Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(x^2\left(x-3\right)+12-4x=x^2\left(x-3\right)-4x+12\)
\(=x^2\left(x-3\right)-4\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2-4\right)\)
\(=\left(x-3\right)\left(x^2-2^2\right)\)
\(=\left(x+3\right)\left(x-2\right)\left(x+2\right)\)
b)\(x^2-4+\left(x-2\right)^2=x^2-2^2+\left(x-2\right)^2\)
\(=\left(x-2\right)\left(x+2\right)+\left(x-2\right)^2\)
\(=\left(x-2\right)\left(x+2+x-2\right)\)
\(=\left(x-2\right)2x\)
c)\(x^3-4x^2-12x+27=x^3+3x^2-7x^2-21x+9x+27\)
\(=x^2\left(x+3\right)-7x\left(x+3\right)+9\left(x+3\right)\)
\(=\left(x+3\right)\left(x^2-7x+9\right)\)
\(=\left(x+3\right)\left(x^2-7x+9\right)\)
a) => x2.(x-3)-4(x-3)=(x-3)(x2-4)=(x-3)(x-2)(x+2)
b) => (x+2)(x-2)+(x-2)2=(x-2)(x+2+x-2)=2x(x-2)
c) => x3+27-(4x2+12x)=(x+3)(x2-3x+3)-4x(x+3)=(x+3)(x2-3x+3-4x)=(x-3)(x2-7x+3)
a, x^2 -4+ (x-2)^2=(x-2)(x+2)+(x-2)^2=(x-2)(x+2+x-2)=(x-2)2x , b, x^3-2x^2+x-xy^2=x(x^2-2x+1-y^2)=x((x-1)^2-y^2)=x(x-1-y)(x-1+y) c,x^3-4x^2-4x^2-12x+27=(x^3+27)-(4x^2+12x)=(x+3)(x^2-3x+9)-4x(x+3)=(x+3)(x^2-7x+9) cách giải đó pn.......
a) x2 - 4 + (x - 2)2
\(=\left(x^2-4\right)+\left(x-2\right)^2\)
\(=\left(x^2-2^2\right)+\left(x-2\right)^2\)
\(=\left(x-2\right)\left(x+2\right)+\left(x-2\right)^2\)
\(=\left(x-2\right)\left[\left(x+2\right)+\left(x-2\right)\right]\)
\(=\left(x-2\right)\left(x+2+x-2\right)\)
\(=\left(x-2\right)2x\)
b) x3 - 2x2 + x - xy2
\(=x\left(x^2-2x+1-y^2\right)\)
\(=x\left[\left(x^2-2x+1\right)-y^2\right]\)
\(=x\left[\left(x-1\right)^2-y^2\right]\)
\(=x\left[\left(x-1-y\right)\left(x-1+y\right)\right]\)
\(=x\left(x-1-1\right)\left(x-1+y\right)\)
c) x3 - 4x2 - 12x + 27
\(=\left(x^3+27\right)-\left(4x^2+12x\right)\)
\(=\left(x^3+3^3\right)-\left(4x^2+12x\right)\)
\(=\left(x+3\right)\left(x^2-3x+9\right)-4x\left(x+3\right)\)
\(=\left(x+3\right)\left[\left(x^2-3x+9\right)-4x\right]\)
\(=\left(x+3\right)\left(x^2-3x+9-4x\right)\)
\(=\left(x+3\right)\left(x^2-7x+9\right)\)
mk làm cho 1) các phần sau cũng z
1) = x2 - 22 + (x-2)2
= (x+2)(x-2) +(x-2)(x-2)
= (x-2)(x+2+x-2)
2x(x-2)
a) x^2 - 4 + ( x - 2 )^2
= ( x- 2 )(x + 2 ) + ( x- 2)^2
= ( x - 2 ) ( x + 2 + x - 2 )
= 2x (x-2)
b) x^3 - 2x^2 + x - xy^2
= x ( x^2 - 2x + 1 - y^2)
= x [ ( x - 1 )^2 - y^2 ]
= x(x - 1 - y)( x - 1 + y )
c) x^3 - 4x^2 - 12x + 27
= x^3 + 3x^2 - 7x^2 - 21x + 9x + 27
= x^2 ( x + 3 ) - 7x ( x+ 3 ) + 9(x + 3 )
Để hai lần nha
= ( x+ 3 )(x^2 - 7x + 9 )
\(x^2-4+\left(x-2\right)^2\)
\(=\left(x-2\right)\left(x+2\right)+\left(x-2\right)^2\)
