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a) ( 4x+1) (12x-1) (3x+2) (x+1) -4
=(4x+1)(3x+2)(12x-1)(x+1)-4
=(12x2+11x+2)(12x2+11x-1)-4
Đặt t=12x2+11x+2 ta được:
t.(t-3)-4
=t2-3t-4
=t2+t-4t-4
=t.(t+1)-4.(t+1)
=(t+1)(t-4)
thay t=12x2+11x+2 ta được:
(12x2+11x+3)(12x2+11x-2)
Vậy ( 4x+1) (12x-1) (3x+2) (x+1) -4=(12x2+11x+3)(12x2+11x-2)
b) (x2+2x)2+9x2+18x+20
=(x2+2x)2+9.(x2+2x)+20
Đặt y=x2+2x ta được:
y2+9y+20
=y2+4y+5y+20
=y.(y+4)+5.(y+4)
=(y+4)(y+5)
thay y=x2+2x ta được:
(x2+2x+4)(x2+2x+5)
Vậy (x2+2x)2+9x2+18x+20=(x2+2x+4)(x2+2x+5)
\(ĐKXĐ:x\ne\pm\frac{3}{2};x\ne1;x\ne0\)
\(A=\left(\frac{2+3x}{2-3x}-\frac{36x^2}{9x^2-4}-\frac{2-3x}{2+3x}\right):\frac{x^2-x}{2x^2-3x^3}\)
\(=\left[\frac{\left(2+3x\right)^2}{\left(2+3x\right)\left(2-3x\right)}+\frac{36x^2}{\left(2-3x\right)\left(2+3x\right)}-\frac{\left(2-3x\right)^2}{\left(2-3x\right)\left(2+3x\right)}\right]:\frac{x\left(x-1\right)}{x^2\left(2-3x\right)}\)
\(=\frac{4+12x+9x^2+36x^2-4+12x-9x^2}{\left(2+3x\right)\left(2-3x\right)}\cdot\frac{x\left(2-3x\right)}{x-1}\)
\(=\frac{36x^2+24x}{\left(2+3x\right)\left(2-3x\right)}\cdot\frac{x\left(2-3x\right)}{x-1}\)
\(=\frac{12x\left(3x+2\right)}{2+3x}\cdot\frac{x}{x-1}\)
\(=\frac{12x^2}{x-1}\)
Để A nguyên dương hay \(\frac{12x^2}{x-1}\) nguyên dương
Mà \(12x^2\ge0\Rightarrow x-1>0\Rightarrow x>1\)
Vậy để A nguyên dương thì x là số nguyên dương lớn hơn 1.
\(x^2-2x-4y^2-4y\)
\(=\left(x^2-4y^2\right)-\left(2x+4y\right)\)
\(=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-2y-2\right)\)
\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) 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).
a)\(x^2+4x-4y^2-8y\)
\(=x^2+2xy+4x-2xy-4y^2-8y\)
\(=x\left(x+2y+4\right)-2y\left(x+2y+4\right)\)
\(=\left(x-2y\right)\left(x+2y+4\right)\)
b)sai đề
c)sai đề tiếp
a)x2+4x-4y2-8y=(x2-4y2)+(4x-8y)
=(x+2y(x-2y)+4(x-2y)
=(x-2y)(x+2y+4)
a, x^5+x^4+x^3-x^3-x²-x+x²+x+1
= x^3(x²+x+1)-x(x²+x+1)+1(x²+x+1)
= (x²+x+1).(x³-x²+1)
f, 9x4-\(\frac{1}{4}\)= (3x2)2-\(\left(\frac{1}{2}\right)^2\)= (3x2-\(\frac{1}{2}\))(3x2+\(\frac{1}{2}\))
h, (x-y)2-(m+n)2=(x-y+m+n)(x-y-m-n)
i, 9(a-b)2-4(x-y)2= (3a-3b)2-(2x-2y)2=(3a-3b+2x-2y)(3a-3b-2x+2y)
x^4 - 2x^3 - 2x^2 - 2x - 3
= ( x^4 - 3x^3 ) + ( x^3 - 3x^2 )+ ( x^2 - 3x ) + ( x - 3)
= x^3 ( x - 3 ) + x^2 ( x - 3 ) + x ( x - 3 ) + ( x - 3 )
= ( x - 3 ) ( x^3 + x^2 + x + 1 )
= ( x - 3 ) [( x^3 + x^2 ) + ( x + 1 )]
= ( x - 3 ) [ x^2 ( x + 1 ) + ( x + 1)]
= ( x - 3 ) ( x + 1 ) ( x^2 + 1 )
\(a,=\left(3x-11\right)\left(3x+11\right)\\ b,=\left(3x+1-x+2\right)\left(3x+1+x-2\right)\\ =\left(2x+3\right)\left(4x-1\right)\\ c,=\left(2x+1-8\right)\left(2x+1+8\right)=\left(2x-7\right)\left(2x+9\right)\)