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\(64x^4+y^4\)
\(=64x^4+16x^2y^2+y^4-16x^2y^2\)
\(=\left(8x^2+y^2\right)^2-16x^2y^2\)
\(=\left(8x^2-4xy+y^2\right)\left(8x^2+4xy+y^2\right)\)
\(a^3+b^3+c^3-3abc\)
\(=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
Ta có :
\(a^3+b^3+c^3-3abc\)
\(\Rightarrow\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)
\(\Rightarrow\left(a+b+c\right)\left[\left(a+b^2\right)-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)
\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
P/s tham khảo nha
hok tốt
a, 4x2 - 12x + 9
= (2x + 3)2
b, 9x4y3 + 3x2y4
= 3x2y3(3x2 + y)
c, ( x - 3 )2 - 2x ( x - 3 )
= (x - 3)(x - 3 - 2x)
= (x - 3)(-x - 3)
d, 3x ( x - 1 ) + 6 ( x - 1 )
= 3(x - 1)(x + 2)
e, 2x ( x + 1 ) - 4x - 4
= 2x(x + 1) - 4(x + 1)
= (x + 1)(2x - 4)
= 2(x + 1)(x - 2)
f, ( 2x - 3 )2 - 4x + 6
= (2x - 3)2 - 2(2x - 3)
= (2x - 3)(2x - 3 - 2)
= (2x - 3)(2x - 5)
\(3x\left(x-5\right)-x\left(4+3x\right)=43\)
\(\Leftrightarrow3x^2-15x-4x-3x^2=43\)
\(\Leftrightarrow-19x=43\)
\(\Leftrightarrow x=\frac{-43}{19}\)
Bài 1:
a, x2-3xy-10y2
=x2+2xy-5xy-10y2
=(x2+2xy)-(5xy+10y2)
=x(x+2y)-5y(x+2y)
=(x+2y)(x-5y)
b, 2x2-5x-7
=2x2+2x-7x-7
=(2x2+2x)-(7x+7)
=2x(x+1)-7(x+1)
=(x+1)(2x-7)
Bài 2:
a, x(x-2)-x+2=0
<=>x(x-2)-(x-2)=0
<=>(x-2)(x-1)=0
<=>\(\orbr{\begin{cases}x-2=0\\x-1=0\end{cases}}\)<=>\(\orbr{\begin{cases}x=2\\x=1\end{cases}}\)
b, x2(x2+1)-x2-1=0
<=>x2(x2+1)-(x2+1)=0
<=>(x2+1)(x2-1)=0
<=>x2+1=0 hoặc x2-1=0
1, x2+1=0 2, x2-1=0
<=>x2= -1(loại) <=>x2=1
<=>x=1 hoặc x= -1
c, 5x(x-3)2-5(x-1)3+15(x+2)(x-2)=5
<=>5x(x-3)2-5(x-1)3+15(x2-4)=5
<=>5x(x2-6x+9)-5(x3-3x2+3x-1)+15x2-60=5
<=>5x3-30x2+45x-5x3+15x2-15x+5+15x2-60=5
<=>30x-55=5
<=>30x=55+5
<=>30x=60
<=>x=2
d, (x+2)(3-4x)=x2+4x+4
<=>(x+2)(3-4x)=(x+2)2
<=>(x+2)(3-4x)-(x+2)2=0
<=>(x+2)(3-4x-x-2)=0
<=>(x+2)(1-5x)=0
<=>\(\orbr{\begin{cases}x+2=0\\1-5x=0\end{cases}}\)<=>\(\orbr{\begin{cases}x=-2\\-5x=-1\end{cases}}\)<=>\(\orbr{\begin{cases}x=-2\\x=\frac{-1}{-5}\end{cases}}\)<=>\(\orbr{\begin{cases}x=-2\\x=\frac{1}{5}\end{cases}}\)
Bài 3:
a, Sắp xếp lại: x3+4x2-5x-20
