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3 tháng 5 2019
a, \(x^4+2013x^2+2012x+2013\)
\(=x^4+2013x^2-x+2013x+2013\)
\(=\left(x^4-x\right)+\left(2013x^2+2013x+2013\right)\)
\(=x\left(x^3-1\right)+2013\left(x^2+x+1\right)\)
\(=x\left(x-1\right)\left(x^2+x+1\right)+2013\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left\{x\left(x-1\right)+2013\right\}\)
\(=\left(x^2+x+1\right)\left(x^2-x+2013\right)\)
QT
2
12 tháng 1 2018
x^4+2013x^2+2012x+2013
=(x^4-x)+(2013x^2+2013x+2013)
=x(x^3-1)+2013(x^2+x+1)
=x(x-1)(x^2+x+1)+2013(x^2+x+1)
=(x^2+x+1)(x^2-x+2013)
chúc bạn học tốt ^ ^
HL
12 tháng 1 2018
\(x^4+2013x^2+2012x+2013\)
=\(x^4+2013x^2+2013x-x+2013\)
=\(\left(x^4-x\right)+\left(2013x^2+2013x+2013\right)\)
=\(x\left(x^3-1\right)+2013\left(x^2+x+1\right)\)
=\(x\left(x-1\right)\left(x^2+x+1\right)+2013\left(x^2+x+1\right)\)
=\(\left(x^2+x+1\right)\left(x^2-x+2013\right)\)
1 tháng 3 2019
b) \(\dfrac{5x-150}{50}+\dfrac{5x-102}{49}+\dfrac{5x-56}{48}+\dfrac{5x-12}{47}+\dfrac{5x-660}{46}=0\)
\(\Leftrightarrow\dfrac{5x-150}{50}-1+\dfrac{5x-102}{49}-2+\dfrac{5x-56}{48}-3+\dfrac{5x-12}{47}-4+\dfrac{5x-660}{46}+10=0\)
\(\Leftrightarrow\dfrac{5x-200}{50}+\dfrac{5x-200}{49}+\dfrac{5x-200}{48}+\dfrac{5x-200}{47}+\dfrac{5x-200}{46}=0\)
\(\Leftrightarrow\left(5x-200\right)\left(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+\dfrac{1}{47}+\dfrac{1}{46}\right)=0\)
\(\Leftrightarrow5x-200=0\)
\(\Leftrightarrow x=40\)
LA
1 tháng 3 2019
b)
\(\dfrac{5x-150}{50}+\dfrac{5x-102}{49}+\dfrac{5x-56}{48}+\dfrac{5x-12}{47}+\dfrac{5x-660}{46}=0\)
\(\Rightarrow\left(\dfrac{5x-150}{50}-1\right)+\left(\dfrac{5x-102}{49}-2\right)+\left(\dfrac{5x-56}{48}-3\right)+\left(\dfrac{5x-12}{47}-4\right)\)
\(+\left(\dfrac{5x-660}{46}+10\right)=0\)
\(\Rightarrow\dfrac{5x-200}{50}+\dfrac{5x-200}{49}+\dfrac{5x-200}{48}+\dfrac{5x-200}{47}+\dfrac{5x-200}{46}=0\)
\(\Rightarrow\left(5x-200\right)\left(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+\dfrac{1}{47}+\dfrac{1}{46}\right)=0\)
\(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+\dfrac{1}{47}+\dfrac{1}{46}\ne0\)
\(\Rightarrow5x-200=0\Rightarrow x=40\)
29 tháng 7 2016
\(x^4+2014x^2+2013x+2014\)
\(=x^4+2014x^2+2014x-x+2014\)
\(=\left(x^4-x\right)+\left(2014x^2+2014x+2014\right)\)
\(=x\left(x^3-1\right)+2014\left(x^2+x+1\right)\)
