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1) \(\left(x+\dfrac{1}{3}\right)^3=x^3+3.x^2.\dfrac{1}{3}+3.x.\left(\dfrac{1}{3}\right)^2+\left(\dfrac{1}{3}\right)^3\)
\(=x^3+x^2+\dfrac{x}{3}+\dfrac{1}{27}\)
2) \(\left(2x+y^2\right)^3=\left(2x\right)^3+3.\left(2x\right)^2.y^2+3.2x.\left(y^2\right)^2+\left(y^2\right)^3\)
\(=8x^3+12x^2y^2+6xy^4+y^6\)
3) \(\left(\dfrac{1}{2}x^2+\dfrac{1}{3}y\right)^3=\left(\dfrac{1}{2}x^2\right)^3+3.\left(\dfrac{1}{2}x^2\right)^2.\dfrac{1}{3}y+3.\dfrac{1}{2}x^2.\left(\dfrac{1}{3}y\right)^2+\left(\dfrac{1}{3}y\right)^3\)
\(=\dfrac{1}{8}x^6+\dfrac{1}{4}x^4y+\dfrac{1}{6}x^2y^2+\dfrac{1}{27}y^3\)
4) \(\left(3x^2-2y\right)^3=\left(3x^2\right)^3-3.\left(3x^2\right)^2.2y+3.3x^2.\left(2y\right)^2-\left(2y\right)^3\)
\(=27x^6-54x^4y+36x^2y^2-8y^3\)
5) \(\left(\dfrac{2}{3}x^2-\dfrac{1}{2}y\right)^3=\left(\dfrac{2}{3}x^2\right)^3-3.\left(\dfrac{2}{3}x^2\right)^2.\dfrac{1}{2}y+3.\dfrac{2}{3}x^2.\left(\dfrac{1}{2}y\right)^2-\left(\dfrac{1}{2}y\right)^3\)
\(=\dfrac{8}{27}x^6-\dfrac{1}{3}x^4y+\dfrac{1}{2}x^2y^2-\dfrac{1}{8}y^3\)
6) \(\left(2x+\dfrac{1}{2}\right)^3=\left(2x\right)^3+3.\left(2x\right)^2.\dfrac{1}{2}+3.2x.\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3\)
\(=8x^3+6x^2+\dfrac{3}{2}x+\dfrac{1}{8}\)
7) \(\left(x-3\right)^3=x^3-3.x^2.3+3.x.3^2-3^3\)
\(=x^3-9x^2+27x-27\)
8) \(\left(x+1\right)\left(x^2-x+1\right)\)
\(=\left(x+1\right)\left(x^2-x.1+1^2\right)\)
\(=x^3+1^3\)
\(=x+1\)
9) \(\left(x-3\right)\left(x^2+3x+9\right)\)
\(=\left(x-3\right)\left(x^2+x.3+3^2\right)\)
\(=x^3-3^3\)
\(=x^3-27\)
10) \(\left(x-2\right)\left(x^2+2x+4\right)\)
\(=\left(x-2\right)\left(x^2+x.2+2^2\right)\)
\(=x^3-2^3\)
\(=x^3-8\)
11) \(\left(x+4\right)\left(x^2-4x+16\right)\)
\(=\left(x+4\right)\left(x^2-x.4+4^2\right)\)
\(=x^3+4^3\)
\(=x^3+64\)
12) \(\left(x-3y\right)\left(x^2+3xy+9y^2\right)\)
\(=\left(x-3y\right)\left[x^2+x.3y+\left(3y\right)^2\right]\)
\(=x^3-\left(3y\right)^3\)
\(=x^3-27y^3\)
13) \(\left(x^2-\dfrac{1}{3}\right)\left(x^4+\dfrac{1}{3}x^2+\dfrac{1}{9}\right)\)
\(=\left(x^2-\dfrac{1}{3}\right)\left[\left(x^2\right)^2+x^2.\dfrac{1}{3}+\left(\dfrac{1}{3}\right)^2\right]\)
\(=\left(x^2\right)^3-\left(\dfrac{1}{3}\right)^3\)
\(=x^6-\dfrac{1}{27}\)
14) \(\left(\dfrac{1}{3}x+2y\right)\left(\dfrac{1}{9}x^2-\dfrac{2}{3}xy+4y^2\right)\)
\(=\left(\dfrac{1}{3}x+2y\right)\left[\left(\dfrac{1}{3}x\right)^2-\dfrac{1}{3}x.2y+\left(2y\right)^2\right]\)
\(=\left(\dfrac{1}{3}x\right)^3+\left(2y\right)^3\)
\(=\dfrac{1}{27}x^3+8y^3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
\(a,x^2-y^2-2x+2y=\left(x-y\right)\left(x+y\right)-2\left(x-y\right)=\left(x-y\right)\left(x+y-2\right)\)
\(b,2x+2y-x^2-xy=2\left(x+y\right)-x\left(x+y\right)=\left(2-x\right)\left(x+y\right)\)
\(c,3a^2-6ab+3b^2-12c^2=3\left(a-b\right)^2-12c^2=3\left[\left(a-b\right)^2-4c^2\right]=3\left(a-b-2c\right)\left(a-b+2c\right)\)
\(d,x^2-25+y^2+2xy=\left(x-y\right)^2-25=\left(x-y-5\right)\left(x-y+5\right)\)
Bài 1:
\(e,a^2+2ab+b^2-ac-bc=\left(a+b\right)^2-c\left(a+b\right)=\left(a+b-c\right)\left(a+b\right)\)
\(f,x^2-2x-4y^2-4y=\left(x-1\right)^2-\left(2y+1\right)^2=\left(x-2y-2\right)\left(x+2y\right)\)
\(g,x^2y-x^3-9y+9x=x^2\left(y-x\right)-9\left(y-x\right)=\left(x-3\right)\left(x+3\right)\left(y-x\right)\)
\(h,x^2\left(x-1\right)+16\left(1-x\right)=\left(x-1\right)\left(x-4\right)\left(x+4\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Quãng đường chưa sửa sau ngày thứ nhất là: \(1245\times\left(1-\dfrac{2}{5}\right)=747\left(m\right)\)
Quãng đường ngày thứ 3 công nhân làm được là: \(747\times\left(1-\dfrac{1}{3}\right)=498\left(m\right)\)
Bài 3:
a: \(x^2+2x+1=\left(x+1\right)^2\)
b: \(4x^2-4x+1=\left(2x-1\right)^2\)
c: \(a^2-6a+9=\left(a-3\right)^2\)
d: \(x^2+10x+25=\left(x+5\right)^2\)