\(=\left(x-2\right)\left(x+2+x-2\right)\)
\(=2x\left(x-2\right)\)
hk tốt
^^
\(x^2-2x-4y^2-4y\)
\(=\left(x^2-4y^2\right)-\left(2x+4y\right)\)
\(=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-2y-2\right)\)
\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} |
a) \(x^3-\frac{1}{4}x=x\left(x^2-\frac{1}{4}\right)=x\left(x-\frac{1}{2}\right)\left(x+\frac{1}{2}\right)\)
b) \(\left(2x-1\right)^2-\left(x+3\right)^2=\left(2x-1-x-3\right)\left(2x-1+x+3\right)=\left(x-4\right)\left(3x+2\right)\)
c) \(x^2-y^2-2y-1=x^2-\left(y^2+2y+1\right)=x^2-\left(y+1\right)^2=\left(x-y-1\right)\left(x+y+1\right)\)
d) \(x^2\left(x-3\right)+12-4x=x^2\left(x-3\right)-4\left(x-3\right)=\left(x-3\right)\left(x^2-2^2\right)=\left(x-3\right)\left(x-2\right)\left(x+2\right)\)
Phép tính b):
Đặt 2x - 1 = a ; x + 3 = b. Từ đầu bài suy ra:
\(\left(2x-1\right)^2-\left(x+3\right)^2\Rightarrow a^2-b^2\)
\(\Rightarrow a^2-b^2-\left(ab-ab\right)\Rightarrow\left(a^2-ab\right)-\left(b^2-ab\right)\)
\(\Rightarrow a\left(a-b\right)-b\left(b-a\right)\Rightarrow a\left(a-b\right)+b\left(a-b\right)\)
\(\Rightarrow\left(a+b\right)\left(a-b\right)\)
Thế lại vào ta có:
\(\orbr{\begin{cases}a+b=\left(2x-1\right)+\left(x+3\right)=\left(2x+x\right)-\left(1-3\right)=3x+2\\a-b=\left(2x-1\right)-\left(x-3\right)=\left(2x-x\right)-\left(1-3\right)=x+2\end{cases}}\)
\(\Rightarrow\left(a+b\right)\left(a-b\right)=\left(3x+2\right)\left(x+2\right)\)
1/ \(x^2+x-90=\left(x^2-10x\right)+\left(9x-90\right)=x\left(x-10\right)+9\left(x-10\right)=\left(x-10\right)\left(x+9\right)\)
2/ \(2x^2+4xy+2y^2=\left(2x^2+2xy\right)+\left(2xy+2y^2\right)=2x\left(x+y\right)+2y\left(x+y\right)=\left(x+y\right)\left(2x+2y\right)\)
3/ \(2y^2-14y+24=2\left(y^2-7y+12\right)=2\left[\left(y^2-4y\right)+\left(12-3y\right)\right]=2\left[y\left(y-4\right)-3\left(y-4\right)\right]\)
\(=2\left(y-4\right)\left(y-3\right)\)
4/ \(x^8+x^4+1=\left(x^8+x^7+x^6\right)-\left(x^7+x^6+x^5\right)+\left(x^5+x^4+x^3\right)-\left(x^3+x^2+x\right)+\left(x^2+x+1\right)\)
\(=x^6\left(x^2+x+1\right)-x^5\left(x^2+x+1\right)+x^3\left(x^2+x+1\right)-x\left(x^2+x+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^6-x^5+x^3-x+1\right)\)
\(=\left(x^2+x+1\right)\left[\left(x^6-x^5+x^4\right)-\left(x^4-x^3+x^2\right)+\left(x^2-x+1\right)\right]\)
\(=\left(x^2+x+1\right)\left[x^4\left(x^2-x+1\right)\right]-x^2\left(x^2-x+1\right)+\left(x^2-x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+1\right)\left(x^4-x^2+1\right)\)