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\(a,\left(3x+1\right)^3=9x^3+9x^2+9x+1\)
\(b,\left(\frac{2}{3}x+1\right)^2=\frac{4}{9}x^2+\frac{4}{3}x+1\)
\(c,\left(x-y\right)^2-\left(x+y\right)^2=\left(x-y-x-y\right)\left(x-y+x+y\right)=-2y\cdot2x=-4xy\)
\(d,\left(x+y\right)^2-\left(x-y\right)^2=\left(x+y-x+y\right)\left(x+y+x-y\right)=2y\cdot2x=4xy\)
\(\left(x+y\right)^2+2\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2=\left[\left(x+y\right)+\left(x-y\right)\right]^2=\left[x+y+x-y\right]^2=\left(2x\right)^2=4x^2\)
\(\left(a-x-y\right)^3-\left(a+x-y\right)^3\)
\(=\left[\left(a-x-y\right)-\left(a+x-y\right)\right]\left[\left(a-x-y\right)^2+\left(a-x-y\right)\left(a+x-y\right)+\left(a+x-y\right)^2\right]\)
\(=-2x.\left[a^2+x^2+y^2-2ax+2xy-2ay+\left(a-y\right)^2-x^2+a^2+x^2+y^2+2ax-2xy-2ay\right]\)
\(=-2x\left[a^2+x^2+y^2-2ax+2xy-2ay+a^2-2ay+y^2-x^2+a^2+x^2+y^2+2ax-2xy-2ay\right]\)
\(=-2x\left(3a^2+x^2+3y^2-4ay\right)\)
a: =-3x^2y*x^2y+3x^2y*2xy
=-3x^4y^2+6x^3y^2
b: =x^3-x^2y+x^2y+y^2=x^3+y^2
c: =x*4x^3-x*5xy+2x*x
=4x^4-5x^2y+2x^2
d: =x^3+x^2y+2x^3+2xy
=3x^3+x^2y+2xy
\(a\text{) }pt\Leftrightarrow\left(y^2+2y+1\right)+\left[\left(2^x\right)^2-2.2^x+1\right]=0\)
\(\Leftrightarrow\left(y+1\right)^2+\left(2^x-1\right)^2=0\)
\(\Leftrightarrow y+1=0\text{ và }2^x-1=0\)
\(\Leftrightarrow y=-1\text{ và }x=0\)
\(b\text{) }pt\Leftrightarrow\left(4x^2+4y^2+8xy\right)+\left(x^2-2x+1\right)+\left(y^2+2y+1\right)=0\)
\(\Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\)
\(\Leftrightarrow x+y=0\text{ và }x-1=0\text{ và }y+1=0\)
\(\Leftrightarrow x=1\text{ và }y=-1\)
\(\left(c^2+b^2-5\right)^2-4\left(ab+2\right)^2\)
\(=\left(a^2+b^2-5-2ab-4\right)\left(a^2+b^2-5+2ab+4\right)\)
\(=\left(a^2+b^2-2ab-9\right)\left(a^2+b^2+2ab-1\right)\)
\(=\left[\left(a-b\right)^2-9\right]\left[\left(a+b\right)^2-1\right]\)
\(=\left(a-b-3\right)\left(a-b+3\right)\left(a+b-1\right)\left(a+b+1\right)\)
Ta có:
\(\left(x^2+y^2\right)^2-4x^2y^2-\left(x+y\right)^2\left(x-y\right)^2.\)
\(=x^4+2.x^2.y^2+y^4-4x^2y^2-\left[\left(x+y\right)\left(x-y\right)\right]^2\)
\(=x^4+2.x^2.y^2+y^4-4x^2y^2-\left[x^2-y^2\right]^2\)
\(=x^4+2x^2y^2+y^4-4x^2y^2-\left(x^4-2x^2y^2+y^4\right)\)
\(=x^4+2x^2y^2+y^4-4x^2y^2-x^4+2x^2y^2-y^4\)
\(=0\)
Vậy \(\left(x^2+y^2\right)^2-4x^2y^2=\left(x+y\right)^2\left(x-y\right)^2.\)