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Ta có : 3x(2x - 7) - (6x + 1)(x - 15) - 2010 = 0
=> 6x2 - 21x - (6x2 + x - 90x - 15) - 2010 = 0
=> 6x2 - 21x - 6x2 + 89x + 15 - 2010 = 0
=> 68x - 1995 = 0
?
b) 2x(x - 2012) - x + 2012 = 0
=> 2x(x - 2012) - (x - 2012) = 0
=> (x - 2012) (2x - 1) = 0
⇔[
x−2012=0 |
2x−1=0 |
⇔[
x=2012 |
2x=1 |
⇔[
x=2012 |
x=12 |
Vậy x = {2012;12 }
Ta có : 3x(2x - 7) - (6x + 1)(x - 15) - 2010 = 0
=> 6x2 - 21x - (6x2 + x - 90x - 15) - 2010 = 0
=> 6x2 - 21x - 6x2 + 89x + 15 - 2010 = 0
=> 68x - 1995 = 0
?
b) 2x(x - 2012) - x + 2012 = 0
=> 2x(x - 2012) - (x - 2012) = 0
=> (x - 2012) (2x - 1) = 0
\(\Leftrightarrow\orbr{\begin{cases}x-2012=0\\2x-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2012\\2x=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2012\\x=\frac{1}{2}\end{cases}}\)
Vậy x = \(\left\{2012;\frac{1}{2}\right\}\)
a) \(x^2-2x-4y^2-4y=\left(x^2-2x+1\right)-\left(4y^2+4y+1\right)\)
\(=\left(x-1\right)^2-\left(2y+1\right)^2=\left(x-1-2y-1\right)\left(x-1+2y+1\right)\)
\(=\left(x-2y-3\right)\left(x+2y\right)\)
b) \(x^2-4x^2y^2+y^2+2xy=\left(x^2+2xy+y^2\right)-4x^2y^2\)
\(=\left(x+y\right)^2-4x^2y^2=\left(x+y-2xy\right)\left(x+y+2xy\right)\)
c) \(x^6-x^4+2x^3+2x^2=\left(x^6+2x^3+1\right)-\left(x^4-2x^2+1\right)\)
\(=\left(x^3+1\right)^2-\left(x^2-1\right)^2=\left(x^3+1-x^2+1\right)\left(x^3+1+x^2-1\right)=x^2\left(x^3-x^2+2\right)\left(x+1\right)\)
d) \(x^3+3x^2+3x+1-8y^3=\left(x+1\right)^3-8y^3=\left(x+1-2y\right)\left(x^2+2x+1+2xy+2y+4y^2\right)\)
a) ( x+2y)(x^2-2xy+4y^2)-8y^3+27=0
x^3+(2y)^3 - (2y)^3 + 3^3 = 0
x^3+3^3 = 0
(x+3)(x^2-3x+9)= 0
=> x+3=0 hoặc x^2 - 3x + 9 = 0
x+3 = 0 => x=-3
x^2-3x+9= 0=> x^2-2x.3/2+9/4-9/4+9=0
(x^2-2x.3/2+9/4)+(-9/4+9)=0
(x-3/2)^2 + 25/4=0
(x-3/2)^2 =-25/4
Vì (x-3/2)^2 >= 0 mà -25/4<0 nên k tìm đc x tỏa mãn đk đề bài
Vậy x=-3
b) x^3 + x^2 - 2x - 8 = 0
(x^3-8)+(x^2-2x)=0
(x-2)(x^2+2x+4)+x(x-2)=0
(x-2)(x^2+2x+4+x)=0
(x-2)(x^2+3x+4)=0
=>x-2=0 hoặc x^2+3x+4=0
x-2=0=>x=2
x^2+3x+4=0
x^2+2x.3/2+9/4-9/4+4=0
(x^2+2x.3/2+9/4)+(-9/4+4)=0
(x+3/2)^2+15/4=0
(x+3/2)^2=-15/4
Vì (x+3/2)^2>=0 mà-15/4<0 nên k tìm đc x thỏa mãn đk đề bài
Vậy x=2
Mình chỉ làm đc như này thui b thông cảm
Tick cho mình nha
\(x^3+8y^3+2xy^2+x^2y\)
\(=x^3+2x^2y-x^2y-2xy^2+4xy^2+8y^3\)
\(=x^2\left(x+2y\right)-xy\left(x+2y\right)+4y^2\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x^2-xy+4y^2\right)\)
\(A=5x^2-3x-x^3+x^2+x^3-62x-10+3x\\ A=6x^2-62x-10\\ B=x^3+x^2+x-x^3-x^2-x+5=5\\ C=3x^2y-15xy^2+15xy^2-10y^3+10y^2-3x^2y-4=-4\)
b: Ta có: \(B=x\left(x^2+x+1\right)-x^2\left(x+1\right)-x+5\)
\(=x^3+x^2+x-x^3-x^2-x+5\)
=5
1/ x^2 +4xy +4y^2 = (x +2y)^2
2/ -x^3 +9x^2 -27x+27= - (x^3 -9x^2+27x-27) = - (x-3)^3
3/ 8x^6 +36x^4y+54^2y^2+27y^3 = (2x^2+3y)^3
4/ x^3 - 6x^2y+12xy^2 -8y^3= (x-2y)^3
a) x^4 - x^3 - x + 1
= x^3 ( x - 1 ) - ( x- 1 )
= ( x^3 - 1 )(x - 1)
= ( x- 1 )^2 (x^2 + x + 1 )
a)x4-x3-x+1
=x3(x-1)-(x-1)
=(x-1)(x3-1)
=(x-1)(x-1)(x2+x+1)
=(x-1)2(x2+x+1)
b)5x2-4x+20xy-8y
(sai đề)
\(C=3.\left(x^2-8y^3-15\right)-3\left(x-2y\right)\left(x^2+2xy+4y^2\right)\)
\(=3x^2-24y^3-45-3\left[x\left(x^2+2xy+4y^2\right)-2y\left(x^2+2xy+4y^2\right)\right]\)
\(=3x^2-24y^3-45-3\left[\left(x^3+2x^2y+4xy^2\right)-\left(2x^2y+4xy^2+8y^3\right)\right]\)
\(=3x^2-24y^3-45-3\left(x^3+2x^2y+4xy^2-2x^2y-4xy^2-8y^3\right)\)
\(=3x^2-24y^3-45-3\left(x^3-8y^3\right)\)
\(=3x^2-24y^3-45-3x^3+24y^3\)
\(=3x^2-3x^3-45\)