Rút gọn:
A=(2x+3)(x-1)-(x-1)x+3x-1/2
B=(2-3x)(x+1)-(x-2)(3+2x)
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a: \(=x^2+2x-8-x^2-2x-1=-9\)
b: \(=\dfrac{x^2+6x+9+3x-9+2x^2-18x}{x\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{3x^2-9x}{x\left(x-3\right)\left(x+3\right)}=\dfrac{3x\left(x-3\right)}{x\left(x-3\right)\left(x+3\right)}=\dfrac{3}{x+3}\)
\(a,=6x^2-4x-x^2-4x-4=5x^2-8x-4\\ b,=x^3+8-2\left(1-x^2\right)=x^3+8-2+2x^2=x^3+2x^2+6\\ c,=\left(2x-1\right)^2-2\left(2x-1\right)\left(2x+1\right)+\left(2x+1\right)^2\\ =\left(2x+1-2x+1\right)^2=4\)
Có thể giúp mình thực hiện cách chi tiết ko ạ ? Gv dạy mik ko hiểu mấy
a: Ta có: \(2x\left(3x-1\right)-\left(x-3\right)\left(6x+2\right)\)
\(=6x^2-2x-6x^2-2x+18x+6\)
=14x+6
b: Ta có: \(\left(2x-3\right)^2-\left(2x+1\right)\left(2x-1\right)+3\left(2x-3\right)\)
\(=4x^2-12x+9-4x^2+1+6x-9\)
\(=-6x+1\)
c: Ta có: \(\left(x+y-1\right)^2-2\left(x+y-1\right)\left(x+y\right)+\left(x+y\right)^2\)
\(=\left(x+y-1-x-y\right)^2\)
=1
a) \(2x\left(3x-1\right)-\left(x-3\right)\left(6x+2\right)=6x^2-2x-6x^2-2x+18x+6=14x+6\)
b) \(\left(2x-3\right)^2-\left(1+2x\right)\left(2x-1\right)+3\left(2x-3\right)=4x^2-12x+9-4x^2+1+6x-9=-6x+1\)
c) \(\left(x+y-1\right)^2-2\left(x+y-1\right)\left(x+y\right)+\left(x+y\right)^2=\left(x+y-1-x-y\right)^2=\left(-1\right)^2=1\)
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: 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: \(\left(2x+1\right)^2+2\left(4x^2-1\right)+\left(2x-1\right)^2\)
\(=\left(2x+1+2x-1\right)^2\)
\(=16x^2\)
\(a,=x^2-4-x^2-2x-1=-2x-5\\ b,=8x^3-1-8x^3-1=-2\\ 3,\\ a,\Rightarrow x^3+8-x^3+2x=15\\ \Rightarrow2x=7\Rightarrow x=\dfrac{7}{2}\\ b,\Rightarrow x^3-3x^2+3x-1-x^3+3x^2+4x=13\\ \Rightarrow7x=14\Rightarrow x=2\)
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) \(\Rightarrow x^3-3x^2+3x-1-x^3+3x^2+4x=13\)
\(\Rightarrow7x=14\Rightarrow x=2\)
a: \(\left(2x-1\right)^2-3\left(x-1\right)\left(x+2\right)-\left(x-3\right)^2\)
\(=4x^2-4x+1-x^2+6x-9-3\left(x^2+x-2\right)\)
\(=3x^2+2x-8-3x^2-3x+6\)
=-x+2
b: \(\left(x-2\right)\left(2x-1\right)-3\left(x+1\right)^2-4x\left(x+2\right)\)
\(=2x^2-x-4x+2-3x^2-6x-3-4x^2-8x\)
\(=-5x^2-19x-1\)
1: \(=6x^2+2x-15x-5-x^2+6x-9+4x^2+20x+25-27x^3-27x^2-9x-1\)
=-27x^3-18x^2+4x+10
2: =4x^2-1-6x^2-9x+4x+6-x^3+3x^2-3x+1+8x^3+36x^2+54x+27
=7x^3+37x^2+46x+33
5:
\(=25x^2-1-x^3-27-4x^2-16x-16-9x^2+24x-16+\left(2x-5\right)^3\)
\(=8x^3-60x^2+150-125+12x^2-x^3+8x-60\)
=7x^3-48x^2+8x-35
a)(a+2b-3c-d)(a+2b+3c+d)=[(a+2b)-(3c+d)][(a+2b)+(3c-d)]
=(a+2b)2-(3c-d)2=a2+4ab+4b2-9c2+6cd-d2
câu b tương tự