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\(\left(x+3\right)^3-x\left(x-9x\right)=27\)
\(\Rightarrow x^3+9x^2+27x-x^3-9x^2=27x\)
\(\Rightarrow\left(x^3-x^3\right)+\left(9x^2-9x^2\right)+27x=27\)
\(\Rightarrow27x=27\)
\(\Rightarrow x=1\)
Vậy x=1
Bài 1 : Khai triển :
a, \(\left(x+5\right)^2=x^2+10x+25\)
b, \(\left(x-3y\right)^2=x^2-6xy+9y^2\)
c, \(\left(x^2-6z\right)\left(x^2+6z\right)=x^4-36z^2\)
d, \(\left(x+3y\right)^3=x^3+9x^2y+27xy^2+27y^3\)
e, \(27x^3-9y^2+y-\frac{1}{27}=\left(3x-\frac{1}{3}\right)^3\)
g, \(8x^6+12x^4y+6x^2y^2+y^3=\left(2x^2+y\right)\)
h, \(4x^2+12x^4y+6x^22y^2+y^3=\left(\sqrt[3]{4x^2}+y\right)\)
\(\left(3x-2\right)^3=\left(3x\right)^3-3.\left(3x\right)^2.2+3.3x.2^2-2^3=27x^3-54x^2+36x-8\)
\(8x^3-27=\left(2x\right)^3-3^3=\left(2x-3\right)\left[\left(2x\right)^2+2x.3+3^2\right]=\left(2x-3\right)\left(4x^2+6x+9\right)\)
\(\left(x^2-3\right)\left(x^4+3x^2+9\right)=\left(x^2-3\right)\left[\left(x^2\right)^2+x^2.3+3^2\right]=x^6-27\)
khai triển :
\(\left(2x^2+3y\right)^3=\left(2x^2\right)^3+3.\left(2x^2\right)^2.3y+3.2x^2.\left(3y\right)^2+\left(3y\right)^3\)
\(=8x^6+3.4x^4.3y+3.2x^2.9y+27y^3=8x^6+36x^4y+54x^2y+27y^3\)
Vậy hệ số của x4y trong khai triển.... là 36
Ta có: \(x^2\cdot\left(x^4+25\right)\cdot\left(x^2-5\right)\cdot\left(x^2+5\right)\cdot\left(x-y\right)\left(x^2+xy+y^2\right)\cdot\left(x^3+y^3\right)\)
\(=x^2\cdot\left(x^4+25\right)\left(x^4-25\right)\cdot\left(x^3-y^3\right)\left(x^3+y^3\right)\)
\(=x^2\cdot\left(x^8-625\right)\cdot\left(x^6-y^6\right)\)
\(25x^2y^2-9x^4y^4=\left(5xy\right)^2-\left(3x^2y^2\right)^2=\left(5xy-3x^2y^2\right)\left(5xy+3x^2y^2\right)=x^2y^2\left(5-3xy\right)\left(5+3xy\right)\)
\(x^4-y^4=\left(x^2\right)^2-\left(y^2\right)^2=\left(x^2-y^2\right)\left(x^2+y^2\right)=\left(x+y\right)\left(x-y\right)\left(x^2+y^2\right)\)
\(64a^2-27=\left(8a\right)^2-25-2=\left(8a-5\right)\left(8a+5\right)-2\)
\(x^4-y^4=\left(x^2+y^2\right)\left(x^2-y^2\right)=\left(x^2+y^2\right)\left(y+x\right)\left(x-y\right)\)