Thực hiện phép chia ta được kết quả là x2-5 dư 0
b, Sau khi thực hiện phép chia ta được :
Để đa thức x3-3x2+5x+a chia hết cho đa thức x-3 thì a+15=0
=>a= -15
\(a.64x^4+y^4\\ =\left(8x^2\right)^2+\left(y^2\right)^2+16x^2y^2-16x^2y^2\\ =\left(8x^2+y^2\right)^2-\left(4xy\right)^2\\= \left(8x^2+y^2-4xy\right)\left(8x^2+y^2+4xy\right)\)
\(b.a^6+a^4+a^2b^2+b^4-b^6\\ =\left(a^2\right)^3-\left(b^2\right)^3+a^4+a^2b^2+b^4\\ =\left(a^2-b^2\right)\left(a^4+a^2b^2+b^4\right)+a^4+a^2b^2+b^4\\ =\left(a^2-b^2+1\right)\left(a^4+a^2b^2+b^4\right)\)
\(c.x^3+3xy+y^3-1\\= x^3+3x^2y+3xy^2+y^3-1-3x^2y-3xy^2+3xy\\ =\left(x+y\right)^3-1-3xy\left(x+y-1\right)\\ =\left(x+y-1\right)\left[\left(x+y\right)^2+x+y+1\right]-3xy\left(x+y-1\right)\\ =\left(x+y-1\right)\left(x^2+y^2+2xy-3xy+x+y+1\right)\\= \left(x+y-1\right)\left(x^2+y^2-xy+x+y+1\right)\)
\(d.4x^4+4x^3+5x^2+2x+1\\ =4x^4+2x^3+2x^3+2x^2+2x^2+x^2+x+x+1\\ =\left(4x^4+2x^3+2x^2\right)+\left(2x^3+x^2+x\right)+\left(2x^2+x+1\right)\\ =2x^2\left(2x^2+x+1\right)+x\left(2x^2+x+1\right)+\left(2x^2+x+1\right)\\ =\left(2x^2+x+1\right)\left(2x^2+x+1\right)\\ =\left(2x^2+x+1\right)^2\)
\(e.x^8+x+1\\= x^8+x^7-x^7+x^6-x^6+x^5-x^5+x^4-x^4+x^3-x^3+x^2-x^2+x+1\\ =\left(x^8+x^7+x^6\right)-\left(x^7+x^6+x^5\right)+\left(x^5+x^4+x^3\right)-\left(x^4+x^3+x^2\right)+\left(x^2+x+1\right)\\ =\left(x^2+x+1\right)\left(x^6-x^5+x^3-x^2+1\right)\)
Mình làm tắt.
a) 64x4 + y4
= (8x2)2 + 16x2y2 + (y2)2 - 16x2y2
= 8x2+y2)2 - (4xy)2
= (8x2 + y2 + 4xy)(8x2 + y2- 4xy)
b) a6 + a4 + a2b2 + b4 - b6
= (a6-b6) + (a4 + a2b2 + b4)
= (a2 - b2)(a4 + a2b2 b4) + (a4 + a2b2 + b4)
= (a2 - b2 + 1)(a4 + a2b2 + b4)
c) x3 + 3xy + y3 - 1
= (x3 + 3x2y + 3xy2 + y2-1)- 3x2y -3xy2 + 3xy
= [ (x+y)3 - 1] - 3xy(x+y-1)
=(x + y -1)(x2 + 2xy + y2 + x + y + 1 - 3xy)
= (x + y - 1)(x2 - xy + y2 + x + y + 1)
d) 4x4 + 4x3 + 5x2 + 2x + 1
= 4x4 + 4x3 + x2 + 4x2 + 2x + 1
= (2x2+x)2 + (2x + 1)2
= x(2x + 1)2 + (2x + 1)2
= (x+1) (2x+1)2
e) x8 + x+ 1
= x8 - x5 + x5 - x2 + x2 +x +1
= x5(x3 - 1) + x2(x3- 1) + (x2 + x + 1)
= x5(x -1)(x2 + x + 1) + x2(x - 1)(x2+x+1) + (x2+x+1)
=(x2 + x + 1)(x6 - x5 + x3 - x2 + 1)