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\(a^3+b^3=\left(a+b\right)\left[\left(a-b\right)^2+ab\right]\)
\(BĐVT,VT=\left(a+b\right)\left[\left(a-b\right)^2+ab\right]\)
\(=\left(a+b\right)\left(a^2-2ab+b^2+ab\right)\)
\(=\left(a+b\right)\left(a^2-ab+b^2\right)\)
\(=a^3+b^3=VP\)
\(\text{Vậy }a^3+b^3=\left(a+b\right)\left[\left(a-b\right)^2+ab\right]\)
Câu hỏi của nguyen cao long - Toán lớp 8 - Học toán với OnlineMath
\(x^2+4y^2+z^2-2x-6z+8y+15\)
\(=\left(x^2-2x+1\right)+\left(4y^2+8y+4\right)+\left(z^2-6z+9\right)+1\)
\(=\left(x-1\right)^2+4\left(y+1\right)^2+\left(z-3\right)^2+1\ge0\)
=>đpcm
x2+4y2+z2-2x-6z+8y+15
=x2+4y2+z2-2x-6z+8y+1+1+4+9
=(x2-2x+1)+(4y2+8y+4)+(z2-6z+9)+1
=(x-1)2+4(y+1)2+(z-3)2+1
Ta thấy:\(\begin{cases}\left(x-1\right)^2\\4\left(y+1\right)^2\\\left(z-3\right)^2\end{cases}\ge0\)
\(\Rightarrow\left(x-1\right)^2+4\left(y+1\right)^2+\left(z-3\right)^2\ge0\)
\(\Rightarrow\left(x-1\right)^2+4\left(y+1\right)^2+\left(z-3\right)^2+1\ge0+1=1>0\)
Đpcm
\(B=\frac{1}{x^4}+\frac{1}{y^4}=\frac{y^4}{x^4.y^4}+\frac{x^4}{x^4.y^4}=\frac{x^4+y^4}{x^4y^4}=\frac{\left(x^2+y^2\right)^2-2x^2y^2}{x^4y^4}=\frac{\left[\left(x+y\right)^2-2xy\right]^2-2x^2y^2}{x^4y^4}\)
\(=\frac{\left[20^2-2.\left(-44\right)\right]^2-2.\left(-44\right)^2}{\left(-44\right)^4}=\frac{488^2+3872}{3748096}=\frac{234272}{3748096}=\frac{7321}{117128}\)
\(x^4-x^3-2x-4\)
\(=x^4-x^3-2x^2+2x^2-2x-4\)
\(=x^2\left(x^2-x-2\right)+2\left(x^2-x-2\right)\)
\(=\left(x^2-x-2\right)\left(x^2+2\right)\)
\(=\left(x^2+x-2x-2\right)\left(x^2+2\right)\)
\(=\left[x\left(x+1\right)-2\left(x+1\right)\right]\left(x^2+2\right)\)
\(=\left(x-2\right)\left(x+1\right)\left(x^2+2\right)\)
