\(M=xy^3+5xyz-4x^2+6-x^3y+2xyz-xy^3+x^2-5\)

     \(=7xyz-3x^2-x^3y+1\)

Câu hỏi cụ thể nhé

1) \(4x^2-7x-2=4x^2-8x+x-2=\left(4x^2-8x\right)+\left(x-2\right)\)

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

2) \(4x^2+5x-6=4x^2+8x-3x-6=\left(4x^2+8x\right)-\left(3x+6\right)\)

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

3) \(5x^2-18x-8=5x^2-20x+2x-8=\left(5x^2-20x\right)+\left(2x-8\right)\)

\(=5x\left(x-4\right)+2\left(x-4\right)=\left(x-4\right)\left(5x+2\right)\)

4) \(xy\left(x+y\right)-yz\left(y+z\right)+xz\left(x-z\right)\)

\(=xy\left(x+y\right)-y^2z-yz^2+x^2z-xz^2\)

\(=xy\left(x+y\right)+\left(x^2z-y^2z\right)-\left(yz^2+xz^2\right)\)

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

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

\(=xy\left(x+y\right)+\left(zx-zy\right)\left(x+y\right)-z^2\left(x+y\right)\)

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

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

1) 4x² - 7x - 2 = 4x² - 8x + x - 2 = 4x( x - 2 ) + ( x - 2 ) = ( x - 2 )( 4x + 1 )

2) 4x² + 5x - 6 = 4x² - 8x + 3x - 6 = 4x( x - 2 ) + 3( x - 2 ) = ( x - 2 )( 4x + 3 )

3) 5x² - 18x - 8 = 5x² - 20x + 2x - 8 = 5x( x - 4 ) + 2( x - 4 ) = ( x - 4 )( 5x + 2 )

4) xy( x + y ) - yz( y + z ) + xz( x - z )

= x²y + xy² - y²z - yz² + xz( x - z )

= ( x²y - yz² ) + ( xy² - y²z ) + xz( x - z )

= y( x² - z² ) + y²( x - z ) + xz( x - z )

= y( x - z )( x + z ) + y²( x - z ) + xz( x - z )

= ( x - z )[ y( x + z ) + y² + xz ]

= ( x - z )( xy + yz + y² + xz )

= ( x - z )[ ( xy + y² ) + ( xz + yz ) ]

= ( x - z )[ y( x + y ) + z( x + y ) ]

= ( x - z )( x + y )( y + z )

5) xy( x + y ) + yz + xz( x + z ) + 2xyz ( đề có thiếu không vậy .-. )

\(4x^2-7x-2=\left(4x^2-8x\right)+\left(x-2\right)=4x\left(x-2\right)+\left(x-2\right)=\left(4x-1\right)\left(x-2\right)\)

\(=4x^2+8x-3x-6=4x\left(x+2\right)-3\left(x+2\right)=\left(4x-3\right)\left(x+2\right)\)

\(=5x^2-18x-8=5x^2-20x+2x-8=5x\left(x-4\right)+2\left(x-4\right)=\left(5x+2\right)\left(x-4\right)\)

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

* Đa thức thu gọn là đa thức không còn hai hạng tử nào đồng dạng

A = (xy⁷- xy⁷) + (x³y⁵-x³y⁵)+x⁸+10

A = x⁸+10

* M + N

= (5xyz -5x² + 8xy + 5)+(5x²+2xyz-8xy-7+y²)

= 5xyz - 5x² +8xy +5+5x² +2 xyz - 8xy -7 + y²

= ( 5xyz + 2xyz ) + ( -5x² +5x²) + ( 8xy - 8xy ) + ( 5-7) +y²

= 7xyz - 2 + y²

* M - N

= ( 5xyz - 5x² +8xy +5) - ( 5x² + 2xyz - 8xy -7 +y²)

= 5xyz - 5x² + 8xy + 5 - 5x² - 2xyz + 8xy + 7 - y²

= ( 5xyz - 2xyz) + ( -5x² - 5x²) + ( 8xy + 8xy) + ( 5+7) -y²

= 3xyz - 10x² +16xy +12 -y²

a) \(-xy\cdot2x^3y^4\cdot-\dfrac{5}{4}x^2y^3\)

\(=\left(-1\cdot2\cdot-\dfrac{5}{4}\right)\cdot\left(x\cdot x^3\cdot x^2\right)\cdot\left(y\cdot y^4\cdot y^3\right)\)

\(=\dfrac{5}{2}x^6y^8\)

Bậc là: \(6+8=14\)

Hệ số: \(\dfrac{5}{2}\)

Biến: \(x^6y^8\)

b) \(5xyz\cdot4x^3y^2\cdot-2x^5y\)

\(=\left(5\cdot4\cdot-2\right)\cdot\left(x\cdot x^3\cdot x^5\right)\cdot\left(y\cdot y^2\cdot y\right)\cdot z\)

\(=-40x^9y^4z\)

Bậc là: \(9+4=13\)

Hệ số: \(-40\)

Biến: \(x^9y^4z\)

c) \(-2xy^5\cdot-x^2y^2\cdot7x^2y\)

\(=\left(-2\cdot-1\cdot7\right)\cdot\left(x\cdot x^2\cdot x^2\right)\cdot\left(y^5\cdot y^2\cdot y\right)\)

\(=14x^6y^8\)

Bậc là: \(6+8=14\)

Hệ số: \(14\)

Biến: \(x^6y^8\)

Bài này x;y;z phải dương chứ nhỉ? Có dấu "=" ở số 0 thế kia thì bối rối quá

Dự đoán dấu "=" xảy ra tại \(x=y=z=\frac{1}{2}\)

Theo nguyên lý Dirichlet, trong 3 số x;y;z luôn tồn tại 2 số nằm cùng phía so với \(\frac{1}{2}\) ; giả sử đó là x và y

\(\Rightarrow\left(x-\frac{1}{2}\right)\left(y-\frac{1}{2}\right)\ge0\Leftrightarrow\frac{1}{2}\left(x+y\right)-xy\le\frac{1}{4}\)

\(\Leftrightarrow x+y-2xy\le\frac{1}{2}\)

Mặt khác:

\(1=2xyz+x^2+y^2+z^2\ge2xyz+2xy+z^2=2xy\left(1+z\right)+z^2\)

\(\Rightarrow1-z^2\ge2xy\left(1+z\right)\Leftrightarrow\left(1-z\right)\left(1+z\right)\ge2xy\left(1+z\right)\)

\(\Leftrightarrow1-z\ge2xy\Rightarrow xy\le\frac{1-z}{2}\)

\(\Rightarrow P=xy+z\left(x+y-2xy\right)\le\frac{1-z}{2}+\frac{z}{2}=\frac{1}{2}\)

Dấu "=" xảy ra khi \(x=y=z=\frac{1}{2}\)

`a)`

`A-B=(6x^2-7xy-4y^2)-(-2x^2+7xy+5y^2)`

      `=6x^2-7xy-4y^2+2x^2-7xy-5y^2`

      `=(6x^2+2x^2)-(7xy+7xy)-(4y^2+5y^2)`

      `=8x^2-14xy-9y^2`

___________________________________________

`b)`

`Q-(3x^4-2xyz)=xy+3x^4-5xyz-713`

`Q=(xy+3x^4-5xyz-713)+(3x^4-2xyz)`

`Q=xy+3x^4-5xyz-713+3x^4-2xyz`

`Q=xy+6x^4-7xyz-713`