5.

\(4x^5y^2+8x^4y^3+4x^3y^4=4x^3y^2(x^2+2xy+y^2)\)

\(=4x^3y^2(x+y)^2\)

9.

\(4x^5y^2+16x^4y^2-6x^3y^2=2x^3y^2(2x^2+4x-3)\)

13.

\(-3x^4y+6x^3y-3x^2y=-3x^2y(x^2-2x+1)=-3x^2y(x-1)^2\)

17.

\(8x^3-8x^2y+2xy^2=2x(4x^2-4xy+y^2)\)

\(=2x[(2x)^2-2.2x.y+y^2]=2x(2x-y)^2\)

21.

\((a^2+4)^2-16a^2b^2=(a^2+4)^2-(4ab)^2\)

\(=(a^2+4-4ab)(a^2+4+4ab)\)

25.

\(100a^2-(a^2+25)^2=(10a)^2-(a^2+25)^2\)

\(=(10a-a^2-25)(10a+a^2+25)\)

\(=-(a^2-10a+25)(a^2+10a+25)=-(a-5)^2(a+5)^2\)

29.

\(25a^2b^2-4x^2+4x-1=25a^2b^2-(4x^2-4x+1)\)

\(=(5ab)^2-(2x-1)^2=(5ab-2x+1)(5ab+2x-1)\)

33.

\(1-2m+m^2-x^2-4x-4=(m^2-2m+1)-(x^2+4x+4)\)

\(=(m-1)^2-(x+2)^2=[(m-1)-(x+2)][(m-1)+(x+2)]\)

\(=(m-x-3)(m+x+1)\)

37.

\(ax^2+bx^2+2xy(a+b)+ay^2+by^2\)

\(=x^2(a+b)+2xy(a+b)+y^2(a+b)\)

\(=(a+b)(x^2+2xy+y^2)=(a+b)(x+y)^2\)

41.

\(5a^2-5=5(a^2-1)=5(a^2-1^2)=5(a-1)(a+1)\)

a) x²+y²-4x+4y+8=0

⇔ (x-2)²+(y+2)²=0

\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\y+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-2\end{matrix}\right.\)

b)5x²-4xy+y²=0

⇔ x²+(2x-y)²=0

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\2x-y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)

c)x²+2y²+z²-2xy-2y-4z+5=0

⇔ (x-y)²+(y-1)²+(z-2)²=0

\(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y-1=0\\z-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y=1\\z=2\end{matrix}\right.\)

b: Ta có: \(5x^2-4xy+y^2=0\)

\(\Leftrightarrow x^2-\dfrac{4}{5}xy+y^2=0\)

\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{2}{5}y+\dfrac{4}{25}y^2+\dfrac{21}{25}y^2=0\)

\(\Leftrightarrow\left(x-\dfrac{2}{5}y\right)^2+\dfrac{21}{25}y^2=0\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)

d)3x²+3y²+3xy-3x+3y+3=0

⇔ 6x²+6y²+6xy-6x+6y+6=0

⇔ 3(x+y)²+3(x-1)²+3(y+1)²=0

\(\Leftrightarrow\left\{{}\begin{matrix}x+y=0\\x-1=0\\y+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)

\(\left\{{}\begin{matrix}x^2+2xy-3y^2=-4\left(1\right)\\2x^2+xy+4y^2=5\left(2\right)\end{matrix}\right.\)\(với\)\(y=0\Rightarrow hpt\Leftrightarrow\left\{{}\begin{matrix}x^2=-4\\2x^2=5\end{matrix}\right.\)\(\left(loại\right)\)

\(y\ne0\) \(đặt:x=t.y\Rightarrow hpt\Leftrightarrow\left\{{}\begin{matrix}t^2y^2+2ty^2-3y^2=-4\left(3\right)\\2t^2y^2+ty^2+4y^2=5\left(4\right)\end{matrix}\right.\)

\(\Leftrightarrow5t^2y^2+10ty^2-15y^2=-8t^2y^2-4ty^2-16y^2\)

\(\Leftrightarrow13t^2y^2+14ty^2+y^2=0\)

\(\Leftrightarrow13t^2+14t+1=0\Leftrightarrow\left[{}\begin{matrix}t=-\dfrac{1}{13}\\t=-1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{13}y\left(5\right)\\x=-y\left(6\right)\end{matrix}\right.\)

\(thay\left(5\right)và\left(6\right)\) \(lên\left(1\right)hoặc\left(2\right)\Rightarrow\left(x;y\right)=\left\{\left(1;-1\right);\left(-1;1\right);\left(-\dfrac{1}{\sqrt{133}};\dfrac{13}{\sqrt{133}}\right)\right\}\)

\(pt:x^4-4x^3+x^2+6x+m+2=0\)

\(\Leftrightarrow x^4-4x^3+4x^2-3x^2+6x+m+2=0\)

\(\Leftrightarrow\left(x^2-2x\right)^2-3\left(x^2-2x\right)+m+2=0\left(1\right)\)

\(đặt:x^2-2x=t\ge-1\)

\(\Rightarrow\left(1\right)\Leftrightarrow t^2-3t=-m-2\)

\(xét:f\left(t\right)=t^2-3t\) \(trên[-1;+\text{∞})\) \(và:y=-m-2\)

\(\Rightarrow f\left(-1\right)=4\)

\(f\left(-\dfrac{b}{2a}\right)=-\dfrac{9}{4}\)

\(\left(1\right)\) \(có\) \(3\) \(ngo\) \(pb\Leftrightarrow-m-2=4\Leftrightarrow m=-6\)

5x²+2xy+y²-4x-40=0

<=>(x+y)²=4(10+x-x²)

<=>x+y=2\(\sqrt{10+x-x^2}\)

( mik k ghi đề nhé bn)

a) (2x)^3 - y^3 + (2x)^3 + y^3 - 16x^3 + 16xy = 16

=>  8x^3 - y^3 + 8x^3 + y^3 - 16x^3 + 16xy = 16

=>  16xy = 16

=>  xy = 1

Vì x, y nguyên => x = 1, y = 1       hoặc x = -1, y = -1

mik xin lỗi nha, mik chỉ bt làm câu a

uk thank bạn