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\(a,2x^2-2xt-5x+5y\)
\(=\left(2x^2-5x\right)-\left(2xy-5y\right)\)
\(=x\left(2x-5\right)-y\left(2x-5\right)\)
\(=\left(2x-5\right)\left(x-y\right)\)
\(b,8x^2+4xy-2ax-ay\)
\(=\left(8x^2-2ax\right)+\left(4xy-ay\right)\)
\(=2x\left(4x-a\right)+y\left(4x-a\right)\)
\(=\left(4x-a\right)\left(2x+y\right)\)
\(c,x^3-4x^2+4x\)
\(=x^3-2x^2-2x^2+4x\)
\(=\left(x^3-2x^2\right)-\left(2x^2-4x\right)\)
\(=x^2\left(x-2\right)-2x\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2-2x\right)=x\left(x-2\right)\left(x-2\right)\)
\(=x\left(x-2\right)^2\)
\(d,2xy-x^2-y^2+16\)
\(=-\left(x^2-2xy+y^2-16\right)\)
\(=-\left[\left(x-y\right)^2-4^2\right]\)
\(=-\left(x-y-4\right)\left(x-y+4\right)\)
\(e,x^2-y^2-2yz-z^2\)
\(=x^2-\left(y^2+2yz+z^2\right)\)
\(=x^2-\left(y+z\right)^2=\left(x-y-z\right)\left(x+y+z\right)\)
a) x3 + 2x2y + xy2– 9x = x(x2 +2xy + y2 – 9)
= x[(x2 + 2xy + y2) – 9]
= x[(x + y)2 – 32]
= x(x + y – 3)(x + y + 3)
b) 2x – 2y – x2 + 2xy – y2 = (2x – 2y) – (x2 – 2xy + y2)
= 2(x – y) – (x – y)2
= (x – y)[2 – (x – y)]
= (x – y)(2 – x + y)
c) x4 – 2x2 = x2(x2 – (√2)2) = x2(x - √2)(x + √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.
a) (x2-y2)+(2x+2y)
= (x-y)(x+y)+2(x+y)
= (x+y)(x-y+2)
b) (3a2-6ab+3b2)-12c2
= 3(a2-2ab+b2)-12c2
= 3(a-b)2-3.(2c)2
= 3[(a-b)2-(2c)2]
= 3(a-b-2c)(a-b+2c)
c) (x2+2xy+y2)-25
= (x+y)2-25=(x+y-5)(x+y+5)
d) 81x2-(z2+6yz+9y2)=(9x)2-(z+3y)2=(9x-z-3y)(9x+z+3y)
Bài dễ muốn chết mà giải không được. Chắc do đến Tết lười nè! Nói chơi thôi chứ ai mà không như vậy.
a) \(x^2-y^2+2x+2y=\left(x+y\right)\left(x-y\right)+2\left(x+y\right)=\left(x+y\right)\left(x-y+2\right)\).
b) \(3a^2-6ab+3b^2-12c^2=3\left(a^2-2ab+b^2-4c^2\right)=3\left[\left(a^2-2ab+b^2\right)-4c^2\right]\)
\(=3\left[\left(a-b\right)^2-\left(2c\right)^2\right]=3\left(a-b+2c\right)\left(a-b-2c\right)\).
c) \(x^2-25+y^2+2xy=\left(x^2+2xy+y^2\right)-25=\left(x+y\right)^2-5^2\)
\(=\left(x+y+5\right)\left(x+y-5\right)\).
d) \(81x^2-6yz-9y^2-z^2=81x^2-\left(9y^2+6yz+z^2\right)\)
\(=81x^2-\left[\left(3y\right)^2+2.3y.z+z^2\right]=\left(9x\right)^2-\left(3y+z\right)^2=\left(9x+3y+z\right)\left(9x-3y-z\right)\).
Mình không biết bạn ở trình độ nào nên mình làm chi tiết như vậy. Khi giải, bạn có thể lược bỏ một số bước nếu bạn thấy không cần thiết.
\(x^3-x^2-5x+125\)
\(=\left(x+5\right)\left(x^2-5x+25\right)-x\left(x+5\right)\)
\(=\left(x+5\right)\left(x^2-5x+25-x\right)\)
\(=\left(x+5\right)\left(x^2-6x+25\right)\)
\(x^6-x^4-9x^3+9x^2\)
\(=x^4\left(x^2-1\right)-9x^2\left(x-1\right)\)
\(=x^4\left(x-1\right)\left(x+1\right)-9x^2\left(x-1\right)\)
\(=x^2\left(x-1\right)\left[x^2\left(x+1\right)-9\right]\)
\(=x^2\left(x-1\right)\left(x^3+x^2-9\right)\)
\(x^4-4x^3+8x^2-16x+16\)
\(=\left(x^2+4\right)^2-4x\left(x^2+4\right)\)
\(=\left(x^2+4\right)\left(x^2+4-4x\right)\)
\(=\left(x^2+4\right)\left(x-2\right)^2\)
\(3a^2-6ab+3b^2-12c^2\)
\(=3\left(a^2-2ab+b^2-4c^2\right)\)
\(=3\left[\left(a-b\right)^2-\left(2c\right)^2\right]\)
\(=3\left(a-b+2c\right)\left(a-b-2c\right)\)
\(x^2-2x-4y^2-4y\)
\(=\left(x^2-4y^2\right)-\left(2x+4y\right)\)
\(=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-2y-2\right)\)
\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} |
a)\(81x^2-6yz-9y^2-z^2\)
\(=81x^2-\left(z-3y\right)^2\)
\(=\left(9x-z+3y\right)\left(9x+z-3y\right)\)
b)\(x^2y-x^3-9y+9x\)
\(=x^2\left(y-x\right)-9\left(y-x\right)\)
\(=\left(y-x\right)\left(x-3\right)\left(x+3\right)\)
c)\(3a^2-6ab+3b^2-12c^2\)
\(=3\left(a^2-2ab+b^2-4z^2\right)\)
\(=3\left[\left(a-b\right)^2-4z^2\right]\)
\(=3\left(a-b-2z\right)\left(a-b+2z\right)\)
a)\(81x^2-6yz-9y^2-z^2=\left(9x\right)^2-\left(9y^2+6yz+z^2\right)=\left(9x\right)^2-\left(3y+z\right)^2=\left(9x-3y-z\right)\left(9x+3y+z\right)\)b)\(x^2y-x^3-9y+9x=x^2\left(y-x\right)-9\left(y-x\right)=\left(x^2-9\right)\left(y-x\right)=\left(x-3\right)\left(x+3\right)\left(y-x\right)\)
c)\(3a^2-6ab+3b^2-12c^2=3\left(a^2-2ab+b^2-4c^2\right)=3\left[\left(a-b\right)^2-\left(2c\right)^2\right]=3\left(a-b-2c\right)\left(a-b+2c\right)\)
a)x2 - y2 - 2x + 2y=(x^2-y^2)-(2x+2y)=(x-y)(x+y)-2(x+y)=(x-y)(x+y-2)
b)2x + 2y - x2 - xy=(2x+2y)-(x^2-xy)=2(x+y)-x(x+y)=(2-x)(x+y)
c,d bạn làm tương tự nha^^