Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1/Tự chép lại đb nha :v
=a2 - 9b2+2ab+3a2-8b2-12ab+6ab-3b2-2a2+ab
= 2a2-3ab-20b2
= (2a2+5ab) - (8ab+20b2)
= a(2a+5b) - 4b(2a+5b)
=(2a+5b)(a-4b)
câu 2 tương tự nhé :)
phân tích đa thức thành nhân tử
a/x2(x+1)-2x(x+1)+(x+1)=(x+1)(x^2-2x+1)=(x+1)(x-1)^2
b/a2+b2+2a-2b-2ab=(a^2-ab)+(b^2-ab)+2(a-b)=a(a-b)-b(a-b)+2(a-b)=(a-b)(a-b+2)
c/ 4x2-8x+3=(2x-2)^2-1=(2x-2-1)(2x-2+1)=(2x-3)(2x-1)
d/25-16x2=5^2-(4x)^2=(5-4x)(5+4x)
a) \(2x\left(x-3\right)^2+5x\left(3-x\right)\)
\(=2x\left(x-3\right)^2-5x\left(x-3\right)\)
\(=\left(x-3\right)\left[2x\left(x-3\right)-5x\right]\)
\(=\left(x-3\right)\left(2x^2-6x-5x\right)\)
\(=\left(x-3\right)\left(2x^2-11x\right)\)
\(=x\left(x-3\right)\left(2x-11\right)\)
b) \(\left(x+3\right)^2-4\left(y^2-2y+1\right)\)
\(=\left(x+3\right)^2-2^2\left(y-1\right)^2\)
\(=\left(x+3\right)^2-\left[2\left(y-1\right)\right]^2\)
\(=\left[\left(x+3\right)-2\left(y-1\right)\right]\left[\left(x+3\right)+2\left(y-1\right)\right]\)
\(=\left(x+3-2y+2\right)\left(x+3+2y-2\right)\)
\(=\left(x-2y+5\right)\left(x+2y+1\right)\)
a) \(2x.\left(x-3\right)^2+5x.\left(-x+3\right)=2x.\left(x-3\right)^2-5x.\left(x-3\right)\)
\(=\left(x-3\right).\left(2x^2-11x\right)=\left(x-3\right).x.\left(2x-11\right)\)
b) \(\left(x+3\right)^2-4.\left(y^2-2y+1\right)=\left(x+3\right)^2-2^2.\left(y-1\right)^2\)
\(=\left(x+3\right)^2-\left[2.\left(y-1\right)\right]^2=\left(x-2y+1\right).\left(x+2y+5\right)\)
4/ a/ Ta có \(x^2-2xy+y^2+a^2=\left(x-y\right)^2+a^2\)
Mà \(\hept{\begin{cases}\left(x-y\right)^2\ge0\\a^2\ge0\end{cases}}\)=> \(\left(x-y\right)^2+a^2\ge0\)
=> \(x^2-2xy+y^2+a^2\ge0\)
Vậy \(x^2-2xy+y^2\)chỉ nhận những giá trị không âm.
b/ Ta có \(x^2+2xy+2y^2+2y+1=\left(x^2+2xy+y^2\right)+\left(y^2+2y+1\right)=\left(x+y\right)^2+\left(y+1\right)^2\)
Mà \(\hept{\begin{cases}\left(x+y\right)^2\ge0\\\left(y+1\right)^2\ge0\end{cases}}\)=> \(\left(x+y\right)^2+\left(y+1\right)^2\ge0\)
=> \(x^2+2xy+2y^2+2y+1\ge0\)
Vậy \(x^2+2xy+2y^2+2y+1\)chỉ nhận những giá trị không âm.
c/ Ta có \(9b^2-6b+4c^2+1=\left(3b-1\right)^2+4c^2\)
Mà \(\hept{\begin{cases}\left(3b-1\right)^2\ge0\\4c^2\ge0\end{cases}}\)=> \(\left(3b-1\right)^2+4c^2\ge0\)
=> \(9b^2-6b+4c^2+1\ge0\)
Vậy \(9b^2-6b+4c^2+1\)chỉ nhận những giá trị không âm.
d/ Ta có \(x^2+y^2+2x+6y+10=\left(x+1\right)^2+\left(y+3\right)^2\)
Mà \(\hept{\begin{cases}\left(x+1\right)^2\ge0\\\left(y+3\right)^2\ge0\end{cases}}\)=> \(\left(x+1\right)^2+\left(y+3\right)^2\ge0\)
=> \(x^2+y^2+2x+6y+10\ge0\)
Vậy \(x^2+y^2+2x+6y+10\)chỉ nhận những giá trị không âm.
