RÚT GỌN MỘT CÁCH NHANH NHẤT
\(\left(y+1\right)^4+\left(y-1\right)^4\)
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Đặt \(3x-1=y,x+2=z\)
\(\Rightarrow y^2-2yz+z^2=\left(y-z\right)^2\)
\(=\left(3x-1-x-2\right)^2=\left(2x-3\right)^2\)
\(\left(x-1\right)^3+4\left(x+1\right)\left(1-x\right)+3\left(x-1\right)\left(x^2+x+1\right).\)
\(=\left(x-1\right)^3+4\left(x+1\right)\left(1-x\right)+3\left(x-1\right)^3.\)
\(=\left(x-1\right)^3+4\left(1-x^2\right)+3\left(x-1\right)^3.\)
\(=\left(x-1\right)^3+3\left(x-1\right)^3+4\left(1-x^2\right)\)
\(=4\left(x-1\right)^3+4\left(1-x^2\right)\)
\(=4\left[\left(x-1\right)^3+\left(1-x^2\right)\right]\)
a) \(x\left(x^2-16\right)-\left(x^2+1\right)\left(x-1\right)\) =\(x^3-16x^2-x^3+x^2-x+1\)
= \(x^2-17x+1\)
b) \(\left(y^2-9\right)\left(y^2+9\right)-\left(y^4-4\right)\) = \(\left(y^4-81\right)-\left(y^4-16\right)\)
=\(-65\)
Ta có \(x-y=1\)
\(=>x+y=\left(x+y\right).\left(x-y\right)\)
\(A=\left(x+y\right).\left(x-y\right).\left(x^2+y^2\right).\left(x^4+y^4\right)\)
\(A=\left(x^2-y^2\right).\left(x^2+y^2\right).\left(x^4+y^4\right)\)
\(A=\left(x^4-y^4\right).\left(x^4+y^4\right)\)
\(A=x^8-y^8\)
= \(-\left[\left(x-y\right)\left(x^2-y^2\right)\left(x^4-y^4\right)\left(x^8-y^8\right)\left(x^{16}-y^{16}\right)\right]\)
= \(-\left[\left(x-y\right)\left(x-y\right)^2\left(x-y\right)^4\left(x-y\right)^8\left(x-y\right)^{16}\right]\)
= \(-\left(1\cdot1^2\cdot1^4\cdot1^8\cdot1^{16}\right)\)
= -1
\(\left(y+1\right)^4+\left(y-1\right)^4=\left(y+1\right)^4+\left[\left(y-1\right)^2\right]^2=\left(y+1\right)^4+\left[\left(y+1\right)\left(y-1\right)\right]^2=\left(y+1\right)^2\left[\left(y+1\right)^2+\left(y-1\right)^2\right]=2\left(y+1\right)^2\left(y^2+1\right)\)