: \(x^2-19x-30=\frac{4x^2-76x-120}{4}\)
\(=\frac{1}{4}.\left[\left(4x^2-76x+361\right)-481\right]\)
\(=\frac{1}{4}.\left[\left(2x-19\right)^2-481\right]\)
\(=\frac{1}{4}.\left(2x-19-\sqrt{481}\right).\left(2x-19+\sqrt{481}\right)\)
10 tháng 9 2020
Nghiệm xấu nên phân tích khó :) Sửa thành x3 - 19x - 30 cho dễ
x3 - 19x - 30
= x3 + 3x2 - 3x2 - 9x - 10x - 30
= ( x3 + 3x2 ) - ( 3x2 + 9x ) - ( 10x + 30 )
= x2( x + 3 ) - 3x( x + 3 ) - 10( x + 3 )
= ( 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 )
TT
1
MT
25 tháng 7 2015
\(a,x^2-5=x^2-\left(\sqrt{5}\right)^2=\left(x-\sqrt{5}\right)\left(x+\sqrt{5}\right)\)
\(b,x^4+x^3+x+1=x^3.\left(x+1\right)+\left(x+1\right)\)
\(=\left(x+1\right).\left(x^3+1\right)=\left(x+1\right)\left(x+1\right)\left(x^2-x+1\right)\)
\(=\left(x+1\right)^2\left(x^2-x+1\right)\)
\(c,x^3-19x-30=x^3-25x+6x-30\)
\(=x.\left(x^2-25\right)+6.\left(x-5\right)\)
\(=x.\left(x-5\right)\left(x+5\right)+6.\left(x-5\right)\)
\(=\left(x-5\right).\left[x\left(x+5\right)+6\right]\)
\(=\left(x-5\right).\left(x^2+5x+6\right)\)
\(=\left(x-5\right).\left(x^2+2x+3x+6\right)\)
\(=\left(x-5\right)\left[x.\left(x+2\right)+3.\left(x+2\right)\right]\)
\(=\left(x-5\right)\left(x+2\right)\left(x+3\right)\)
TT
1
TN
27 tháng 5 2016
a) 4x*(x+y)*(x+y+z)*(x+z)+y^2+z^2
=4*x*y*z^2+4*x^2*z^2+z^2+4*x*y^2*z+12*x^2*y*z+8*x^3*z+4*x^2*y^2+y^2+8*x^3*y+4*x^4
b) x^3-19x-30
=(x-5)*(x+2)*(x+3)
A
1
S
10 tháng 12 2018
\(x^2-x-6=x^2+2x-3x-6=x\left(x+2\right)-3\left(x+2\right)=\left(x-3\right)\left(x+2\right)\)
\(x^3-19x-30=x^3+6x-25x-30=x\left(x^2-25\right)+6x-30=x\left(x^2-25\right)+6\left(x-5\right)\)
\(=x\left(x-5\right)\left(x+5\right)+6\left(x-5\right)=\left(x-5\right)\left[\left(x\right)\left(x+5\right)+6\right]\)
\(\text{x3 - 19x - 30}=x^3+2x^2-4x-15x-30\)
\(=x^2\left(x+2\right)-2x\left(x+2\right)-15\left(x+2\right)\)
\(=\left(x^2-2x-15\right)\left(x+2\right)\)
\(=\left[x^2-5x+3x-15\right]\left(x+2\right)\)
\(=\left[x\left(x-5\right)+3\left(x-5\right)\right]\left(x+2\right)\)
\(=\left(x+3\right)\left(x-5\right)\left(x+2\right)\)
\(x^3-19x+30\)
\(=x^3-9x-10x+30\)
\(=x\left(x^2-9\right)-10\left(x-3\right)\)
\(=x\left(x-3\right)\left(x+3\right)-10\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2+3x-10\right)\)
\(=\left(x-3\right)\left(x^2-2x+5x-10\right)\)
\(=\left(x-3\right)\left[x\left(x-2\right)+5\left(x-2\right)\right]\)
\(=\left(x-2\right)\left(x-3\right)\left(x+5\right)\)