\(=x\left(x-1\right)\left(x^2+x+1\right)+2014\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+2014\right)\)
b)\(x^8+7x^4+6\)
\(=x^8+x^4+6x^4+6\)
\(=x^4\left(x^4+1\right)+6\left(x^4+1\right)\)
\(=\left(x^4+1\right)\left(x^4+6\right)\)
NN
30 tháng 11 2017
b) \(x^8+7x^4+16\)
\(=\left(x^8+8x^4+16\right)-x^4\)
\(=\left[\left(x^4\right)^2+2.x^4.4+4^2\right]-x^4\)
\(=\left(x^4+4\right)^2-\left(x^2\right)^2\)
\(=\left(x^4+4-x^2\right)\left(x^4+4+x^2\right)\)
AH
Akai Haruma
Giáo viên
13 tháng 7 2018
c) \(xy(x-y)+yz(y-z)+xz(z-x)\)
\(=xy(x-y)-yz[(x-y)+(z-x)]+zx(z-x)\)
\(=(xy-yz)(x-y)+(zx-yz)(z-x)\)
\(=y(x-z)(x-y)+z(x-y)(z-x)\)
\(=(x-y)(z-x)(z-y)\)
d) \(x^4+4a^4=(x^2)^2+(2a^2)^2\)
\(=(x^2)^2+(2a^2)^2+2x^2.2a^2-4x^2a^2\)
\(=(x^2+2a^2)^2-(2xa)^2\)
\(=(x^2+2a^2-2ax)(x^2+2a^2+2ax)\)
AH
Akai Haruma
Giáo viên
13 tháng 7 2018
e)
\(x^5+x+1=x^5-x^2+x^2+x+1\)
\(=x^2(x^3-1)+x^2+x+1\)
\(=x^2(x-1)(x^2+x+1)+(x^2+x+1)\)
\(=(x^2+x+1)[x^2(x-1)+1]=(x^2+x+1)(x^3-x^2+1)\)
f)
\(x^4+2013x^2+2012x+2013\)
\(=x^4-x+2013x^2+2013x+2013\)
\(=x(x^3-1)+2013(x^2+x+1)\)
\(=x(x-1)(x^2+x+1)+2013(x^2+x+1)\)
\(=(x^2+x+1)[x(x-1)+2013]=(x^2+x+1)(x^2-x+2013)\)
NT
2
25 tháng 10 2018
a) \(x^4+4\)
\(=\left(x^2\right)^2+2\cdot x^2\cdot2+2^2-2\cdot x^2\cdot2\)
\(=\left(x^2+2\right)^2-\left(2x\right)^2\)
\(=\left(x^2-2x+2\right)\left(x^2+2x+2\right)\)
b) \(4x^8+1\)
\(=\left(2x^4\right)^2+2\cdot2x^4\cdot1+1^2-2\cdot2x^4\cdot1\)
\(=\left(2x^4+1\right)^2-\left(2x^2\right)^2\)
\(=\left(2x^4-2x^2+1\right)\left(2x^4+2x^2+1\right)\)
25 tháng 10 2018
\(x^4+4\)
\(=\left(x^2\right)^2+2.x^2.2+2^2-2.x^2.2\)
\(=\left(x^2+2\right)^2-4x^2\)
\(=\left(x^2+2+2x\right)\left(x^2+2-2x\right)\)
20 tháng 10 2018
\(\left(x-y+4\right)^2-\left(2x+3y-1\right)^2\)\(=\left(x-y+4-2x-3y+1\right)\left(x-y+4+2x+3y-1\right)\)
\(=\left(-x-4y+5\right)\left(3x+2y+3\right)\)
16 tháng 10 2018
\(x^2-10x+16\)
\(=x^2-2x-8x+16\)
\(=x\left(x-2\right)-8\left(x-2\right)=\left(x-2\right)\left(x-8\right)\)
\(x^2-8x+15\)
\(=x^2-3x-5x+15\)
\(=x\left(x-3\right)-5\left(x-3\right)=\left(x-3\right)\left(x-5\right)\)
16 tháng 10 2018
c) x2 -10x + 16
= x2 - 2x - 8x + 16
= x.(x-2) - 8.(x-2)
= (x-2).(x-8)
d) x2 - 8x + 15
= x2 - 3x - 5x + 15
= x.(x-3) - 5.(x-3)
= (x-3).(x-5)
x4+2013x2+2012x+2013= (x4-x)+(2013x2+2013x+2013)
=x(x3-1)+2013(x2+x+1)
=x(x-1)(x2+x+1)+2013(x2+x+1)
=(x2+x+1)(x2-x+2013)