1/
a/ \(x^4-y^4=\left(x^2-y^2\right)\)
b/ \(\left(a+b\right)^3-\left(a-b\right)^3=\left(a+b-a+b\right)\left[\left(a+b\right)^2-\left(a+b\right)\left(a-b\right)+\left(a-b\right)^2\right]\)
\(=2b\left[a^2+2ab+b^2-\left(a^2-b^2\right)+\left(a^2-2ab+b^2\right)\right]\)
\(=2b\left(a^2+b^2\right)\)
c/ \(\left(a^2+2ab+b^2\right)+\left(a+b\right)\)
= \(\left(a+b\right)^2+\left(a+b\right)\)
= \(\left(a+b\right)\left(a+b+1\right)\)
a)Bt = (x2-a2)-(2x-2a)
=....
b)Bấm máy tìm nghiệm đi rồi phân tích
c);d);e);f)Nhóm số đầu vs số thứ 2, số thứ 3 vs số thứ 4
a. \(2a^2+5ab-3b^2-7b-2\)
\(=\left(2a^2+6ab+2a\right)-\left(ab+3b^2+b\right)-\left(2a+6b+2\right)\)
\(=2a\left(a+3b+1\right)-b\left(a+3b+1\right)-2\left(a+3b+1\right)\)
\(=\left(2a-b-2\right)\left(a+3b+1\right)\)
b. \(2x^2-7xy+x+3y^2-3y\)
\(=\left(2x^2-xy\right)-\left(6xy-3y^2\right)+\left(x-3y\right)\)
\(=x\left(2x-y\right)-3y\left(2x-y\right)+\left(x-3y\right)\)
\(=\left(x-3y\right)\left(2x-y\right)+\left(x-3y\right)\)
\(=\left(x-3y\right)\left(2x-y+1\right)\)
c. \(6x^2-xy-2y^2+3x-2y\)
\(=\left(6x^2+3xy\right)-\left(4xy-2y^2\right)+\left(3x-2y\right)\)
\(=3x\left(2x+y\right)-2y\left(2x+y\right)+\left(3x-2y\right)\)
\(=\left(3x-2y\right)\left(2x+y\right)+\left(3x-2y\right)\)
\(=\left(3x-2y\right)\left(2x+y+1\right)\)
c) = \(x^4+4x^2+4-x^2-2x-1\)
= \(\left(x^2+2\right)^2-\left(x+1\right)^2\)
= \(\left(x^2+x+3\right)\left(x^2-x+1\right)\)
Chúc bạn làm bài tốt
a2 + b2 + 2ab + 2a + 2b + 1
= ( a2 + 2ab + b2 ) + ( 2a + 2b ) + 1
= ( a + b )2 + 2( a + b ) + 12
= ( a + b + 1 )2
3x( x - 2y ) - 6y( 2y - x )
= 3x( x - 2y ) + 6y( x - 2y )
= 3( x - 2y )( x + 2y )
x2 + 2x - 3
= x2 - x + 3x - 3
= x( x - 1 ) + 3( x - 1 )
= ( x - 1 )( x + 3 )
a) \(a^2+b^2+2ab+2a+2b+1\)
\(=\left(a^2+2ab+b^2\right)+\left(2a+2b\right)+1\)
\(=\left(a+b\right)^2+2\left(a+b\right)+1\)
\(=\left(a+b+1\right)^2\)
b) \(3x\left(x-2y\right)-6y\left(2y-x\right)\)
\(=3x\left(x-2y\right)+6y\left(x-2y\right)\)
\(=3\left(x-2y\right)\left(x+2y\right)\)
c) \(x^2+2x-3=x^2-x+3x-3\)
\(=\left(x^2-x\right)+\left(3x-3\right)\)
\(=x\left(x-1\right)+3\left(x-1\right)\)
\(=\left(x-1\right)\left(x+3\right)\)