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\(\left(x^2+\frac{1}{x}+\frac{1}{9}\right)\left(x-\frac{1}{3}\right)-\left(x-\frac{1}{3}\right)^3\)
\(=\left[x^3-\left(\frac{1}{3}\right)^3\right]-\left(x-\frac{1}{3}\right)^3\)
\(=\left(x-\frac{1}{3}\right)^3-\left(x-\frac{1}{3}\right)^3\)
\(=\left(x-\frac{1}{3}\right)\left[x^2+\frac{1}{x}+\frac{1}{9}-\left(x-\frac{1}{3}\right)^2\right]\)
\(=\left(x-\frac{1}{3}\right)\left(\frac{1}{x}+\frac{2x}{3}\right)\)
\(=\frac{3x-1}{3}\times\frac{3+2x^2}{3x}\)
\(=\frac{9x+6x^2-3-2x^2}{9x}\)
\(=\frac{4x^2+9x-3}{9x}\)
1) \(4x^2-y^2=\left(2x-y\right)\left(2x+y\right)\)
2) \(8x^3-27=\left(2x-3\right)\left(4x^2+6x+9\right)\)
3) \(x^3+27y^3=\left(x+3y\right)\left(x^2-3xy+9y^2\right)\)
4) \(x^2-25y^2=\left(x-5y\right)\left(x+5y\right)\)
5) \(8x^3+\frac{1}{27}=\left(2x+\frac{1}{3}\right)\left(4x^2-\frac{2}{3}x+\frac{1}{9}\right)\)
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) \(2x^2+x-3=2x^2-4x+2+5x-5=2\left(x^2-2x+1\right)+5\left(x-1\right)=2\left(x-1\right)^2+5\left(x-1\right)=\left(x-1\right)\left(2x+3\right)\)
a) \(\left(x-2\right)^3-\left(x+4\right)^2\)
\(=x^3-6x^2+12x-8-\left(x^2+8x+16\right)\)
\(=x^3-6x^2+12x-8-x^2-8x-16\)
\(=x^3-7x^2+4x-24\)
b) \(\left(x-3\right)^3+\left(x+3\right)^3\)
\(=x^3-9x^2+27x-27+x^3+9x^2+27x+27\)
\(=2x^3+54x\)
\(=2x\left(x^2+27\right)\)
c) \(\left(x-2\right)^2-\left(x+2\right)^2=\left(x^2-4x+4\right)-\left(x^2+4x+4\right)\)
\(=x^2-4x+4-x^2-4x-4=-8x\)
d) \(\frac{x^2-25}{x+5}=\frac{\left(x-5\right)\left(x+5\right)}{x+5}=x-5\)
e) \(\frac{x^3-6x^2+12x-8}{x-2}=\frac{\left(x-2\right)^3}{x-2}=\left(x-2\right)^2\)
g) \(\frac{x^3-125}{x-5}=\frac{x^3-5^3}{x-5}=\frac{\left(x-5\right)\left(x^2+5x+25\right)}{x-5}=x^2+5x+25\)
1)\(8x^6-\frac{1}{125}y^3=\left(2x^2\right)^3-\left(\frac{1}{5}y\right)^3\)
Bạn tự lm tiếp.AD HĐT số (7)
2)\(\left(x+4\right)^3-64=\left(x+4\right)^3-4^3\)
AD HĐT số (7).Tự lm tiếp
3)\(x^6+1=\left(x^2\right)^3+1\)
AD HĐT số (7).Tự lm tiếp
4)\(x^9+1=\left(x^3\right)^3+1\)
AD HĐT số (7).Tự lm tiếp
5,\(x^{12}-y^4=\left(x^6\right)^2-\left(y^2\right)^2\)
AD HĐT số (3).Tự lm tiếp
6)\(x^3+6x^2+12x+8=\left(x+2\right)^3\)
AD HĐT số (4)
7)\(x^3-15x^2+75x-125=\left(x-5\right)^3\)
AD HĐT số (5)
8)\(27a^3-54a^2b+36ab^2-8b^3\)
\(=\left(3a\right)^3-3.\left(3a\right)^2.2b+3.3a.\left(2b\right)^2-\left(2b\right)^3\)
\(=\left(3a-2b\right)^3\)
AD HĐT số (5)
\(\left(x-1\right)-\left(x-2\right)\left(x+2\right)\)
\(=\left(x-1\right)-\left(x^2-2^2\right)\)
\(=\left(x-1\right)-x^2+2^2\)
\(=x-1-x^2+2^2\)
\(=x-x^2+\left(2-1\right)\left(2+1\right)\)
\(=x-x^2+3\)
a/ (x-1)2-(x-2)(x+2)
=(x-1)-(x2-22)
=(x-1)-x2-22
=x-x2 +(2-1)(2+1)
=x-x2+3
\(S=1^3+2^3+3^3+...+n^3=\left(1+2+3+...+n\right)^2\)
\(=\left[\dfrac{n\left(n+1\right)}{2}\right]^2=\dfrac{n^2\cdot\left(n+1\right)^2}{4}\)