\(VT=\left[\left(x-2\right)^2+4\left(x+y+1\right)\right]\left[\left(y-2\right)^2+4\left(x+y+1\right)\right]\)

\(VT=\left(x-2\right)^2\left(y-2\right)^2+4\left(x+y+1\right)\left[\left(x-2\right)^2+\left(y-2\right)^2\right]+16\left(x+y+1\right)^2\)

\(VP=\left[4\left(x+y+1\right)-\left(x-y\right)\right]\left[4\left(x+y+1\right)+\left(x-y\right)\right]\)

\(VP=16\left(x+y+1\right)^2-\left(x-y\right)^2\)

Ta có \(VT=VP\)

\(\Leftrightarrow\left(x-2\right)^2\left(y-2\right)^2+4\left(x+y+1\right)\left[\left(x-2\right)^2+\left(y-2\right)^2\right]=-\left(x-y\right)^2\)

\(\Leftrightarrow\left(x-2\right)^2\left(y-2\right)^2+4\left(x+y+1\right)\left[\left(x-2\right)^2+\left(y-2\right)^2\right]+\left(x-y\right)^2=0\) (1)

Nhận xét:

\(\left\{{}\begin{matrix}\left(x-y\right)^2\ge0\\\left(x-2\right)^2\left(y-2\right)^2\ge0\\x;y\ge0\Rightarrow4\left(x+y+1\right)>0\Rightarrow4\left(x+y+1\right)\left[\left(x-2\right)^2+\left(y-2\right)^2\right]\ge0\end{matrix}\right.\)

Vậy (1) xảy ra khi và chỉ khi:

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

Vậy phương trình đã cho có nghiệm duy nhất \(x=y=2\)

Bài 2:

1: \(\left(2x-1\right)^2-4\left(2x-1\right)=0\)

=>\(\left(2x-1\right)\left(2x-1-4\right)=0\)

=>(2x-1)(2x-5)=0

=>\(\left[{}\begin{matrix}2x-1=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)

2: \(9x^3-x=0\)

=>\(x\left(9x^2-1\right)=0\)

=>x(3x-1)(3x+1)=0

=>\(\left[{}\begin{matrix}x=0\\3x-1=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)

3: \(\left(3-2x\right)^2-2\left(2x-3\right)=0\)

=>\(\left(2x-3\right)^2-2\left(2x-3\right)=0\)

=>(2x-3)(2x-3-2)=0

=>(2x-3)(2x-5)=0

=>\(\left[{}\begin{matrix}2x-3=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)

4: \(\left(2x-5\right)\left(x+5\right)-10x+25=0\)

=>\(2x^2+10x-5x-25-10x+25=0\)

=>\(2x^2-5x=0\)

=>\(x\left(2x-5\right)=0\)

=>\(\left[{}\begin{matrix}x=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{5}{2}\end{matrix}\right.\)

Bài 1:

1: \(3x^3y^2-6xy\)

\(=3xy\cdot x^2y-3xy\cdot2\)

\(=3xy\left(x^2y-2\right)\)

2: \(\left(x-2y\right)\left(x+3y\right)-2\left(x-2y\right)\)

\(=\left(x-2y\right)\cdot\left(x+3y\right)-2\cdot\left(x-2y\right)\)

\(=\left(x-2y\right)\left(x+3y-2\right)\)

3: \(\left(3x-1\right)\left(x-2y\right)-5x\left(2y-x\right)\)

\(=\left(3x-1\right)\left(x-2y\right)+5x\left(x-2y\right)\)

\(=(x-2y)(3x-1+5x)\)

\(=\left(x-2y\right)\left(8x-1\right)\)

4: \(x^2-y^2-6y-9\)

\(=x^2-\left(y^2+6y+9\right)\)

\(=x^2-\left(y+3\right)^2\)

\(=\left(x-y-3\right)\left(x+y+3\right)\)

5: \(\left(3x-y\right)^2-4y^2\)

\(=\left(3x-y\right)^2-\left(2y\right)^2\)

\(=\left(3x-y-2y\right)\left(3x-y+2y\right)\)

\(=\left(3x-3y\right)\left(3x+y\right)\)

\(=3\left(x-y\right)\left(3x+y\right)\)

6: \(4x^2-9y^2-4x+1\)

\(=\left(4x^2-4x+1\right)-9y^2\)

\(=\left(2x-1\right)^2-\left(3y\right)^2\)

\(=\left(2x-1-3y\right)\left(2x-1+3y\right)\)

8: \(x^2y-xy^2-2x+2y\)

\(=xy\left(x-y\right)-2\left(x-y\right)\)

\(=\left(x-y\right)\left(xy-2\right)\)

9: \(x^2-y^2-2x+2y\)

\(=\left(x^2-y^2\right)-\left(2x-2y\right)\)

\(=\left(x-y\right)\left(x+y\right)-2\left(x-y\right)\)

\(=\left(x-y\right)\left(x+y-2\right)\)

a) \(3x\left(x-2\right)-5x\left(1-x\right)-8\left(x^2-3\right)\)

\(=3x^2-6x-5x+5x^2-8x^2+24\)

