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rut gon bieu thuc :
\(\left(x+1\right)^4-6\left(x+1\right)^2-\left(x^2-2\right).\left(x^2+2\right)\)
\(\left(x+1\right)^4-6\left(x+1\right)^2-\left(x^2-2\right)\left(x^2+2\right)\\ =x^4+4x^3+6x^2+4x+1-6x^2-12x-6-x^4+4\\ =4x^3-8x+5\)
a/\(\frac{10x}{5x^2}=\frac{2}{x}\)
b/\(\frac{x\left(x^2-y^2\right)}{x^2\left(x+y\right)}=\frac{\left(x-y\right)\left(x+y\right)}{x\left(x+y\right)}=\frac{x-y}{x}\)
\(A-1=\left(x+1\right)\left(x^2+1\right)...\left(x^{256}+1\right)\)
\(\Rightarrow\left(A-1\right)\left(x-1\right)=\left(x-1\right)\left(x+1\right)\left(x^2+1\right)...\left(x^{256}+1\right)\)
\(\Rightarrow\left(A-1\right)\left(x-1\right)=\left(x^2-1\right)\left(x^2+1\right)...\left(x^{256}+1\right)\)
\(\Rightarrow\left(A-1\right)\left(x-1\right)=\left(x^4-1\right)\left(x^4+1\right)...\left(x^{256}+1\right)\)
\(\Rightarrow\left(A-1\right)\left(x-1\right)=\left(x^{256}-1\right)\left(x^{256}+1\right)=x^{512}-1\)
\(\Rightarrow A-1=\dfrac{x^{512}-1}{x-1}\)
\(\Rightarrow A=\dfrac{x^{512}-1}{x-1}+1=\dfrac{x^{512}+x-2}{x-1}\)
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