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b) (1 + 2x)(1- 2x) - x(x+2)(x-2)
= (1- 4x2) - x(x2 - 4)
= 1 - 4x2- x3- 4x
= (1 - x3) + (4x - 4x2)
= (1- x) (1 + x + x2) + 4x(1 -x)
= (1-x)(1+5x + x2)
2x( x - 1 ) - x( 1 - x )2 - ( 1 - x )3
= 2x( x - 1 ) - x( x - 1 )2 + ( x - 1 )3
= ( x - 1 )[ 2x - x( x - 1 ) + ( x - 1 )2 ]
= ( x - 1 )( 2x - x2 + x + x2 - 2x + 1 )
= ( x - 1 )( x + 1 )
Ta có: \(2x\left(x-1\right)-x\left(1-x\right)^2-\left(1-x\right)^3\)
\(=\left(x-1\right)\left(2x-x^2+x+x^2-2x+1\right)\)
\(=\left(x-1\right)\left(x+1\right)\)
Ta có: 3x2 - 3y2 - 12x + 12y
= (3x2 - 3y2) - (12x - 12y)
= 3.(x2 - y2) - 12.(x - y)
= 3.(x - y).(x + y) - 4.3(x - y)
= 3.(x - y).(x + y - 4)
\(16^4+y^4=\left[\left(y^2\right)^2+2.y^2.16^2+\left(16^2\right)^2\right]-2.y^2.16^2=\left(y^2+16^2\right)^2-2.y^2.16^2\)
b tự tính tiếp nhé
ý b tương tự. ( gợi ý: thêm bớt hạng tử 16y^4 )
\(y^8+64\)
\(=\left(y^4\right)^2+2\cdot y^4\cdot8+8^2-2\cdot y^4\cdot8\)
\(=\left(y^4+8\right)^2-16y^4\)
\(=\left(y^4+8\right)^2-\left(4y^2\right)^2\)
\(=\left(y^4+8-4y^2\right)\left(y^4+8+4y^2\right)\)
a kudo shinichi làm rồi đó
Bài 1:
a) \(3x^2-9x=3x\left(x-3\right)\)
b) \(x^2-4x+4=\left(x-2\right)^2\)
c) \(x^2+6x+9-y^2=\left(x+3\right)^2-y^2=\left(x-y+3\right)\left(x+y+3\right)\)
Bài 2:
a) \(101^2-1=\left(101-1\right)\left(101+1\right)=102.100=10200\)
b) \(67^2+66.67+33^2=67^2+2.33.67+33^2\)
\(=\left(67+33\right)^2=100^2=10000\)
Bài 3:
\(x\left(x-3\right)+2\left(x+3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)
Vậy \(x=-2\)hoặc \(x=3\)
B1:
a) \(3x^2-9x=3x.\left(x-3\right)\)
b) \(x^2-4x+4=\left(x-2\right)^2\)
c) \(x^2+6x+9-y^2=\left(x+3\right)^2-y^2=\left(x+3+y\right).\left(x+3-y\right)\)
B2:
a) \(101^2-1=\left(101+1\right).\left(101-1\right)=102.100=10200\)
b) \(67^2+66.67+33^2=67^2+2.33.67+33^2=\left(67+33\right)^2=100^2=10000\)
B3:
\(x\left(x-3\right)+2\left(x-3\right)=0\)
\(\left(x-3\right).\left(x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\x+2=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)
x8 + x4 + 1
=x8+2x4+1-x4
=(x4+1)2-x4
=(x4-x2+1)(x4+x2+1)
=(x4-x2+1)(x4+2x2+1-x2)
=(x4-x2+1)[(x2+1)2-x2]
=(x4-x2+1)(x2-x+1)(x2+x+1)
x8 + x4 + 1
= ( x4 )2 + 2x4 + 1 - x4
= ( x4 + 1 )2 - x4
= ( x4 + 1 - x2 ) ( x4 + 1 + x2 )
\(2x^2-4xy-xy+2y^2=2x\left(x-2y\right)-y\left(x-2y\right)=\left(x-2y\right)\left(2x-y\right)\)
Biểu thức này không phân tích được nhé.
\(64x^4+1\)
\(=64x^4+16x^2+1-16x^2\)
\(=\left(8x^2-4x+1\right)\left(8x^2+4x+1\right)\)