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2x2 - 5x + 3
= 2x2 - 2x - 3x + 3
= 2x( x - 1 ) - 3( x - 1 )
= ( x - 1 )( 2x - 3 )
= ( x + 1 - 2 )[ 2( x + 1 ) - 5 ] (*)
Đặt y = x + 1
(*) trở thành
( y - 2 )( 2y - 5 )
= 2y2 - 5y - 4y + 10
= 2y2 - 9y + 10
(2x+3y)2 + 2 ( 2x + 3 y) + 1
=(2x+3y)2 + 2 ( 2x + 3 y).1 + 12
=[(2x+3y)+1]2
=(2x+3y+1)2
1) \(2x^2-5x+3=2x^2-2x-3x+3=2x\left(x-1\right)-3\left(x-1\right)\)
\(=\left(2x-3\right)\left(x-1\right)=\left(2x+2-5\right)\left(x+1-2\right)=\left(2\left(x+1\right)-5\right)\left(x+1-2\right)\)
\(=\left(2y-5\right)\left(y-2\right)\)
a) \(27x^3+8^3\)
\(=\left(3x\right)^3+2^3\)
\(=\left(3x+2\right)\left[\left(3x\right)^2+6x+2^2\right]\)
\(=\left(3x+2\right)\left(9x^2-6x+4\right)\)
b) \(8x^3-y^3\)
\(=\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)
c) \(x^2+4xy+4y^2\)
\(=\left(x+2y\right)^2\)
\(27x^3+8\)
\(=\left(3x\right)^3+2^3\)
\(=\left(3x+2\right)\left(9x^2-6x+4\right)\)
\(8x^3-y^3\)
\(=\left(2x\right)^3-y^3\)
\(=\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)
\(x^2+4xy+4y^2\)
\(=x^2+2.x.2y+\left(2y\right)^2\)
\(=\left(x+2y\right)^2\)
_Minh ngụy_
Bài 1 : \(\left(y+a\right)^3=y^3+3y^2a+3ya^2+a^3\)
Bài 2:
1. \(x^2-2x+1=\left(x-1\right)^2\)
2. \(x^2+2x+1=\left(x+1\right)^2\)
3. \(x^2-6x+9=\left(x-3\right)^2\)
4. \(x^2-10x+25=\left(x-5\right)^2\)
5. \(x^2+14x+49=\left(x+7\right)^2\)
6. \(x^2-22x+121=\left(x-11\right)^2\)
7. \(4x^2-4x+1=\left(2x-1\right)^2\)
8. \(x^2-4x+4=\left(x-2\right)^2\)
9. \(x^2-2xy+y^2=\left(x-y\right)^2\)
10. \(4x^2-4xy+y^2=\left(2x-y\right)^2\)
Bài 1 :
\(\left(y+a\right)^3=y^3+3y^2a+3ya^2+a^3\)
Bài 2 : mk lm tiếp phần còn lại thôi, mấy câu mk ko lm có ở bài trc rồi
\(x^2+14x+49=\left(x+7\right)^2\)
\(x^2-22x+121=\left(x-11\right)^2\)
\(4x^2-4x+1=\left(2x-1\right)^2\)
\(x^2-4x+4=\left(x-2\right)^2\)
\(x^2-2xy+y^2=\left(x-y\right)^2\)
\(4x^2-4xy+y^2=\left(2x-y\right)^2\)
3. \(P=2x^2-8\)
\(P=2x^2-2.4\)
\(P=2\left(x^2-4\right)\)
\(P=2\left(x^2-2^2\right)\)
\(P=2\left(x-2\right)\left(x+2\right)\)
Sửa thành \(x^2+2x\left(y+1\right)+y\left(y+2\right)+1\)
\(=x^2+2xy+2x+y^2+2y+1\)
\(=x^2+2xy+y^2+2\left(x+y\right)+1\)
\(=\left(x+y\right)^2+2\left(x+y\right)+1\)
\(=\left(x+y+1\right)^2\)