4*x^4 +4*x^3 +5*x^2 +2*x+1
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NL
Nguyễn Lê Phước Thịnh
CTVHS
26 tháng 8 2023
a: =4(x-2)(x+1)+4(x-2)^2+(x+1)^2
=(2x-4)^2+2*(2x-4)(x+1)+(x+1)^2
=(2x-4+x+1)^2=(3x-3)^2=9(x-1)^2
b: =x^7(x^2-1)-x^5(x+1)+x^3(x+1)+(x^2-1)
=(x+1)[x^7(x-1)-x^5+x^3+x-1]
=(x+1)[x^7(x-1)-x^3(x-1)(x+1)+(x-1)]
=(x+1)(x-1)(x^7-x^4-x^3+1)
=(x+1)(x-1)(x^3-1)(x^4-1)
=(x+1)(x-1)^2*(x^2+x+1)(x^2+1)(x-1)(x+1)
=(x+1)^2*(x-1)^3*(x^2+1)(x^2+x+1)
NV
0
KD
1
NL
Nguyễn Lê Phước Thịnh
CTVHS
29 tháng 9 2025
Đặt A=\(\left(x+3\right)^4+\left(x+5\right)^4-2\)
\(=\left\lbrack\left(x+4\right)-1\right\rbrack^4+\left\lbrack\left(x+4\right)+1\right\rbrack^4-2\)
Đặt b=x+4
=>\(A=\left(b-1\right)^4+\left(b+1\right)^4-2\)
\(=\left(b^2-2b+1\right)^2+\left(b^2+2b+1\right)^2-2\)
\(=\left(b^2+1\right)^2-4b\left(b^2+1\right)+4b^2+\left(b^2+1\right)^2+4b\left(b^2+1\right)+4b^2-2\)
\(=2\left(b^2+1\right)^2+8b^2-2\)
\(=2\left\lbrack\left(b^2+1\right)^2+4b^2-1\right\rbrack\)
\(=2\cdot\left\lbrack b^4+2b^2+1+4b^2-1\right\rbrack=2\left(b^4+6b^2\right)=2b^2\left(b^2+6\right)\)
\(=2\left(x+4\right)^2\left\lbrack\left(x+4\right)^2+6\right\rbrack\)
KD
1
NL
Nguyễn Lê Phước Thịnh
CTVHS
29 tháng 9 2025
Đặt A=\(\left(x+3\right)^4+\left(x+5\right)^4-2\)
\(=\left\lbrack\left(x+4\right)-1\right\rbrack^4+\left\lbrack\left(x+4\right)+1\right\rbrack^4-2\)
Đặt b=x+4
=>\(A=\left(b-1\right)^4+\left(b+1\right)^4-2\)
\(=\left(b^2-2b+1\right)^2+\left(b^2+2b+1\right)^2-2\)
\(=\left(b^2+1\right)^2-4b\left(b^2+1\right)+4b^2+\left(b^2+1\right)^2+4b\left(b^2+1\right)+4b^2-2\)
\(=2\left(b^2+1\right)^2+8b^2-2\)
\(=2\left\lbrack\left(b^2+1\right)^2+4b^2-1\right\rbrack\)
\(=2\cdot\left\lbrack b^4+2b^2+1+4b^2-1\right\rbrack=2\left(b^4+6b^2\right)=2b^2\left(b^2+6\right)\)
\(=2\left(x+4\right)^2\left\lbrack\left(x+4\right)^2+6\right\rbrack\)
L
2
T
16 tháng 9 2023
Mình bổ sung nhé:
\(=\left(x+1\right)\left(x^4+x^3+x^2-x^3+1\right)\)
\(=\left(x+1\right)\left[x^2\left(x^2+x+1\right)-\left(x^3-1\right)\right]\)
\(=\left(x+1\right)\left[x^2\left(x^2+x+1\right)-\left(x-1\right)\left(x^2+x+1\right)\right]\)
\(=\left(x+1\right)\left(x^2+x+1\right)\left(x^2-x+1\right)\)
NL
Nguyễn Lê Phước Thịnh
CTVHS
16 tháng 9 2023
=x^3(x^2+x+1)+(x^2+x+1)
=(x^2+x+1)(x^3+1)
=(x^2+x+1)(x+1)(x^2-x+1)
DM
1
15 tháng 9 2015
(x+2).(x+3)(x+4)(x+5)-24
=(x+2)(x+5)(x+3)(x+4)-24
=(x2+7x+10)(x2+7x+12)-24
=(x2+7x+11-1)(x2+7x+11+1)-24
Đặt x2+7x+11=a thì
=(a-1)(a+1)-24
=a2-1-24=a2-25=a2-52
=(a+5)(a-5)
=(x2+7x+16)(x2+7x+6)
25 tháng 10 2021
\(=\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]-24\\ =\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\\ =\left(x^2+7x+11\right)^2-1-24\\ =\left(x^2+7x+11\right)^2-25\\ =\left(x^2+7x+11-5\right)\left(x^2+7x+11+5\right)\\ =\left(x^2+7x+6\right)\left(x^2+7x+16\right)\\ =\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)
AQ
4
MT
22 tháng 7 2015
( x+3) (x+2) (x+4) (x+5) -24
=(x+3)(x+4)(x+2)(x+5)-24
=(x2+7x+12)(x2+7x+10)-24
Đặt t=x2+7x+10 ta được:
(t+2)t-24
=t2+2t-24
=t2+4t-6t-24
=t.(t+4)-6.(t+4)
=(t+4)(t-6)
thay t=x2+7t+10 ta được:
(x2+7x+14)(x2+7+4)
Vậy ( x+3) (x+2) (x+4) (x+5) -24=(x2+7x+14)(x2+7x+4)
#)Giải :
\(4x^4+4x^3+5x^2+2x+1\)
\(=\left(4x^4+4x^3+x^2\right)+2\left(2x^2+x\right)+1\)
\(=\left(2x^2+x\right)^2+2\left(2x^2+x\right)+1\)
\(=\left(2x^2+x+1\right)^2\)
\(4x^4+4x^3+5x^2+2x+1=\left(4x^4+4x^3+x^2\right)+2\left(2x^2+x\right)+1\)
\(=\left(2x^2+x\right)^2+2\left(2x^2+x\right)+1\)
\(=\left(2x^2+x+1\right)^2\)
~ Rất vui vì giúp đc bn ~