Phân tích đa thức thành nhân tử : x^3+10x-y^2+25
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\(x^4+2x^3+10x-25\)
\(=x^4+5x^2+2x^3+10x-5x^2-25\)
\(=\left(x^2+5\right)\left(x^2+2x-5\right)\)
d: \(=\left(x+1-5\right)\left(x+1+5\right)=\left(x-4\right)\left(x+6\right)\)
a.\(x^2y-xz+z-y=\)\(\left(x^2y-y\right)-\left(xz-z\right)=\)\(y\left(x^2-1\right)-z\left(x-1\right)\)
\(y\left(x+1\right)\left(x-1\right)-z\left(x-1\right)\)=\(\left(x-1\right)\left(xy+y-z\right)\)
b.\(x^4-x^3+x^2-1=x^3\left(x-1\right)+\left(x+1\right)\left(x-1\right)\)=\(\left(x-1\right)\left(x^3+x+1\right)\)
c.\(x^4-x^2+10x-25=x^4-\left(x^2-10x+25\right)\)=\(\left(x^2\right)^2-\left(x-5\right)^2=\left(x^2+x-5\right)\left(x^2-x+5\right)\)
a,
\(y^2-x^2+10x-25\)
\(=y^2-\left(x^2-10x+25\right)\)
\(=y^2-\left(x-5\right)^2\)
\(=\left(y+x-5\right)\left(y-x+5\right)\)
a) \(y^2-x^2+10x-25=y^2-\left(x^2-10x+25\right)=y^2-\left(x^2-2.x.5+5^2\right)\)
\(=y^2-\left(x-5\right)^2=\left(y-x+5\right).\left(y+x-5\right)\)
b) \(\left(3x+1\right)^2=3x+1\Rightarrow\left(3x-1\right)^2-\left(3x+1\right)=0\)
\(\Rightarrow\left(3x+1\right)\left(3x+1-1\right)=0\Rightarrow\left(3x+1\right).3x=0\)
\(\Rightarrow\orbr{\begin{cases}3x+1=0\\3x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{-1}{3}\\x=0\end{cases}}}\)