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a) (x2-1)(x2+4)(x2-4)=(x2-1)(x4-16)
b) 9x2+6x+1+4-9x2= 6x+5
a) x(x + 1) - 2x(x - 2) = x2 + x - 2x2 + 4x = -x2 + 5x
b) -3x(x - 1) + (x - 1)(x + 1) = -3x2 + 3x + x2 - 1 = -2x2 + 3x - 1
c) (3x - 2)(3x + 2) - (x - 1)(x + 2) = 9x2 - 4 - x2 - x + 2
= 8x2 - x - 2
a, x(x+1) - 2x(x -2 )
= x2 +x - 2x2 + 4x = -x2 + 5x
b, -3x( x - 1 ) + ( x -1 ) ( x+1 )
= -3x2 + 3x + x2 -1
= -2x2 + 3x -1
c, ( 3x-2 ) ( 3x + 2 ) - ( x -1 ) ( x +2 )
= 9x2 - 4 - ( x2 + 2x -x -2 )
= 9x2 -4 - x2 -2x + x + 2
= 8x2 -x -2
*Sxl
\(a,2\left(x-1\right)\left(x+1\right)+\left(x-1\right)^2+\left(x+1\right)^2\)
\(=2\left(x^2-1\right)+x^2-2x+1+x^2+2x+1\)
\(=2x^2-2+2x^2+2=4x^2\)
\(b,\left(x-y+1\right)^2+\left(1-y\right)^2+2\left(x-y+1\right)\left(y-1\right)\)
\(=\left(x-y+1\right)^2+2\left(x-y+1\right)\left(y-1\right)+\left(y-1\right)^2\)
\(=\left[\left(x-y+1\right)+\left(y-1\right)\right]^2\)
\(=\left[x-y+1+y-1\right]^2=x^2\)
đề cuối phải sửa cái cuối thành \(\left(3x+5\right)^2\)
\(c,\left(3x+1\right)^2-2\left(3x+1\right)\left(3x+5\right)+\left(3x+5\right)^2\)
\(=\left[\left(3x+1\right)-\left(3x+5\right)\right]^2=\left[3x+1-3x-5\right]^2=16\)
Sửa đề: (sửa sai thì em làm lại:v) \(A=\left(x^2+3x+1\right)^2+\left(3x-1\right)^2-2\left(x^2+3x+1\right)\left(3x-1\right)\)
Đặt \(x^2+3x+1=a;3x-1=b\) cho nó dễ nhìn!
\(A=a^2+b^2-2ab=\left(a-b\right)^2\)
\(=\left(x^2+3x+1-\left(3x-1\right)\right)^2=\left(x^2+2\right)^2=x^4+4x^2+4\)
=> B= (x-1)(x^2-x+1).2(x+1)3(x^2+x+1)
=> B= 6(x-1)(x^2+x+1).(x+1)(x^2-x+1)
=>B =6(x^3-1)(x^3+1)
=> B 6x^6-6
1) \(\left(x+1\right)\left(x+2\right)-3x\left(x-4\right)=x^2+3x+2-3x^2+12x=-2x^2+15x+2\)
2) \(\left(3x-4\right)\left(x-2\right)=3x\left(x-9\right)\)
\(\Leftrightarrow3x^2-10x+8=3x^2-27x\)
\(\Leftrightarrow17x=-8\Leftrightarrow x=-\dfrac{8}{17}\)
3) \(-3\left(x-4\right)\left(x-2\right)-x^2\left(-3x+18\right)+24x-25\)
\(=-3x^3+6x^2+12x^2-24x+3x^3-18x^2+24x-25=-25\)
Thu gọn:
(x-1)^2 - (x+2)(x-2)
4x(x-3) -3x(2+x)
2x(5x+2) + (2x-3)(3x-1)
\(\left(x-1\right)^2-\left(x+2\right)\left(x-2\right)\)
\(=x^2-2x+1-x^2+4\)
\(=-2x+5\)
\(4x\left(x-3\right)-3x\left(2+x\right)\)
\(=4x^2-12x-6x-3x^2\)
\(=x^2-18x\)
\(2x\left(5x+2\right)+\left(2x-3\right)\left(3x-1\right)\)
\(=10x^2+4x+6x^2-11x+3\)
\(=16x^2-7x+3\)
a) (3x + 1)^2 - 2(3x + 1)(3x - 5) + (3x - 5)^2
= 9x^2 + 6x + 1 - 18x^2 + 24x + 10 + 9x^2 - 30x + 25
= 36
b) (3x^2 - y)^2
= 9x^4 - 6x^2y + y^2
c) (3x + 5)^2 + (3x - 5)^2 - (3x + 2)(3x - 2)
= 9x^2 + 30x + 25 + 9x^2 - 30x + 25 - 9x^2 + 4
= 9x^2 + 54
d) 2x(2x - 1)^2 - 3x(x + 3)(x - 3) - 4x(x + 1)^2
= 8x^3 - 8x^2 + 2x - 3x^2 + 27x - 4x^3 - 8x^2 - 4x
= x^3 - 16x^2 + 25x
e) (x - 2)(x^2 + 2x + 4) - (x + 1)^2 + 3(x - 1)(x + 1)
= x^3 - 8 - x^2 - 2x - 1 + 3x^2 - 2
= x^3 + 2x^2 - 2x - 12
f) (x^4 - 5x^2 + 25)(x^2 + 5) - (2 + x^2)^2 + 3(1 + x^2)^2
= x^6 + 125 - 4 - 4x^2 - x^2 + 3 + 6x^2 + 3x^4
= x^6 + 2x^4 + 2x^2 + 124
1.
a) \(=x^2-6x+9+3x^2-15x=4x^2-21x+9\)
b) \(=9x^2+12x+4-x^2+9=8x^2+12x+13\)
2.
a) \(\Leftrightarrow x^2+8x+16-x^2+4-5=0\\ \Leftrightarrow8x=-15\\ \Leftrightarrow x=-\dfrac{15}{8}\)
b) \(\Leftrightarrow9x^2-6x+1-8x^2+12x-2x+3-5-x^2=0\\ \Leftrightarrow4x=1\\ \Leftrightarrow x=\dfrac{1}{4}\)
mình nghĩ để sai. biểu thức ở giữa phải là (3x-1)^2 mới đúng