\((x^2-4x+3):(x-3)\)
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c: \(=\dfrac{x^3+2x+2x^2+2x+x^2-x+1}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{x^3+3x^2+3x+1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{x^2+2x+1}{x^2-x+1}\)
{4x-2(x-3)-3[x-3(4-2x)+8]}.(-3x)
={4x-2x+6-3[x-12+6x+8]}.(-3x)
={2x+6-3x+36-18x-24}.(-3x)
=(-19x+18).(-3x)
=57x2-54x
\(A=4x^2+12xy+9y^2\)
\(B=25x^2-10xy+y^2\)
\(C=8x^3+12x^2y^2+6xy^4+y^6\)
\(D=\left(x^2\right)^2-\left(\dfrac{2}{5}y\right)^2=x^4-\dfrac{4y^2}{25}\)
\(E=x^3-27y^3\)
\(F=x^6-27\)
1.
\(\sqrt{50}-3\sqrt{8}+\sqrt{32}=5\sqrt{2}-6\sqrt{2}+4\sqrt{2}=3\sqrt{2}\)
2.
a, ĐK: \(x\in R\)
\(pt\Leftrightarrow\sqrt{\left(x-2\right)^2}=1\)
\(\Leftrightarrow\left|x-2\right|=1\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
b, ĐK: \(x\ge3\)
\(pt\Leftrightarrow\sqrt{x-3}\left(\sqrt{x}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-3}=0\\\sqrt{x}-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=1\left(l\right)\end{matrix}\right.\)
Dễ
(\(x^2-4x+3):\left(x-3\right)\)
= [\(\left(x^2-x\right)+\left(3x-3\right)\)] : (\(x\) - 3)
= [\(x\)(\(x\) - 1) + 3(\(x\) - 1)] : (\(x\) - 3)
= (\(x\) - 1)(\(x\) + 3) : (\(x\) - 3)
= (\(x\) - 1).[(\(x\) + 3) : (\(x\) - 3)]
= \(x\) - 1