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1,
b,a2+4ab+4b2
=a2+2.a.2b+(2b)2
=(a+2b)2
c,5x.(x-2y)+2(x-2y)2
=5x(x-2y)+2.(x-2y).(x-2y)
=(x-2y).[5x+2.(x-2y)]
=(x-2y).(5x+2x-4y)
=(x-2y).(7x-4y)
nhớ t*** mình nha mỏi tay quá!!!
còn câu a sử dụng hằng đẳng thức hiệu hai bình phương nha
a/ \(x^3-5x^2+8x-4\)
= \(\left(x^3-x^2\right)-\left(4x^2-4x\right)+\left(4x-4\right)\)
= \(x^2\left(x-1\right)-4x\left(x-1\right)+4\left(x-1\right)\)
= \(\left(x-1\right)\left(x^2-4x+4\right)\)
= \(\left(x-1\right)\left(x-2\right)^2\)
b/ \(x^3-x^2+x-1\)
= \(\left(x^3-x^2\right)+\left(x-1\right)\)
= \(x^2\left(x-1\right)+\left(x-1\right)\)
= \(\left(x-1\right)\left(x^2+1\right)\)
\(x^2-2x-4y^2-4y\)
\(=\left(x^2-4y^2\right)-\left(2x+4y\right)\)
\(=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-2y-2\right)\)
\begin{array}{l} a){\left( {ab - 1} \right)^2} + {\left( {a + b} \right)^2}\\ = {a^2}{b^2} - 2ab + 1 + {a^2} + 2ab + {b^2}\\ = {a^2}{b^2} + 1 + {a^2} + {b^2}\\ = {a^2}\left( {{b^2} + 1} \right) + \left( {{b^2} + 1} \right)\\ = \left( {{a^2} + 1} \right)\left( {{b^2} + 1} \right)\\ c){x^3} - 4{x^2} + 12x - 27\\ = {x^3} - 27 + \left( { - 4{x^2} + 12x} \right)\\ = \left( {x - 3} \right)\left( {{x^2} + 3x + 9} \right) - 4x\left( {x - 3} \right)\\ = \left( {x - 3} \right)\left( {{x^2} + 3x + 9 - 4x} \right)\\ = \left( {x - 3} \right)\left( {{x^2} - x + 9} \right)\\ b){x^3} + 2{x^2} + 2x + 1\\ = {x^3} + 2{x^2} + x + x + 1\\ = x\left( {{x^2} + 2x + 1} \right) + \left( {x + 1} \right)\\ = x{\left( {x + 1} \right)^2} + \left( {x + 1} \right)\\ = \left( {x + 1} \right)\left( {x\left( {x + 1} \right) + 1} \right)\\ = \left( {x + 1} \right)\left( {{x^2} + x + 1} \right)\\ d){x^4} - 2{x^3} + 2x - 1\\ = {x^4} - 2{x^3} + {x^2} - {x^2} + 2x - 1\\ = {x^2}\left( {{x^2} - 2x + 1} \right) - \left( {{x^2} - 2x + 1} \right)\\ = \left( {{x^2} - 2x + 1} \right)\left( {{x^2} - 1} \right)\\ = {\left( {x - 1} \right)^2}\left( {x - 1} \right)\left( {x + 1} \right)\\ = {\left( {x - 1} \right)^3}\left( {x + 1} \right)\\ e){x^4} + 2{x^3} + 2{x^2} + 2x + 1\\ = {x^4} + 2{x^3} + {x^2} + {x^2} + 2x + 1\\ = {x^2}\left( {{x^2} + 2x + 1} \right) + \left( {{x^2} + 2x + 1} \right)\\ = \left( {{x^2} + 2x + 1} \right)\left( {{x^2} + 1} \right)\\ = {\left( {x + 1} \right)^2}\left( {{x^2} + 1} \right) \end{array} |
1:
a) \(x^3+2x^2+x=x\left(x^2+2x+1\right)=x\left(x+1\right)^2\)
b) \(25-x^2+4xy-4y^2=25-\left(x-2y\right)^2=\left(5-x+2y\right)\left(5+x-2y\right)\)
2
\(-2x^2-4x+6=0\)
\(\Leftrightarrow-2\left(x^2+2x-3\right)=0\)
\(\Leftrightarrow x^2-x+3x-3=0\)
\(\Leftrightarrow x\left(x-1\right)+3\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=0\\x+3=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=-3\end{array}\right.\)
1,
a) x( x2 + 2x +1) = x(x+1)2
b)25 - (x-2y)2 = (5-x+2y)(5+x-2y)
2,
(x-1)(x+3)=0
<=>x=1 hoặc x=-3
1.x2-9
= (x-3)(x+3)
2. -2x2+2x+12
= -2x2+6x-4x+12
= -2x(x+2)+6(x+2)
= (x+2)(-2x+6)
4. -2x2+2x+24
= -2x2+8x-6x+24
= -2x(x+3)+8(x+3)
= (x+3)(-2x+8)
6. x2-5x+4
= x2-4x-x+4
= x(x-1) -4(x-1)
= (x-1)(x-4)
8. x2-7x+6
= x2-6x-x+6
= x(x-1)-6(x-1)
= (x-1)(x-6)
9. x2+5x+4
= x2+4x+x+4
= x(x+1)+4(x+1)
=(x+1)(x+4)
10. x2+7x+6
= x2 +x+6x+6
= x(x+1)+6(x+1)
= (x+6)(x+1)
K nhé
8x3 - 27y3 = 23 . x3 - 33 . y3 = ( 2x )3 - ( 3y )3 = ( 2x - 3y ) [(2x)2 + 12xy + (3y)2 ].
Bài 1:
a) \(x.\left(x^2-2xy+1\right)=x^3-2x^2y+x\)
b) \(\left(2x-3\right).\left(x+2\right)=2x^2+4x-3x-6=2x^2-x-6\)
Bài 2:
a) \(x^3-2x^2+x=x.\left(x^2-2x+1\right)=x.\left(x-1\right)^2\)
b) \(x^2-xy+2x-2y=\left(x^2-xy\right)+\left(2x-2y\right)=x.\left(x-y\right)+2.\left(x-y\right)=\left(x-y\right).\left(x+2\right)\)
c) Đề sai.
1) \(x^2-5x-6=x^2-6x+x-6\)
\(=\left(x^2-6x\right)+\left(x-6\right)\)
\(=x\left(x-6\right)+\left(x-6\right)\)
\(=\left(x-6\right)\left(x+1\right)\)
2) \(2x^2-4xy+2y^2=2x^2-2xy-2xy+2y^2\)
\(=\left(2x^2-2xy\right)-\left(2xy-2y^2\right)\)
\(=2x\left(x-y\right)-2y\left(x-y\right)\)
\(=\left(x-y\right)\left(2x-2y\right)\)
\(=\left(x-y\right).2\left(x-y\right)\)
\(=2\left(x-y\right)^2\)
HỌC TỐT NHA !!
\(x^2-6x+x-6\)
\(\left(x^2-6x\right)+\left(x-6\right)\)
\(x\left(x-6\right)+\left(x-6\right)\)
\(\left(x-6\right)\left(x+1\right)\)
\(x-6=0\) hoặc \(x+1=0\)
\(x=6\) hoặc \(x=-1\)