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21 tháng 10 2021

\(=5\left(x^2+2x+1-y^2\right)=5\left[\left(x+1\right)^2-y^2\right]\\ =5\left(x+y+1\right)\left(x-y+1\right)\)

21 tháng 10 2021

5x2+10x+5-5y2=5(x2+2x+1-y2)

16 tháng 10 2021

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

\(x^2+6x-5x-30=\left(x-5\right)\left(x+6\right)\)

 

31 tháng 10 2021

1.\(=5\left(x^2-2xy+y^2-4z^2\right)=5\left[\left(x+y\right)^2-\left(2z\right)^2\right]=5\left(x+y-2z\right)\left(x+y+2z\right)\)

2. \(=\left(-5x^2+15x\right)+\left(x-3\right)=-5x\left(x-3\right)+\left(x-3\right)=\left(1-5x\right)\left(x-3\right)\)

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

4.\(=3\left(x^2-2xy+y^2-4z^2\right)=3\left[\left(x-y\right)^2-\left(2z\right)^2\right]=3\left(x-y-2z\right)\left(x-y+2z\right)\)

5. \(=\left(x^2+x\right)+\left(3x+3\right)=x\left(x+1\right)+3\left(x+1\right)=\left(x+1\right)\left(x+3\right)\)

6. \(=\left(x^2-2x+1\right)\left(x^2+2x+1\right)=\left(x-1\right)^2\left(x+1\right)^2\)

7. \(=\left(x^2+x\right)-\left(5x+5\right)=x\left(x+1\right)-5\left(x+1\right)=\left(x-5\right)\left(x+1\right)\)

31 tháng 10 2021

\(1,=5\left[\left(x-y\right)^2-4z^2\right]=5\left(x-y-2z\right)\left(x-y+2z\right)\\ 2,=-5x^2+15x+x-3=\left(x-3\right)\left(1-5x\right)\\ 3,=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)=\left(x-y\right)\left(x+y-5\right)\\ 4,=3\left[\left(x-y\right)^2-4z^2\right]=3\left(x-y-2z\right)\left(x-y+2z\right)\\ 5,=x^2+x+3x+3=\left(x+3\right)\left(x+1\right)\\ 6,=\left(x^2+2x+1\right)\left(x^2-2x+1\right)=\left(x-1\right)^2\left(x+1\right)^2\\ 7,=x^2+x-5x-5=\left(x+1\right)\left(x-5\right)\)

30 tháng 6 2019

5 x 2 - 10 x y + 5 y 2 - 20 z 2 =   5 x 2   –   2 x y   +   y 2   –   4 z 2 =   5 x   –   y 2   –   2 z 2     =   5 x   –   y   +   2 z x   –   y   –   2 z  

21 tháng 10 2021

\(5x^2-10xy+5y^2-20z^2\)

\(=5\left(x^2-2xy+y^2-4z^2\right)\)

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

21 tháng 10 2021

5(x2-2xy+y2-4-4z2)

14 tháng 12 2021

\(a,=\left(3x+1-2x-1\right)\left(3x+1+2x+1\right)=x\left(5x+2\right)\\ b,=5\left[4z^2-\left(x-y\right)^2\right]=5\left(2z-x+y\right)\left(2z+x-y\right)\)

14 tháng 12 2021

\(b,\left(3x+1\right)^2-\left(2x+1\right)^2\\ =\left[\left(3x+1\right)+\left(2x+1\right)\right]\left[\left(3x+1\right)-\left(2x+1\right)\right]\)

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

\(c,-5x^2+10xy-5y^2+20z^2\\ =-5\left(x^2-2xy+y^2-4z^2\right)\\ =-5\left[\left(x^2-2xy+y^2\right)-\left(2z\right)^2\right]\\ =-5\left[\left(x-y\right)^2-\left(2z\right)^2\right]\\ =-5\left(x-y+2z\right)\left(x-y-2z\right)\)

8 tháng 9 2021

\(a,\Leftrightarrow\left(9x^2-18x+9\right)+\left(y^2-6y+9\right)+\left(2z^2+4z+2\right)=0\\ \Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\\z=-1\end{matrix}\right.\)

\(b,\Leftrightarrow\left(4x^2+8xy+4y^2\right)+\left(x^2-2x+1\right)+\left(y^2+2y+1\right)=0\\ \Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=-y\\x=1\\y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)

\(c,\Leftrightarrow\left(4x^2+4xy+y^2\right)+\left(x^2-2x+1\right)+\left(y^2+4y+4\right)=0\\ \Leftrightarrow\left(2x+y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}2x=-y\\x=1\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

 

8 tháng 9 2021

a,9x^2+y^2+2z^2−18x+4z−6y+20=0

⇔9(x−1)^2+(y−3)^2+2(z+1)^2=0

⇔x=1;y=3;z=−1

b,5x^2+5y^2+8xy+2y−2x+2=0

⇔4(x+y)2+(x−1)2+(y+1)2=0

⇔x=−y;x=1y=−1⇔x=1y=−1

c,5x^2+2y^2+4xy−2x+4y+5=0

⇔(2x+y)^2+(x−1)^2+(y+2)^2=0

⇔2x=−y;x=1;y=−2

⇔x=1;y=−2

d,x^2+4y^2+z^2=2x+12y−4z−14

⇔(x−1)^2+(2y−3)^2+(z+2)^2=0

⇔x=1;y=3/2;z=−2

e: Ta có: x^2−6x+y2+4y+2=0

⇔x^2−6x+9+y^2+4y+4−11=0

⇔(x−3)^2+(y+2)^2=11

Dấu '=' xảy ra khi x=3 và y=-2

 

13 tháng 12 2023

Bài 1:

\(a)5x^2-10xy+5y^2-20z^2\\=5(x^2-2xy+y^2-4z^2)\\=5[(x^2-2xy+y^2)-4z^2]\\=5[(x-y)^2-(2z)^2]\\=5(x-y-2z)(x-y+2z)\\---\\b)x^2+4x+3\\=x^2+x+3x+3\\=x(x+1)+3(x+1)\\=(x+1)(x+3)\)

Bài 2:

\(a.(2x-3y)^2\\=(2x)^2-2\cdot2x\cdot3y+(3y)^2\\=4x^2-12xy+9y^2\)

25 tháng 12 2021

\(a,=5\left(x^2+2xy+y^2\right)-10y^2+5=5\left(x+y\right)^2-10y^2+5\\ =5\left(1+2\right)^2-10\cdot4+5=45-40+5=10\\ b,=7\left(x-y\right)-\left(x-y\right)^2=\left(x-y\right)\left(7-x+y\right)\\ =\left(2-2\right)\left(7-2+2\right)=0\)

25 tháng 12 2021

b: \(=7\left(x-y\right)-\left(x-y\right)^2\)

\(=\left(x-y\right)\left(7-x+y\right)=0\)

2 tháng 1 2022

a) x3 + 2x2y +  xy2 - 9x

= x(x2 + 2xy + y2 - 9)

= x(x+y)2 - 9

= x(x + y - 3) ( x + y + 3).

b) 5x2 - 10xy + 5y2

= 5(x2 - 2xy + y2)

= 5(x-y)2

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