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\(S=1^3+2^3+3^3+...+n^3=\left(1+2+3+...+n\right)^2\)
\(=\left[\dfrac{n\left(n+1\right)}{2}\right]^2=\dfrac{n^2\cdot\left(n+1\right)^2}{4}\)
Cái này áp dụng hằng đẳng thức 100%
a, \(\left(3x-1\right)^3=27x^3-3.9x^2+3.3x-1=27x^3-27x^2+9x-1\)
b, \(\left(4x-\dfrac{1}{2}\right)^2=16x^2-2.4x.\dfrac{1}{2}+\dfrac{1}{4}=16x^2-4x+\dfrac{1}{4}\)
c, \(\left(\dfrac{1}{3}x+1\right)^3=\dfrac{1}{27}x^3+3.\dfrac{1}{9}x^2+3.\dfrac{1}{3}x+1=\dfrac{1}{27}x^3+\dfrac{1}{3}x^2+x+1\)
d, \(\left(\dfrac{2}{3}x+\dfrac{1}{2}\right)^2=\dfrac{4}{9}x^2+2.\dfrac{2}{3}x.\dfrac{1}{2}+\dfrac{1}{4}=\dfrac{4}{9}x^2+\dfrac{2}{3}x+\dfrac{1}{4}\)
e, \(x^6-1=\left(x^3\right)^2-1=\left(x^3-1\right)\left(x^3+1\right)\)
f, \(27x^3+8=\left(3x\right)^3+2^3=\left(3x+2\right)\left(9x^2-6x+4\right)\)
g, \(9x^2-4=\left(3x\right)^2-2^2=\left(3x-2\right)\left(3x+2\right)\)
a) \(\left(3x-1\right)^3=21x^3-27x^2+9x-1\)
b) \(\left(4x-\dfrac{1}{2}\right)^2=16x^2-4x+\dfrac{1}{4}\)
c) \(\left(\dfrac{1}{3}x+1\right)^3=\dfrac{1}{27}x^3+\dfrac{1}{3}x^2+x+1\)
d) \(\left(\dfrac{2}{3}x+\dfrac{1}{2}\right)^2=\dfrac{4}{9}x^2+\dfrac{2}{3}x+\dfrac{1}{4}\)
e) \(x^6-1=\left(x^3\right)^2-1=\left(x^3-1\right)\times\left(x^3+1\right)\)
\(=\left(x-1\right)\left(x^2+x+1\right)\left(x+1\right)\left(x^2-x+1\right)\)
f) \(27x^3+8=\left(3x+2\right)\left(9x^2-6x+4\right)\)
g) \(9x^2-4=\left(3x-2\right)\left(3x+2\right)\)
1) \(\left(x-1\right)^3-125\)
\(=\left(x-1-5\right)\left[\left(x-1\right)^2+5\left(x-1\right)+25\right]\)
\(=\left(x-6\right)\left(x^2-2x+1+5x-5+25\right)\)
=\(=\left(x-6\right)\left(x^2+3x+21\right)\)
2)\(=3^3\left(x+3\right)^3-2^3\)
\(=\left(3+x+3\right)^3-2^3\)
\(=\left(x+6\right)^3-2^3\)
\(=\left(x+6-2\right)\left[\left(x-6\right)^2+2\left(x+6\right)+2^2\right]\)(phá xong rút gọn như câu 1)
Các câu còn lại đều giống nhau là hiệu hai lập phương, bạn cứ làm như trên là đc
d) \(\left(4x^2-2x+1\right)\left(2x+1\right)\)
\(=\left(2x+1\right)\left[\left(2x\right)^2-2x.1+1^2\right]\)
\(=\left(2x\right)^3+1\)
\(=8x^3+1\)
a) \(\left(x+2y\right)^3=x^3+3.x^2.2y+3.x.\left(2y\right)^2+\left(2y\right)^3\)
\(=x^3+6x^2y+12xy^2+8y^3\)
b) \(\left(2x-y\right)^3=\left(2x\right)^3-3.\left(2x\right)^2.y+3.2x.y^2-y^3\)
\(=8x^3-12x^2y+6xy^2-y^3\)
c) \(\left(x^2+x+1\right).\left(x-1\right)=x^3-x^2+x^2-x+x-1\)
\(=\left(x^3-1\right)\)
#Câu này mình làm chi tiết 1 tí :) Bạn có thể tự làm gọn cho lẹ luôn nha :)
d) \(\left(4x^2-2x+1\right)\left(2x+1\right)=\left(2x+1\right)\left(4x^2-2x+1\right)\)
\(=\left(2x+1\right)\left(\left(2x\right)^2-2x.1+1^2\right)\)
\(=\left(2x\right)^3+1\)
\(=8x^3+1\)
a) \(x^2+4x+3=\left(x^2+4x+4\right)-1=\left(x+2\right)^2-1^2=\left(x+1\right)\left(x+3\right)\) (mình sửa lại)
b) \(x^2+8x-9=\left(x^2+8x+16\right)-25=\left(x+4\right)^2-5^2=\left(x-1\right)\left(x+9\right)\)
c) \(3x^2+6x-9=3\left[\left(x^2+2x+1\right)-4\right]=3\left[\left(x+1\right)^2-2^2\right]=3\left(x-1\right)\left(x+3\right)\)
d) \(2x^2+x-3=2x^2-4x+2+5x-5=2\left(x^2-2x+1\right)+5\left(x-1\right)=2\left(x-1\right)^2+5\left(x-1\right)=\left(x-1\right)\left(2x+3\right)\)
