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![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(2x+1\right)^2\left(x-1\right)-2\left(x-2\right)^3+x\left(3-2x\right)\left(3+x\right)-\left(3x-3\right)^2\)
\(=\left(4x^2+4x+1\right)\left(x-1\right)-2\left(x^3-6x^2+12x-8\right)+x\left(9+3x-6x-2x^2\right)-\left(9x^2-18x+9\right)\)
\(=4x^3+4x^2+x-4x^2-4x-1-2x^3+12x^2-24x+16+9x+3x^2-6x^2-2x^3-9x^2+18x+9\)
\(=\left(4x^3-2x^2-2x^3\right)+\left(4x^2-4x^2+12x^2+3x^2-6x^2-9x^2\right)+\left(x-4x-24x+9x+18x\right)+\left(-1+16+9\right)\)
\(=24\)
Vậy...........
Chúc bạn học tốt!!!
![](https://rs.olm.vn/images/avt/0.png?1311)
câu 2:
a(b-c)-b(a+c)+c(a-b)=-2bc
ta có:
a( b-c ) - b ( a +c )+ c(a-b)
=ab-ac-(ba+bc)+(ca-cb)
=ab-ac-ba-bc+ca-cb
=ab-ba-ac+ca-bc-cb
=0-0-bc-cb
=bc+(-cb)
=-2cb hay -2bc
b)a(1-b)+a(a^2-1)=a(a^2-b)
Ta có:
a(1-b) + a(a^2-1)
=a-ab+(a^3-a)
=a-ab+a^3-a
=a-a-ab+a^3
=0-ab+a^3
=-ab+a^3
=a(-b +a^2) hay a(a^2-b)
![](https://rs.olm.vn/images/avt/0.png?1311)
1: \(=8x^3+12x^2+6x+1-8x^3+12x^2-6x+1-2\left(4x+3\right)^2+8\left(x+3\right)^2\)
\(=24x^2+2-2\left(16x^2+24x+9\right)+8\left(x^2+6x+9\right)\)
\(=24x^2+2-32x^2-48x-18+8x^2+48x+72\)
=56
2: \(=\left(4x^2+4x+1\right)\left(x-1\right)-2\left(x^3-6x^2+12x-8\right)+x\left(3-2x\right)\left(3+x\right)-\left(3x-3\right)^2\)
\(=4x^3-3x-1-2x^3+12x^2-24x+16+x\left(9-3x-2x^2\right)-\left(3x-3\right)^2\)
\(=2x^3+12x^2-27x+15+9x-3x^2-2x^3-9x^2+18x-9\)
\(=6\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1> 3x(x-2)-2x(2x-1)=(1-x)(1+x)
⇔\(3x^2\)-6x-\(4x^2\)+2x=1-\(x^2\)
⇔-1\(x^2\) - 4x= 1- \(x^2\)
⇔ -1\(x^2\) -4x+ \(x^2\) = 1
⇔-4x=1
⇔ x = \(\dfrac{-1}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\left(a^2+b^2-c^2\right)^2-\left(a^2-b^2+c^2\right)^2-4a^2b^2\)
\(=\left(a^2+b^2-c^2+a^2-b^2+c^2\right)\left(a^2+b^2-c^2-a^2+b^2-c^2\right)-4a^2b^2\)
\(=2a^2.2b^2-4a^2b^2=0\)
\(C=\left(2-6x\right)^2+\left(2-5x\right)^2+2\left(6x-2\right)\left(2-5x\right)\)
\(=\left[\left(2-6x\right)+\left(2-5x\right)\right]^2\)
\(=\left[4-11x\right]^2\)
\(=16-88x+121x^2\)
chúc bn học tốt
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(2x+1\right)^3-\left(2x-1\right)^3-2\left(4x+3\right)^2+8\left(x+3\right)^2\)
\(=8x^3+12x^2+6x+1-\left(8x^3-12x^2+6x-1\right)-2\left(16x^2+24x+9\right)+8\left(x^2+6x+9\right)\)
\(=8x^3+12x^2+1-8x^3+12x^2-6x+1-32x^2-48x-18+8x^2+48x+72\)
\(=56\)
Chúc bạn học tốt!!!
\(1,\left(2x+1\right)^3-\left(2x-1\right)^3-2\left(4x+3\right)^2+8\left(x+3\right)^2\)\(=8x^3+12x^2+6x+1-8x^3+12x^2-6x+1-2\left(16x^2+24x+9\right)+8\left(x^2+6x+9\right)\)\(=8x^3+12x^2+6x+1-8x^3+12x^2-6x+1-32x^2-48x-18+8x^2+48x+72\)\(=56\)
vậy:.............
\(VT=x\left(x+1\right)\left(x+2\right)=\left(x^2+x\right)\left(x+2\right)\)
\(=x^3+3x^2+2x=VP\)
\(\Rightarrowđpcm\)
chỗ kia phải là \(\left(x^2+x\right)\left(x+2\right)\) mới đúng nha tú