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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
a) \(\left(x^2-1\right)^3-\left(x^4+x^2+1\right)\left(x^2-1\right)=\left(x^2-1\right)\left[\left(x^2-1\right)^2-\left(x^4+x^2+1\right)\right]\)
\(=\left(x^2-1\right)\left(x^4-2x^2+1-x^4-x^2-1\right)=\left(x^2-1\right)\left(-3x^2\right)\)
\(=-3x^4+3x^2=3\left(x^2-x^4\right)=3\left(x-x^2\right)\left(x+x^2\right)=\left(3x-3x^2\right)\left(x+x^2\right).\)
b)\(\left(x^4-3x^2+9\right)\left(x^2+3-\left(3+x^2\right)\right)^3=\left(x^4-3x^2+9\right).0^3=0\)
c)\(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=\left(x-3\right)^3-\left(x^3-3^3\right)+6\left(x^2+2x+1\right)\)
\(=\left(x-3\right)^3-\left[\left(x-3\right)^3+3.x.3.\left(x-3\right)\right]+6x^2+12x+6\)
\(=6x^2+12x+6-9x\left(x-3\right)=6x^2+12x+6-9x^2+27x\)
\(=39x-3x^2+6=3\left(13x-x^2+2\right).\)
b/ \(4\left(x-1\right)\left(x+1\right)-5x\left(x-2\right)+x^2\)
= \(4\left(x^2-1\right)-5x^2+10x+x^2\)
= \(4x^2-4-5x^2+10x+x^2\)
= \(10x-4\)
= \(2\left(5x-2\right)\)
c/ \(\left(3-2x\right)\left(x-2\right)+4\left(x-1\right)\left(x-3\right)-2\left(x-2\right)\left(x+2\right)\)
= \(3x-6-2x^2+4x+4\left(x^2-4x-4\right)-2\left(x^2-4\right)\)
= \(3x-6-2x^2+4x+4x^2-16x-16-8x^2-18\)
= \(-9x-12\)
= \(-3\left(3x+4\right)\)
a: \(\frac{2x}{x-2}+\frac{x+1}{x+2}+\frac{3x^2-1}{4-x^2}\)
\(=\frac{2x}{x-2}+\frac{x+1}{x+2}-\frac{3x^2-1}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{2x\left(x+2\right)+\left(x+1\right)\left(x-2\right)-3x^2+1}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{2x^2+4x+x^2-x-2-3x^2+1}{\left(x-2\right)\left(x+2\right)}=\frac{3x-1}{\left(x-2\right)\left(x+2\right)}\)
Ta có: \(P=\left(\frac{2x}{x-2}+\frac{x+1}{x+2}+\frac{3x^2-1}{4-x^2}\right):\frac{1}{x+2}\)
\(=\frac{3x-1}{\left(x-2\right)\left(x+2\right)}\cdot\left(x+2\right)=\frac{3x-1}{x-2}\)
b: P<3
=>P-3<0
=>\(\frac{3x-1-3x+6}{x-2}<0\)
=>\(\frac{5}{x-2}<0\)
=>x-2<0
=>x<2
Kết hợp ĐKXĐ, ta được: x<2 và x<>-2
(3x - 1)2 + (x + 3)(2x - 1)
= 9x2 - 6x + 1 + 2x2 - x + 6x - 3
= 11x2 - x - 2
(x - 2)(x2 + 2x + 4) - x(x2 - 2)
= x3 - 8 - x3 + 2x
= 2x - 8
b) B = ( x - 2)(x2 + 2x + 4) - x ( x2 -2 )
= x3 - 8 - x3 + 2x
= 2x - 8
a) \(\left(x+2\right)\left(x-2\right)-\left(x-3\right)\left(x+1\right)\)
\(=x^2-2^2-\left(x^2+x-3x-3\right)\)
\(=x^2-4-x^2-x+3x+3\)
\(=2x-1\)
b) \(\left(2x+1\right)^2+\left(3x-1\right)^2+2\left(2x+1\right)\left(3x-1\right)\)
\(=\left(2x+1\right)^2+2\left(2x+1\right)\left(3x-1\right)+\left(3x-1\right)^2\)
\(=\left[\left(2x+1\right)+\left(3x-1\right)\right]^2\)
\(=\left(2x+1+3x-1\right)^2\)
\(=\left(5x\right)^2=25x^2\)
Có bài nào khó nữa hỏi mình nha Đạt :v
Mình sai chỗ nào bạn nói đi
vậy bạn Dương Hải Đăng sửa chỗ sai của mình được không
Nếu bạn sửa được thì mình sẽ tiếp nhận lỗi sai mà nếu không sửa được thì cậu quấy rối diễn đàn
a.\(x\left(x-1\right)\left(x+1\right)-\left(x+1\right)\left(x^2-x+1\right)\)
\(=x\left(x^2-1\right)-\left(x^3+1\right)\)
\(=x^3-x-x^3-1\)
\(=-x-1\)
b.\(3x^2\left(x+1\right)\left(x-1\right)-\left(x^2-1\right)\left(x^4+x^2+1\right)+\left(x^2-1\right)^3\)
\(=3x^2\left(x^2-1\right)-\left(x^2-1\right)^3+\left(x^2-1\right)^3\)
\(=3x^2\left(x^2-1\right)+\left(x^2-1\right)^3\)
Chắc là vậy!
a) \(x\left(x-1\right)\left(x+1\right)-\left(x+1\right)\left(x^2-x+1\right)\)
\(=x.\left(x^2-1^2\right)-\left(x^3+1^3\right)\)
\(=x^3-1x-x^3+1^3\)
\(=-x+1\)
x(x-1)(x+1)-(x+1)(x^2-x+1)
=x(x^2-1)-(x^3+1)
=x^3-x-x^3-1
=-x-1