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a) \(=x^4-2x^3-3x^2+4x+4+x^2-4x+4\)
\(=x^4-2x^3-2x^2+8\)
\(=x^3\left(x-2\right)-2x\left(x-2\right)-4\left(x-2\right)\)
\(=\left(x^3-2x-4\right)\left(x-2\right)\)
\(=\left[x^2\left(x-2\right)+2x\left(x-2\right)+2\left(x-2\right)\right]\left(x-2\right)\)
\(=\left(x-2\right)^2\left(x^2+2x+2\right)\)
b) \(=x^4-x+2019\left(x^2+x+1\right)\)
\(=x\left(x^3-1\right)+2019\left(x^2+x+1\right)\)
\(=x\left(x-1\right)\left(x^2+x+1\right)+2019\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+2019\right)\)\
Bài 1 :
a ) \(2x\left(x+1\right)+2\left(x+1\right)=\left(x+1\right)\left(2x+2\right)=2\left(x+1\right)^2\)
b ) \(y^2\left(x^2+y\right)-zx^2-zy=y^2\left(x^2+y\right)-z\left(x^2+y\right)=\left(x^2+y\right)\left(y^2-z\right)\)
c ) \(4x\left(x-2y\right)+8y\left(2y-x\right)=4x\left(x-2y\right)-8y\left(x-2y\right)=4\left(x-2y\right)^2\)
d ) \(3x\left(x+1\right)^2-5x^2\left(x+1\right)+7\left(x+1\right)=\left(x+1\right)\left(3x^2+3x-5x^2+7\right)=\left(x+1\right)\left(3x-2x^2+7\right)\)
e ) \(x^2-6xy+9y^2=\left(x-3x\right)^2\)
Bài 1 :
f ) \(x^3+6x^2y+12xy^2+8y^3=\left(x+2y\right)^3\)
g ) \(x^3-64=\left(x-4\right)\left(x^2+4x+16\right)\)
h ) \(125x^3+y^6=\left(5x+y^2\right)\left(25x^2-5xy^2+y^4\right)\)
\(1.\)
\(a.\)
\(\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2}{x^2+3}+\dfrac{1}{x+1}\)
\(=\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2\left(x^2-1\right)}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{1\left(x-1\right)\left(x^2+3\right)}{\left(x^2-1\right)\left(x^2+3\right)}\)
\(=\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2x^2-2}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{x^3-x^2+3x-3}{\left(x^2-1\right)\left(x^2+3\right)}\)
\(=\dfrac{8+2x^2-2+x^3-x^2+3x-3}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{x^3+x^2+3x+3}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{x^2\left(x+1\right)+3\left(x+1\right)}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{\left(x^2+3\right)\left(x+1\right)}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=x-1\)
\(b.\)
\(\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{2y^2}{x^2-y^2}\)
\(=\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{2y^2}{\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{\left(x+y\right)^2}{2\left(x^2-y^2\right)}-\dfrac{\left(x-y\right)^2}{2\left(x^2-y^2\right)}+\dfrac{4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{x^2+2xy+y^2}{2\left(x^2-y^2\right)}-\dfrac{x^2-2xy+y^2}{2\left(x^2-y^2\right)}+\dfrac{4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{x^2+2xy+y^2-x^2+2xy-y^2+4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{4xy+4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{4y\left(x+y\right)}{2\left(x^2-y^2\right)}\)