\(=24-11x\)

b) \(\left(4x^2-3y\right)\cdot2y-\left(3x^2-4y\right)\cdot3y\)

\(=8x^2y-6y^2-9x^2y+12y^2\)

\(=6y^2-x^2y\)

c) \(3y^2\left[\left(2x-1\right)+y+1\right]-y\left(1-y-y^2\right)+y\)

\(=3y^2\cdot\left(2x-1+y+1\right)-y\cdot\left(1-y-y^2\right)+y\)

\(=6xy^2-3y^2+3y^3+3y^2-y+y^2+y^3+y\)

\(=4y^3+y^2+6xy^2\)

a: \(\Leftrightarrow\left\{{}\begin{matrix}8x-4y+12-3x+6y-9=48\\9x-12y+9+16x-8y-36=48\end{matrix}\right.\)

=>5x+2y=48-12+9=45 và 25x-20y=48+36-9=48+27=75

=>x=7; y=5

b: \(\Leftrightarrow\left\{{}\begin{matrix}6x+6y-2x+3y=8\\-5x+5y-3x-2y=5\end{matrix}\right.\)

=>4x+9y=8 và -8x+3y=5

=>x=-1/4; y=1

c: \(\Leftrightarrow\left\{{}\begin{matrix}-4x-2+1,5=3y-6-6x\\11,5-12+4x=2y-5+x\end{matrix}\right.\)

=>-4x-0,5=-6x+3y-6 và 4x-0,5=x+2y-5

=>2x-3y=-5,5 và 3x-2y=-4,5

=>x=-1/2; y=3/2

e: \(\Leftrightarrow\left\{{}\begin{matrix}x\cdot2\sqrt{3}-y\sqrt{5}=2\sqrt{3}\cdot\sqrt{2}-\sqrt{5}\cdot\sqrt{3}\\3x-y=3\sqrt{2}-\sqrt{3}\end{matrix}\right.\)

=>\(x=\sqrt{2};y=\sqrt{3}\)

a, \(3x\left(x-2\right)-5x\left(1-x\right)-8\left(x^2-2\right)\)

\(=3x^2-6x-5x+5x^2-8x^2+16\)

\(=-11x+16\)

b, \(\left(4x^2-3y\right)2y-\left(3x^2-4y\right)3y\)

\(=8x^2y-6y^2-\left(9x^2y-12y^2\right)\)

\(=8x^2y-6y^2-9x^2y+12y^2=-x^2y+6y^2\)

c, \(3y^2\left[\left(2y-1\right)+y+1\right]-y\left(1-y-y^2\right)+y\)

\(=3y^2.3y-y+y^2+y^3+y\)

\(=9y^3+y^2+y^3=10y^3+y^2\)

Chúc bạn học tốt!!!

a, \(3x\left(x-2\right)-5x\left(1-x\right)-8\left(x^2-2\right)\)

\(=3x^2-6x-5x+5x^2-8x^2+16\)

\(=-11x+16\)

b, \(\left(4x^2-3y\right)2y-\left(3x^2-4y\right)3y\)

\(=8x^2y-6y^2-9x^2y+12y^2\)

\(=-x^2y+6y^2\)

c, \(3y^2\left[\left(2y-1\right)+y+1\right]-y\left(1-y-y^2\right)+y\)

\(=3y^2.3y-y\left(1-y-y^2-1\right)\)

\(=9y^3-y\left(-y-y^2\right)\)

\(=9y^3+y^2+y^3=10y^3+y^2\)

a) \(3x\left(x-2\right)-5x\left(1-x\right)-8\left(x^2-2\right)\)

\(=3x^2-6x-5x+5x^2-8x^2+16\)

\(=3x^2+5x^2-8x^2-6x-5x+16\)

\(=-11x+16\)

b) \(\left(4x^2-3y\right)2y-\left(3x^2-4y\right).3y\)

\(=9x^2y-6y^2-\left(9x^2y-12y^2\right)\)

\(=9x^2y-6y^2-9x^2y+12y^2\)

\(=9x^2y-9x^2y-6y^2+12y^2\)

\(=6y^2\)

c) \(3y^2\left[\left(2y-1\right)+y+1\right]-y\left(1-y-y^2\right)+y\)

\(=3y^2\left(2y-1\right)+3y^3+3y^2-y+y^2+y^3+y\)

\(=6y^3-3y^2+3y^3+3y^2-y+y^2+y^3+y\)

\(=6y^3+3y^3+y^3-3y^2+3y^2+y^2-y+y\)

\(=10y^3+y^2\)

a)\(\left\{{}\begin{matrix}8x+2y=4\\8x+3y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=1\\4x+1=2\end{matrix}\right.\Leftrightarrow}\left\{{}\begin{matrix}y=1\\x=\frac{1}{4}\end{matrix}\right.\)b)

\(\left\{{}\begin{matrix}12x-8y=44\\12x-15y=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7y=35\\4x-5y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=5\\4x-5.5=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=5\\x=7\end{matrix}\right.\)c)\(\left\{{}\begin{matrix}9x=-18\\4x+3y=13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\4.\left(-2\right)+3y=13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=7\end{matrix}\right.\)

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