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23 tháng 8 2021

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

23 tháng 8 2021

\(B=a\left(b^3-c^3\right)+b\left(c^3-a^3\right)+c\left(a^3-b^3\right)=ab^3-ac^3+bc^3-a^3b+a^3c-b^3c=ab\left(b^2-a^2\right)-c^3\left(a-b\right)+c\left(a^3-b^3\right)=-ab\left(a-b\right)\left(a+b\right)-c^3\left(a-b\right)+c\left(a-b\right)\left(a^2+ab+b^2\right)=\left(a-b\right)\left(-a^2b-ab^2-c^3+a^2c+abc+b^2c\right)\)

17 tháng 7 2021

a) \(x^4+2x^3-4x-4=\left(x^4+2x^3+x^2\right)-\left(x^2+4x+4\right)\)

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

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

 

a) Ta có: \(x^4+2x^3-4x-4\)

\(=\left(x^4+2x^3+x^2\right)-\left(x^2+4x+4\right)\)

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

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

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

5 tháng 9 2021

\(a,9x^2+y^2+2z^2-18x+4z-6y+20=0\\ \Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\\z=-1\end{matrix}\right.\)

\(b,5x^2+5y^2+8xy+2y-2x+2=0\\ \Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=-y\\x=1\\y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)

\(c,5x^2+2y^2+4xy-2x+4y+5=0\\ \Leftrightarrow\left(2x+y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}2x=-y\\x=1\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

\(d,x^2+4y^2+z^2=2x+12y-4z-14\\ \Leftrightarrow\left(x-1\right)^2+\left(2y-3\right)^2+\left(z+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{3}{2}\\z=-2\end{matrix}\right.\)

\(e,x^2+y^2-6x+4y+2=0\\ \Leftrightarrow\left(x-3\right)^2+\left(y+2\right)^2=11\)

Pt vô nghiệm do ko có 2 bình phương số nguyên có tổng là 11

 

 

e: Ta có: \(x^2-6x+y^2+4y+2=0\)

\(\Leftrightarrow x^2-6x+9+y^2+4y+4-11=0\)

\(\Leftrightarrow\left(x-3\right)^2+\left(y+2\right)^2=11\)

Dấu '=' xảy ra khi x=3 và y=-2

a: \(=4xy\left(1-5x^2y\right)\)

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

c: \(=x\left(x-a\right)+y\left(x-a\right)=\left(x-a\right)\left(x+y\right)\)

d: \(=\left(x+2y\right)^2-36=\left(x+2y+6\right)\left(x+2y-6\right)\)

a: Ta có: \(x^2-4y^2-2x-4y\)

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

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

c: Ta có: \(x^3+2x^2y-x-2y\)

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

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

d: Ta có: \(3x^2-3y^2-2\cdot\left(x-y\right)^2\)

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

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

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

e: Ta có: \(x^3-4x^2-9x+36\)

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

\(=\left(x-4\right)\left(x-3\right)\left(x+3\right)\)

f: Ta có: \(x^2-y^2-2x-2y\)

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

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

a: \(x^2-9-x^2\left(x^2-9\right)\)

\(=\left(x^2-9\right)-x^2\left(x^2-9\right)\)

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

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

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

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

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

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

c: \(x^3+27+\left(x+3\right)\left(x-9\right)\)

\(=\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)\)

\(=\left(x+3\right)\left(x^2-3x+9+x-9\right)\)

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

d: \(x^2+5x+6\)

\(=x^2+2x+3x+6\)

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

e: \(3x^2-4x-4\)

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

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

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

g: \(x^4+64y^4\)

\(=x^4+16x^2y^2+64y^4-16x^2y^2\)

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

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

 

h: \(a^2+b^2+2a-2b-2ab\)

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

\(=\left(a-b\right)^2+2\left(a-b\right)=\left(a-b\right)\left(a-b+2\right)\)

i: \(\left(x+1\right)^2-2\left(x+1\right)\left(y-3\right)+\left(y-3\right)^2\)

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

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

k: \(x^2\left(x+1\right)-2x\left(x+1\right)+\left(x+1\right)\)

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

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

5 tháng 11 2017

a) Cách 1.

Ta có 2xy + 3z + 6y + xz = (2xy + xz) + (3z + 6y)

= x(2 y + z)+3(z + 2 y) = (z + 2y)(x + 3).

Cách 2.

Ta có 2xy + 3z + 6y + xz = (2x1/ + 6y) + (3z + xz)

= 2y(x + 3) + z(3 + x) = (z + 2y)(x + 3).

b) Biến đổi được a 4   -   9 rt 3   +   a 2 -9a = (a- 9)a( a 2  +1).

c) Biến đổi được 3 x 2  + 5y - 3xy + (-5x) = (x - y)(3x - 5).

d) Biến đổi được  x 2  - (a + b)x + ab = (x- a)(x - b).

e) Ta có 4 x 2 - 4xy + y 2   –   9 t 2 =  ( 2 x   -   y ) 2   -   ( 3 t ) 2

= (2x - y - 3t )(2x - y + 31).

g) Ta có  x 3   -   3 x 2 y   +   3 xy 2   -   y 3   -   z 3

= ( x   -   y ) 3   -   z 3 = (x - y - z)( x 2   +   y 2   +   z 2  - 2xy + xz - yz).

h) Ta có x 2   -   y 2 + 8x + 6y+ 7 = ( x 2  +8x + 16) - ( y 2  - 6y+ 9)

= ( x   +   4 ) 2   - ( y - 3 ) 2  =(x-y + 7)(x + y + l).

20 tháng 10 2017

x2 + 4x – 2xy – 4y + y2 = (x2-2xy+ y2) + (4x – 4y) → bạn Việt dùng phương pháp nhóm hạng tử

= (x - y)2 + 4(x – y) → bạn Việt dùng phương pháp dùng hằng đẳng thức và đặt nhân tử chung

= (x – y)(x – y + 4) → bạn Việt dùng phương pháp đặt nhân tử chung