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a: \(=x^2+2x-8-x^2-2x-1=-9\)
b: \(=\dfrac{x^2+6x+9+3x-9+2x^2-18x}{x\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{3x^2-9x}{x\left(x-3\right)\left(x+3\right)}=\dfrac{3x\left(x-3\right)}{x\left(x-3\right)\left(x+3\right)}=\dfrac{3}{x+3}\)
a: Ta có: \(x^2-4-\left(x+2\right)^2\)
\(=x^2-4-x^2-4x-4\)
=-4x-8
b: Ta có: \(\left(x+2\right)\left(x-2\right)-\left(x-3\right)\left(x+1\right)\)
\(=x^2-4-x^2+2x+3\)
=2x-1
c: ta có: \(\left(x-2\right)\left(x+2\right)-\left(x-2\right)\left(x+5\right)\)
\(=\left(x-2\right)\left(x+2-x-5\right)\)
\(=-3x+6\)
d: Ta có: \(\left(6x+1\right)^2-2\left(6x+1\right)\left(6x-1\right)+\left(6x-1\right)^2\)
\(=\left(6x+1-6x+1\right)^2\)
=4
e: ta có: \(7a\left(3a-5\right)+\left(2a-3\right)\left(4a+1\right)-\left(6a-2\right)^2\)
\(=21a^2-35a+8a^2+2a-12a-3-\left(36a^2-24a+4\right)\)
\(=29a^2-45a-3-36a^2+24a-4\)
\(=-7a^2-21a-7\)
g: ta có: \(\left(5y-3\right)\left(5y+3\right)-\left(5y-4\right)^2\)
\(=25y^2-9-25y^2+40y-16\)
=40y-25
h: Ta có: \(\left(3x+1\right)^3-\left(1-2x\right)^3\)
\(=27x^3+27x^2+9x+1-1+6x-12x^2+8x^3\)
\(=35x^3+15x^2+15x\)
i: Ta có: \(\left(2x+1\right)^2+2\left(4x^2-1\right)+\left(2x-1\right)^2\)
\(=\left(2x+1+2x-1\right)^2\)
\(=16x^2\)
a: \(A=\left(2x-5\right)^2-4x\left(x-5\right)\)
\(=4x^2-20x+25-4x^2+20x\)
=25
b: \(B=\left(4-3x\right)\left(4+3x\right)+\left(3x+1\right)^2\)
\(=16-9x^2+9x^2+6x+1\)
=6x+17
c: \(C=\left(x+1\right)^3-x\left(x^2+3x+3\right)\)
\(=x^3+3x^2+3x+1-x^3-3x^2-3x\)
=1
d: \(D=\left(2021x-2020\right)^2-2\left(2021x-2020\right)\left(2020x-2021\right)+\left(2020x-2021\right)^2\)
\(=\left(2021x-2020-2020x+2021\right)^2\)
\(=\left(x+1\right)^2\)
\(=x^2+2x+1\)
\(a,=6x^2-4x-x^2-4x-4=5x^2-8x-4\\ b,=x^3+8-2\left(1-x^2\right)=x^3+8-2+2x^2=x^3+2x^2+6\\ c,=\left(2x-1\right)^2-2\left(2x-1\right)\left(2x+1\right)+\left(2x+1\right)^2\\ =\left(2x+1-2x+1\right)^2=4\)
Có thể giúp mình thực hiện cách chi tiết ko ạ ? Gv dạy mik ko hiểu mấy
Bài 2:
a: Ta có: \(A=\left(x+1\right)^3+\left(x-1\right)^3\)
\(=x^3+3x^2+3x+1+x^3-3x^2+3x-1\)
\(=2x^3+6x\)
b: Ta có: \(B=\left(x-3\right)^3-\left(x+3\right)\left(x^2-3x+9\right)+\left(3x-1\right)\left(3x+1\right)\)
\(=x^3-9x^2+27x-27-x^3-27+9x^2-1\)
\(=27x-55\)
a, \(\left(5x-4\right)\left(5x+4\right)-\left(5x-4\right)^2=\left(25x^2-16\right)-\left(25x^2-40x+16\right)=40x-32\)
b,\(\left(5x+3\right)^2-\left(4x-1\right)^2-\left(9x^2+8\right)=\left(x+4\right)\left(9x-2\right)-\left(9x^2+8\right)\)
\(=9x^2+34x-8-\left(9x^2+8\right)=34x\)
c,\(2\left(x-5y\right)\left(x+5y\right)+\left(x+5y\right)^2+\left(x-5y\right)^2=\left(2x\right)^2=4x^2\)
\(a,=x^2-4-x^2-2x-1=-2x-5\\ b,=8x^3-1-8x^3-1=-2\\ 3,\\ a,\Rightarrow x^3+8-x^3+2x=15\\ \Rightarrow2x=7\Rightarrow x=\dfrac{7}{2}\\ b,\Rightarrow x^3-3x^2+3x-1-x^3+3x^2+4x=13\\ \Rightarrow7x=14\Rightarrow x=2\)
Bài 2:
a) \(=x^2-4-x^2-2x-1=-2x-5\)
b) \(=8x^3-1-8x^3-1=-2\)
Bài 3:
a) \(\Rightarrow x^3+8-x^3+2x=15\)
\(\Rightarrow2x=7\Rightarrow x=\dfrac{7}{2}\)
b) \(\Rightarrow x^3-3x^2+3x-1-x^3+3x^2+4x=13\)
\(\Rightarrow7x=14\Rightarrow x=2\)
Bài 2:
a) \(=x^2-36y^2\)
b) \(=x^3-8\)
Bài 3:
a) \(=x^2+2x+1-x^2+2x-1-3x^2+3=-3x^2+4x+3\)
b) \(=6\left(x-1\right)\left(x+1\right)=6x^2-6\)