\(=\dfrac{2y}{\left(x-y\right)}\)
Tương tự các câu còn lại
a. 25 - \(x^2\) = (5-x) (5+x)
b) -196 + 4\(x^2\) = 196 - 4\(x^2\) = (14- 2x) (14+2x)
c)\(5^4-81x^4\) = \(\left[\left(5^2\right)^2\right]-\left[\left(81x^2\right)^2\right]\) = (\(\left(5^2-81x^2\right)\left(5^2+81x^2\right)\)
\(a,25-e=\left(5-\sqrt{e}\right)\left(5+\sqrt{e}\right)\)
\(b,-196+g=-\left(196-g\right)=-\left(14-\sqrt{g}\right)\left(14+\sqrt{g}\right)\)
\(c,2^6-47^2=\left(2^3\right)^2-47^2=\left(2^3-47\right)\left(2^3+47\right)\)
\(d,5^4-81x^4=\left(5^2\right)^2-\left(9x^2\right)^2=\left(5^2-9x^2\right)\left(5^2+9x^2\right)=\left(25-9x^2\right)\left(25+9x^2\right)\)
\(i,\dfrac{25}{16}-9y^2=\left(\dfrac{5}{4}-3y\right)\left(\dfrac{5}{4}+3y\right)\)
a)
$A=(1^2-2^2)+(3^2-4^2)+....+(2003^2-2004^2)+2005^2$
$=(1-2)(1+2)+(3-4)(3+4)+....+(2003-2004)(2003+2004)+2005^2$
$=-(1+2)-(3+4)-...-(2003+2004)+2005^2$
$=-(1+2+3+...+2004)+2005^2=-\frac{2004.2005}{2}+2005^2$
$=2005^2-1002.2005=2005(2005-1002)=2011015$
b)
$B=(2-1)(2+1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)(2^{32}+1)-2^{64}$
$=(2^2-1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)(2^{32}+1)-2^{64}$
$=(2^4-1)(2^4+1)(2^8+1)(2^{16}+1)(2^{32}+1)-2^{64}$
$=(2^8-1)(2^8+1)(2^{16}+1)(2^{32}+1)-2^{64}$
$=(2^{16}-1)(2^{16}+1)(2^{32}+1)-2^{64}$
$=(2^{32}-1)(2^{32}+1)-2^{64}$
$=2^{64}-1-2^{64}=-1$
c) Do $x=16$ nên $x-16=0$
$R(x)=x^4-17x^3+17x^2-17x+20$
$=(x^4-16x^3)-(x^3-16x^2)+x^2-16x-x+20$
$=x^3(x-16)-x^2(x-16)+x(x-16)-x+20$
$=x^3.0-x^2.0+x.0-x+20=-x+20=-16+20=4$
d) Do $x=12$ nên $x-12=0$. Khi đó:
$S(x)=(x^{10}-12x^9)-(x^9-12x^8)+(x^8-12x^7)-....+(x^2-12x)-x+10$
$=x^9(x-12)-x^8(x-12)+x^7(x-12)-....+x(x-12)-x+10$
$=(x-12)(x^9-x^8+x^7-....+x)-x+10$
$=0-x+10=-x+10=-12+10=-2$
a) \(\left(x^2+x\right)^2-2\left(x^2+x\right)-15\)
Đặt \(x^2+x=t\), đa thức trở thành : \(t^2-2t-15\)
= \(\left(t+3\right)\left(t-5\right)\)
\(=\left(x^2+x+3\right)\left(x^2+x-5\right)\)
b) \(\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=a^3+b^3+c^3+2ab+2ac+2bc-a^3-b^3-c^3\)
\(=2ab+2ac+2bc=2\left(ab+ac+bc\right)\)
c) \(x-1+x^{n+3}-x^n\)
\(=x-1+x^n\left(x^3-1\right)\)
\(=x-1+x^n\left(x-1\right)\left(x^2+x+1\right)\)
\(=\left(x-1\right)\left(x^{n+2}+x^{n+1}+x^n+1\right)\)
d) \(2x^4-7x^3-2x^2+13x+6\)
\(=\left(2x^4+2x^3\right)-\left(9x^3+9x^2\right)+\left(7x^2+7x\right)+\left(6x+6\right)\)
\(=\left(x+1\right)\left(2x^3-9x^2+7x+6\right)\)
\(=\left(x+1\right)\left[\left(2x^3+x^2\right)-\left(10x^2+5x\right)+\left(12x+6\right)\right]\)
\(=\left(x+1\right)\left(2x+1\right)\left(x^2-5x+6\right)\)
\(=\left(x+1\right)\left(2x+1\right)\left(x-2\right)\left(x-3\right)\)
a) Ta có: \(\left(x-3\right)\left(x^2+3x+9\right)\)
