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\(C=\left(x^2-1\right)\left(x^2+1\right)\left(x^4+1\right)\left(x^8+1\right)\left(x^{16}+1\right)\left(x^{32}+1\right)-x^{64}\\ =\left(x^4-1\right)\left(x^4+1\right)\left(x^8+1\right)\left(x^{16}+1\right)\left(x^{32}+1\right)-x^{64}\\ =\left(x^8-1\right)\left(x^8+1\right)\left(x^{16}+1\right)\left(x^{32}+1\right)-x^{64}\\ =\left(x^{16}-1\right)\left(x^{16}+1\right)\left(x^{32}+1\right)-x^{64}\\ =\left(x^{32}-1\right)\left(x^{32}+1\right)-x^{64}\\ =\left(x^{64}-1\right)-x^{64}\\ =-1\)
Vậy đa thức ko phụ thuộc vào x
\(C=(x^2-1)(x^2+1)(x^4+1)(x^8+1)(x^{16}+1)(x^{32}+1)-x^{64}\\=(x^4-1)(x^4+1)(x^8+1)(x^{16}+1)(x^{32}+1)-x^{64}\\=(x^8-1)(x^8+1)(x^{16}+1)(x^{32}+1)-x^{64}\\=(x^{16}-1)(x^{16}+1)(x^{32}+1)-x^{64}\\=(x^{32}-1)(x^{32}+1)-x^{64}\\=x^{64}-1-x^{64}\\=-1\)
⇒ Giá trị của C không phụ thuộc vào giá trị của biến
\(a,=x+x^2-x^3+x^4-x^5+1+x-x^2+x^3-x^4-x-x^2+x^3-x^4+x^5+1+x-x^2+x^3-x^4\\ =2x-2x^2+2x^3-2x^4\)
1: \(=\dfrac{x-1}{x^2+x+1}+\dfrac{x+1}{x-1}\)
\(=\dfrac{x^2-2x+1+x^3+x^2+x^2+x+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{x^3+3x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}\)
2: \(=\dfrac{\left(x^2-y^2\right)^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}=\dfrac{\left(x-y\right)\left(x+y\right)^2}{x^2+xy+y^2}\)
1)
a) \(=3x^2\left(x^2-1\right)-\left(x^3-1\right)+x^8-3x^4+3x^2-1\)
\(=3x^4-3x^2-x^3+1+x^8-3x^4+3x^2-1=x^8-x^3\)
2)
\(=\left(x^2+5x-6\right)\left(x^2+5x+6\right)-6\left(x^2+5x\right)+45\)
\(=\left(x^2+5x\right)^2-6\left(x^2+5x\right)-36+45\)
\(=\left(x^2+5x\right)^2-6\left(x^2+5x\right)+9=\left(x^2+5x-3\right)^2\)
Bài 2:
a.
\(3x(x-4y)-\frac{12}{5}y(y-5x)=3x^2-12xy-\frac{12}{5}y^2+12xy\)
\(=3x^2-\frac{12}{5}y^2=3.4^2-\frac{12}{5}.(-5)^2=-12\)
b.
\(u=\frac{-1}{3}; v=\frac{-2}{3}\Rightarrow u+v+1=0\)
\(2u(1+u-v)-v(1-2u+v)=2u(1+u+v-2v)+v(1+u+v-3u)\)
\(=2u.(-2v)+v(-3u)=-4uv-3uv=-7uv=-7.\frac{-1}{3}.\frac{-2}{3}=\frac{-14}{9}\)
Bài 1:
\(A=x^6-(x^6-x^5)-(x^5+x^4)+(x^4-x^3)+(x^3+x^2)-(x^2+x)+1\)
\(=-x+1=-(x-1)=-(999-1)=-998\)
\(\left(1-x\right)\left(1+x\right)\left(1+x^2\right)\left(1+x^4\right)\left(1+x^8\right)\left(1+x^{16}\right)\left(1+x^{32}\right)\left(1+x^{64}\right)\)
\(=\left(1-x^2\right)\left(1+x^2\right)\left(1+x^4\right)\left(1+x^8\right)\left(1+x^{16}\right)\left(1+x^{32}\right)\left(1+x^{64}\right)\)
\(=\left(1-x^4\right)\left(1+x^4\right)\left(1+x^8\right)\left(1+x^{16}\right)\left(1+x^{32}\right)\left(1+x^{64}\right)\)
\(=\left(1-x^8\right)\left(1+x^8\right)\left(1+x^{16}\right)\left(1+x^{32}\right)\left(1+x^{64}\right)\)
\(=\left(1-x^{16}\right)\left(1+x^{16}\right)\left(1+x^{32}\right)\left(1+x^{64}\right)\)
\(=\left(1-x^{32}\right)\left(1+x^{32}\right)\left(1+x^{64}\right)\)
\(=\left(1-x^{64}\right)\left(1+x^{64}\right)\)
\(=1-x^{128}\)