Rút gọn:
\(\frac{\left(2a-1\right)^3+\left(a-1\right)^3}{a^3-\left(a-1\right)^3}\)
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\(\frac{x^4+x^3+x+1}{x^4-x^3+2x^2-x+1}=\frac{x^3\left(x+1\right)+\left(x+1\right)}{\left(x^4-x^3+x^2\right)+\left(x^2-x+1\right)}\)
\(=\frac{\left(x+1\right)\left(x^3+1\right)}{x^2\left(x^2-x+1\right)+\left(x^2-x+1\right)}\)
\(=\frac{\left(x+1\right)\left(x+1\right)\left(x^2-x+1\right)}{\left(x^2-x+1\right)\left(x^2+1\right)}=\frac{\left(x+1\right)^2}{x^2+1}\)
\(5x^2+y^2-4xy+2y+5=0\)
\(\Rightarrow\left(4x^2-4xy+y^2\right)-2\left(2x-y\right)+1+x^2+4x+4=0\)
\(\Rightarrow\left(2x-y\right)^2-2\left(2x-y\right)+1+\left(x+2\right)^2=0\)
\(\Rightarrow\left(2x-y-1\right)^2+\left(x+2\right)^2=0\)
\(\Rightarrow\hept{\begin{cases}2x-y-1=0\\x+2=0\end{cases}\Rightarrow}\hept{\begin{cases}y=-5\\x=-2\end{cases}}\)
Mình nghĩ bạn viết hơi sai đề bài.
\(x^2+xz-y^2-yz=\left(x^2-y^2\right)+xz-yz=\left(x-y\right)\left(x+y\right)+z\left(x-y\right)=\left(x-y\right)\left(x+y+z\right)\)
Tương tự: \(y^2+xy-z^2-xz=\left(y-z\right)\left(x+y+z\right)\)
\(z^2+yz-x^2-xy=\left(x+y+z\right)\left(z-x\right)\)
Khi đó:
\(P=\frac{1}{\left(y-z\right)\left(x-y\right)\left(x+y+z\right)}+\frac{1}{\left(z-x\right)\left(y-z\right)\left(x+y+z\right)}+\frac{1}{\left(x-y\right)\left(x+y+z\right)\left(z-x\right)}\)
\(=\frac{z-x+x-y+y-z}{\left(x-y\right)\left(y-z\right)\left(z-x\right)\left(x+y+z\right)}=0\)
\(\left(5x+3\right)^2-2\left(5x+3\right)\left(x+3\right)+\left(x+3\right)^2\)
Dễ thấy đây là hằng đẳng thức thứ hai với 5x + 3 là A và x + 3 là B
Do đó : \(\left(5x+3\right)^2-2\left(5x+3\right)\left(x+3\right)+\left(x+3\right)^2\)
\(=\left(5x+3-x-3\right)^2\)
\(=\left(4x\right)^2\)
\(=16x^2\)
\(\frac{\left(2a-1\right)^3+\left(a-1\right)^3}{a^3-\left(a-1\right)^3}=\frac{8a^3-12a^2+6a-1+a^3-3a^2+3a-1}{a^3-\left(a^3-3a^2+3a-1\right)}\)
\(=\frac{9a^3-15a^2+9a-2}{3a^2-3a+1}=\frac{3a\left(3a^2-3a+1\right)-2\left(3a^2-3a+1\right)}{3a^2-3a+1}=3a-2\)