\(=\left(x-3\right)\left(x^2+x\cdot3+3^2\right)\)
\(=x^3-3^3=x^3-27\)
b) Ta có: \(\left(x-2\right)\left(x^2+2x+4\right)\)
\(=\left(x-2\right)\left(x^2+x\cdot2+2^2\right)\)
\(=x^3-2^3=x^3-8\)
c) Ta có: \(\left(x+4\right)\left(x^2-4x+16\right)\)
\(=\left(x+4\right)\left(x^2-x\cdot4+4^2\right)\)
\(=x^3+4^3=x^3+64\)
d) Ta có: \(\left(x-3y\right)\left(x^2+3xy+9y^2\right)\)
\(=\left(x-3y\right)\left[x^2+x\cdot3y+\left(3y\right)^2\right]\)
\(=x^3-\left(3y\right)^3=x^3-27y^3\)
e) Ta có: \(\left(x^2-\frac{1}{3}\right)\left(x^4+\frac{1}{3}x^2+\frac{1}{9}\right)\)
\(=\left(x^2-\frac{1}{3}\right)\left[\left(x^2\right)^2+x^2\cdot\frac{1}{3}+\left(\frac{1}{3}\right)^2\right]\)
\(=\left(x^2\right)^3-\left(\frac{1}{3}\right)^3\)
\(=x^6-\frac{1}{27}\)
f) Ta có: \(\left(\frac{1}{3}x+2y\right)\left(\frac{1}{9}x^2-\frac{2}{3}xy+4y^2\right)\)
\(=\left(\frac{1}{3}x+2y\right)\left[\left(\frac{1}{3}x\right)^2-\frac{1}{3}x\cdot2y+\left(2y\right)^2\right]\)
\(=\left(\frac{1}{3}x\right)^3+\left(2y\right)^3\)
\(=\frac{1}{27}x^3+8y^3\)
a)
\((x^2+x)^2+3(x^2+x)+2\)
\(=(x^2+x)^2+(x^2+x)+2(x^2+x)+2\)
\(=(x^2+x)(x^2+x+1)+2(x^2+x+1)\)
\(=(x^2+x+2)(x^2+x+1)\)
b) \(x(x+1)(x+2)(x+3)+1\)
\(=[x(x+3)][(x+1)(x+2)]+1\)
\(=(x^2+3x)(x^2+3x+2)+1\)
\(=(x^2+3x)^2+2(x^2+3x)+1\)
\(=(x^2+3x+1)^2\)
c) \((x^2+x+1)(x^2+3x+1)+x^2\)
\(=(x^2+x+1)[(x^2+x+1)+2x]+x^2\)
\(=(x^2+x+1)^2+2x(x^2+x+1)+x^2\)
\(=(x^2+x+1+x)^2\)
\(=(x^2+2x+1)^2=[(x+1)^2]^2=(x+1)^4\)
d) \((x^2+1)^2-4x(1-x^2)\)
\(=(x^2+1)^2+4x(x^2-1)\)
\(=(x^2+1)^2+(x-1)(4x^2+4x)\)
\(=(x^2+1)^2+(x-1)[4x^2+4+(4x-4)]\)
\(=(x^2+1)^2+(4x^2+4)(x-1)+(4x-4)(x-1)\)
\(=(x^2+1)^2+2(x^2+1)(2x-2)+(2x-2)^2\)
\(=(x^2+1+2x-2)^2=(x^2+2x-1)^2\)
a) \(A=\left(x^2+x-2\right)\left(x+7\right)-16\)
\(=x^3+8x^2+5x-14-16\)
\(=x^3+8x^2+5x-30\)
\(=x^3+3x^2+5x^2+15x-10x-30\)
\(=x^2\left(x+3\right)+5x\left(x+3\right)-10\left(x+3\right)\)
\(=\left(x^2+5x-10\right)\left(x+3\right)\)
b) \(A=x^4-2x^3-3x^2+4x+4+x^2-4x+4\)
\(=x^4-2x^3-2x^2+8\)
\(=x^3\left(x-2\right)-2\left(x^2-4\right)\)
\(=\left(x-2\right)\left(x^3-2x-4\right)\)
\(=\left(x-2\right)\left[x^2\left(x+2\right)+2x\left(x+2\right)-2\left(x+2\right)\right]\)
\(=\left(x-2\right)\left(x+2\right)\left(x^2+2x-2\right)\)
c) \(81x^4+4=81x^4+36x^2+4-36x^2\)
\(=\left(9x^2+2\right)^2-\left(6x\right)^2\)
\(=\left(9x^2-6x+2\right)\left(9x^2+6x+2\right)\)
d) \(\left(x^2-3\right)^2+16=x^4-6x^2+25\)
\(=\left(x^4+10x^2+25\right)-16x^2\)
\(=\left(x^2+5\right)^2-\left(4x\right)^2\)
\(=\left(x^2-4x+5\right)\left(x^2+4x+5\right)\)
sửa câu b) xíu nha!
\(A=\left(x-2\right)\left(x^3-2x-4\right)\)
\(=\left(x-2\right)\left[x^2\left(x-2\right)+2x\left(x-2\right)+2\left(x-2\right)\right]\)
\(=\left(x-2\right)^2\left(x^2+2x+2\right)\)