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![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 2:
1: \(A=\left(x+2\right)\left(x^2-2x+4\right)+2\left(x+1\right)\left(1-x\right)\)
\(=\left(x+2\right)\left(x^2-x\cdot2+2^2\right)-2\left(x+1\right)\left(x-1\right)\)
\(=x^3+2^3-2\left(x^2-1\right)\)
\(=x^3+8-2x^2+2=x^3-2x^2+10\)
\(B=\left(2x-y\right)^2-2\left(4x^2-y^2\right)+\left(2x+y\right)^2+4\left(y+2\right)\)
\(=\left(2x-y\right)^2-2\cdot\left(2x-y\right)\left(2x+y\right)+\left(2x+y\right)^2+4\left(y+2\right)\)
\(=\left(2x-y-2x-y\right)^2+4\left(y+2\right)\)
\(=\left(-2y\right)^2+4\left(y+2\right)\)
\(=4y^2+4y+8\)
2: Khi x=2 thì \(A=2^3-2\cdot2^2+10=8-8+10=10\)
3: \(B=4y^2+4y+8\)
\(=4y^2+4y+1+7\)
\(=\left(2y+1\right)^2+7>=7>0\forall y\)
=>B luôn dương với mọi y
Bài 1:
5: \(x^2\left(x-y+1\right)+\left(x^2-1\right)\left(x+y\right)\)
\(=x^3-x^2y+x^2+x^3+x^2y-x-y\)
\(=2x^3-x+x^2-y\)
6: \(\left(3x-5\right)\left(2x+11\right)-6\left(x+7\right)^2\)
\(=6x^2+33x-10x-55-6\left(x^2+14x+49\right)\)
\(=6x^2+23x-55-6x^2-84x-294\)
=-61x-349
![](https://rs.olm.vn/images/avt/0.png?1311)
bài 1:
a) x(x-2)-5y-(x-2)=(x-5y)(x-2)
b) =(2x-3-4x)(2x-3+4x)=(-2x-3)(6x-3)
bài 2 bạn tự luyện nhé
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 2:
c: \(=x^2\left(x-3\right)-4\left(x-3\right)\)
\(=\left(x-3\right)\left(x-2\right)\left(x+2\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
B1: a)\(xy\left(3x-2y\right)-2xy^2=3x^2y-2y^2x-2xy^2=3x^2y-4xy^2\)
b) \(\left(x^2+4x+4\right):\left(x+2\right)=\left(x+2\right)^2:\left(x+2\right)=\left(x+2\right)\)
\(\dfrac{2\left(x-1\right)}{x^2}.\dfrac{x}{\left(x-1\right)}=\dfrac{2\left(x-1\right)x}{x^2\left(x-1\right)}=\dfrac{2}{x}\)
B2:
a)\(2x^2-4x+2=2\left(x^2-2x+1\right)=2\left(x-1\right)^2\)
b)\(x^2-y^2+3x-3y=\left(x-y\right)\left(x+y\right)+3\left(x-y\right)=\left(x-y\right)\left(x+y+3\right)\)
Mấy bài này là mấy bài rất rất rất cơ bản, học sinh TB cũng phải tự làm được, mấy bài kiểu này đừng nên đăng lên hỏi nha:vv
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 2:
a: \(\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 3:
3: \(6x\left(x-y\right)-9y^2+9xy\)
\(=6x\left(x-y\right)+9xy-9y^2\)
\(=6x\left(x-y\right)+9y\left(x-y\right)\)
\(=\left(x-y\right)\left(6x+9y\right)\)
\(=3\left(2x+3y\right)\left(x-y\right)\)
Bài 4:
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(5x^2-10xy+5y^2=5\left(x^2-2xy+y^2\right)=5.\left(x-y\right)^2\)
b, \(x^2-4x+4-y^2=\left(x^2-4x+4\right)-y^2=\left(x-2\right)^2-y^2\)
\(=\left(x-2-y\right)\left(x-2+y\right)\)
c, \(3x^2-2x-5=3x^2-5x+3x-5=x\left(3x-5\right)+3x-5\)
