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1) \(x^6+1\)
\(=x^6+x^4-x^4+x^2-x^2+1\)
\(=\left(x^6-x^4+x^2\right)+\left(x^4-x^2+1\right)\)
\(=x^2\left(x^4-x^2+1\right)+\left(x^4-x^2+1\right)\)
\(=\left(x^2+1\right)\left(x^4-x^2+1\right)\)
2) \(x^6-y^6\)
\(=\left(x^3+y^3\right)\left(x^3-y^3\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)\left(x-y\right)\left(x^2+xy+y^2\right)\)
a) \(\left(x+2y\right)^2=x^2+4xy+4y^2\)
b) \(\left(3x-\frac{1}{8}y\right)^2=9x^2-\frac{3}{4}xy+\frac{1}{64}y^2\)
c) \(\left(-6x-\frac{2}{5}\right)^2=36x^2+\frac{24}{5}x+\frac{4}{25}\)
d) \(\left(xy^2+1\right)\left(xy^2-1\right)=x^2y^4-1\)
e) \(\left(x-y\right)^2\left(x+y\right)^2=\left(x^2-y^2\right)^2=x^4-2x^2y^2+y^4\)
f) \(\left(\frac{1}{2}x-\frac{1}{3}y-1\right)^2=\frac{1}{4}x^2+\frac{1}{9}y^2+1-\frac{1}{3}xy-x+\frac{2}{3}y\)
1) Ta có: \(\left(x+y+2\right)^2\)
\(=x^2+y^2+4+2xy+2\cdot x\cdot2+2\cdot y\cdot2\)
\(=x^2+y^2+4+2xy+4x+4y\)
2) Ta có: \(\left(x-2y+3\right)^2\)
\(=x^2+4y^2+9-2\cdot x\cdot2y+2\cdot x\cdot3-2\cdot2y\cdot3\)
\(=x^2+4y^2+9-4xy+6x-12y\)
3) Ta có: \(\left(x^2-y-4\right)^2\)
\(=x^4+y^2+16+2\cdot x^2\cdot\left(-y\right)+2\cdot x^2\cdot\left(-4\right)+2\cdot\left(-y\right)\cdot\left(-4\right)\)
\(=x^4+y^2+16-2x^2y-8x^2+8y\)
4) Ta có: \(100x^2-\left(x^2+25\right)\)
\(=100x^2-x^2-25\)
\(=99x^2-25\)
5) Ta có: \(\left(x-3\right)^2-16\)
\(=x^2-6x+9-16\)
\(=x^2-6x-7\)
a, \(\left(x+2\right)^2=x^2+4x+2^2=x^2+4x+4\)
b, \(\left(x-1\right)^2=x^2-2x+1\)
c, \(\left(x^2+y^2\right)^2=x^4+2x^2y^2+y^4\)
Dựa vào công thức làm nốt nhé
a) ( x + 2 )2 = x2 + 4x + 4
b) ( x - 1 )2 = x2 - 2x + 1
c) ( x2 + y2 )2 = x4 + 2x2y2 + y4
d) ( x3 + 2y2 )2 = x6 + 4x3y2 + 4y4
e) ( x2 - y2 )2 = x4 - 2x2y2 + y4
f) ( x - y2 )2 = x2 - 2xy2 + y4
Ta có: \(x^2\cdot\left(x^4+25\right)\cdot\left(x^2-5\right)\cdot\left(x^2+5\right)\cdot\left(x-y\right)\left(x^2+xy+y^2\right)\cdot\left(x^3+y^3\right)\)
\(=x^2\cdot\left(x^4+25\right)\left(x^4-25\right)\cdot\left(x^3-y^3\right)\left(x^3+y^3\right)\)
\(=x^2\cdot\left(x^8-625\right)\cdot\left(x^6-y^6\right)\)