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a) \(\sqrt{a^3}-\sqrt{b^3}+\sqrt{a^2b}-\sqrt{ab^2}\)
\(=a\sqrt{a}-b\sqrt{b}+a\sqrt{b}-b\sqrt{a}\)
\(=\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)-\left(\sqrt{a}-\sqrt{b}\right)\sqrt{ab}\)
\(=\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b-\sqrt{ab}\right)\)
\(=\left(\sqrt{a}-\sqrt{b}\right)\left(a+b\right)\)
b) \(x-y+\sqrt{xy^2}-\sqrt{y^3}\)
\(=\left(x-y\right)+\left(y\sqrt{x}-y\sqrt{y}\right)\)
\(=\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)+y\left(\sqrt{x}-\sqrt{y}\right)\)
\(=\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}+y\right)\)
\(2x^2-3x\sqrt{x+3}+\left(x+3\right)\)
\(=2x^2-2x\sqrt{x+3}-x\sqrt{x+3}+\left(\sqrt{x+3}\right)^2\)
\(=2x\left(x-\sqrt{x+3}\right)-\sqrt{x+3}\left(x-\sqrt{x+3}\right)\)
\(=\left(2x-\sqrt{x+3}\right)\left(x-\sqrt{x+3}\right)\)
\(2x^2-3x\sqrt{x+3}+\left(x+3\right)\)
\(=2x^2-x\sqrt{x+3}-2x\sqrt{x+3}+\left(\sqrt{x+3}\right)^2\)
\(=x\left(2x-\sqrt{x+3}\right)-\sqrt{x+3}\left(2x-\sqrt{x+3}\right)\)
\(=\left(x-\sqrt{x+3}\right)\left(2x-\sqrt{x+3}\right)\)
1) \(x\sqrt{x}+y\sqrt{y}=\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)\)
2) \(x-3=\left(\sqrt{x}-\sqrt{3}\right)\left(\sqrt{x}+\sqrt{3}\right)\)
3) \(a+b=a-\left(-b\right)=\left(\sqrt{a}-\sqrt{-b}\right)\left(\sqrt{a}+\sqrt{-b}\right)\)
p/s: chúc bạn học tốt
\(x^3+y^3+z^3-3xyz=\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3xyz\)
\(=\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2-3xy\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)
ĐKXĐ : \(x\ne0\)
Câu a :
\(A=\sqrt{\dfrac{\left(x^2-3\right)^2+12x^2}{x^2}}+\sqrt{\left(x+2\right)^2-8x}\)
\(=\sqrt{\dfrac{x^4-6x^2+9+12x^2}{x^2}}+\sqrt{x^2+4x+4-8x}\)
\(=\sqrt{\dfrac{x^4+6x^2+9}{x^2}}+\sqrt{x^2-4x+4}\)
\(=\sqrt{\dfrac{\left(x^2+3\right)^2}{x^2}}+\sqrt{\left(x-2\right)^2}\)
\(=\left|\dfrac{x^2+3}{x}\right|+\left|x-2\right|\)
\(=\left|\dfrac{x^2}{x}+\dfrac{3}{x}\right|+\left|x-2\right|\)
Câu b :
Để \(A\in Z\Leftrightarrow\left|\dfrac{x^2}{x}+\dfrac{3}{x}\right|+\left|x-2\right|\in Z\)
\(\Leftrightarrow\dfrac{3}{x}\in Z\) ( Vì \(x^2⋮x\) )
\(\Leftrightarrow\left\{{}\begin{matrix}x=-3\\x=-1\\x=1\\x=3\end{matrix}\right.\)
Vậy \(x=-3;x=-1;x=1;x=3\) thì A đạt giá trị nguyên .
Chúc bạn học tốt !!
\(x^2-16+2\left(x+4\right)\)
\(=\left(x+4\right)\left(x-4\right)+2\left(x+4\right)\)
\(=\left(x+4\right)\left(x-4+2\right)\)
\(=\left(x+4\right)\left(x-2\right)\)
\(\left(x+y+z\right)^5-x^5-y^5-z^5\)
Xét phương trình: \(\left(x+y+z\right)^5-x^5-y^5-z^5=0\)
Có nghiệm: \(x=-y;x=-z;y=-z\)
Hệ số của mũ là: 5
\(\Rightarrow\left(x+y+z\right)^5-x^5-y^5-z^5\)
\(=5\left(x+y\right)\left(y+z\right)\left(z+x\right)\left(x^2+y^2+z^2+xy+yz+xz\right)\)
Hok Tốt!!!
ta có : \(x^3-12x-y^3+6y^2-16\)
\(=x^3-\left(y^3-6y^2+12y-8\right)-12x+12y-24\)
\(=x^3-\left(y-2\right)^3-12\left(x-y+2\right)\)
\(=\left(x-y+2\right)\left(x^2+x\left(y-2\right)+\left(y-2\right)^2\right)-12\left(x-y+2\right)\)
\(=\left(x-y+2\right)\left(x^2+xy-2x+y^2-4y+4\right)-12\left(x-y+2\right)\)
\(=\left(x-y+2\right)\left(x^2+y^2+xy-2x-4y-8\right)\)