\(=\left(3x-5\right)\left(x+1\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) 2( x - 1 )2 - 4( 3 + x )2 + 2x( x - 5 )
= 2( x2 - 2x + 1 ) - 4( 9 + 6x + x2 ) + 2x2 - 10x
= 2x2 - 4x + 2 - 36 - 24x - 4x2 + 2x2 - 10x
= ( 2x2 - 4x2 + 2x2 ) + ( -4x - 24x - 10x ) + ( 2 - 36 )
= -38x - 34
b) 2( 2x + 5 )2 - 3( 4x + 1 )( 1 - 4x )
= 2( 4x2 + 20x + 25 ) + 3( 4x + 1 )( 4x - 1 )
= 8x2 + 40x + 50 + 3( 16x2 - 1 )
= 8x2 + 40x + 50 + 48x2 - 3
= 56x2 + 40x + 47
c) ( x - 1 )3 - x( x - 3 )2 + 1
= x3 - 3x2 + 3x - 1 - x( x2 - 6x + 9 ) + 1
= x3 - 3x2 + 3x - x3 + 6x2 - 9x
= 3x2 - 6x
d) ( x + 2 )3 - x2( x + 6 )
= x3 + 6x2 + 12x + 8 - x3 - 6x2
= 12x + 8
e) ( x - 2 )( x + 2 ) - ( x + 1 )3 - 2x( x - 1 )2
= x2 - 4 - ( x3 + 3x2 + 3x + 1 ) - 2x( x2 - 2x + 1 )
= x2 - 4 - x3 - 3x2 - 3x - 1 - 2x3 + 4x2 - 2x
= -3x3 + 2x2 - 5x - 5
f) ( a + b - c )2 - ( b - c )2 - 2a( b - c )
= [ ( a + b ) - c ]2 - ( b2 - 2bc + c2 ) - 2ab + 2ac
= [ ( a + b )2 - 2( a + b )c + c2 ] - b2 + 2bc - c2 - 2ab + 2ac
= a2 + 2ab + b2 - 2ac - 2bc + c2 - b2 + 2bc - c2 - 2ab + 2ac
= a2
a) \(2\left(x-1\right)^2-4\left(3+x\right)^2+2x\left(x-5\right)\)
Dùng hẳng đẳng thức thứ nhất + hai :
= \(2\left(x^2-2\cdot x\cdot1+1^2\right)-4\left(3^2+2\cdot3\cdot x+x^2\right)+2x^2-10x\)
= \(2\left(x^2-2x+1\right)-4\left(9+6x+x^2\right)+2x^2-10x\)
= \(2x^2-4x+2-36-24x-4x^2+2x^2-10x\)
= \(-38x-34\)
b) 2(2x + 5)2 - 3(4x + 1)(1 - 4x)
Dùng đẳng thức thứ 1 + 3
= 2[(2x)2 + 2.2x.5 + 52 ] - (-3)[(4x)2 - 12 ]
= 2(4x2 + 20x + 25) - (-3).(16x2 - 1)
= 8x2 + 40x + 50 - (3 - 48x2)
= 8x2 + 40x + 50 - 3 + 48x2
= 56x2 + 40x + 47
c) (x - 1)3 - x(x - 3)2 + 1
Dùng đẳng thức 2 + 5:
= x3 - 3.x2.1 + 3.x.12 - 13 - x(x2 - 2.x.3 + 32) + 1
= x3 - 3x2 + 3x - 1 - x3 + 6x2 - 9x + 1
= (x3 - x3) + (-3x2 + 6x2) + (3x - 9x) + (-1 + 1)
= 3x2 - 6x
d) (x + 2)3 - x2(x + 6)
= x3 + 3.x2.2 + 3.x.22 + 23 - x3 - 6x2
= x3 + 6x2 + 12x + 8 - x3 - 6x2
= (x3 - x3) + (6x2 - 6x2) + 12x + 8 = 12x + 8
e) Dùng đẳng thức thứ 3,4 và 2
= x2 - 4 - (x3 + 3.x2.1 + 3.x.12 + 13) - 2x(x2 - 2.x.1 + 12)
= x2 - 4 - (x3 + 3x2 + 3x + 1) - 2x3 + 4x2 - 2x
= x2 - 4 - x3 - 3x2 - 3x - 1 - 2x3 + 4x2 - 2x
= (x2 - 3x2 + 4x2) + (-4 - 1) + (-x3 - 2x3) + (-3x - 2x)
= 2x2 - 5 - 3x3 - 5x
f) Đặt \(a+b-c=A\)
\(b-c=B\)
= \(A^2-B^2-2AB\)
= \(A^2-2AB+\left(-B\right)^2\)
\(=A^2-2AB+B^2\)
= (A - B)2
= (a + b - c - (b - c))2
= (a + b - c - b + c)2
= a2
2) = ab(a-b) - bc(c-b) + ca(c-a)
= ab(a-b) - bc(c-a+a-b) + ca(c-a)
= ab(a-b) - bc(c-a) - bc(a-b) + ca(c-a)
= b(a-b)(a-c) + c(c-a)(a-b)
= b(a-b)(a-c) - c(a-c)(a-b)
= (a-b)(a-c)(b-c)
nhớ k cho mình
mọi người bảo mk bài 1 vs