0,2:x=1,03+3,97

a: A=-2xy+xy+xy^2=-xy+xy^2

Bậc là 3

b: \(B=xy^2z+2xy^2z-3xy^2z+xy^2z-xyz=-xyz+xy^2z\)

Bậc là 4

c: \(C=4x^2y^3-x^2y^3+x^4+6x^4-2x^2=3x^2y^3+7x^4-2x^2\)

Bậc là 5

d: \(D=\dfrac{3}{4}xy^2-\dfrac{1}{2}xy^2+xy=\dfrac{1}{4}xy^2+xy\)

bậc là 3

e: \(E=2x^2-4x^2+3z^4-z^4-3y^3+2y^3\)

=-2x^2+2z^4-y^3

Bậc là 4

f: \(=3xy^2z+xy^2z+2xy^2z-4xyz=6xy^2z-4xyz\)

Bậc là 4

a: =xy(1/3+4-2)=7/3xy

b: =xy^2(-1+3/2+4/3)=(1/3+3/2)xy^2=11/6xy^2

c: =4x^2y^2+2/3x^2y^2-4/3x^2y=-4/3x^2y+14/3x^2y^2

d: =3x^2y^2z+4x^2y^2z-8x^2y^2z=-x^2y^2z

1) 2x + 2y - x(x+y)

= 2(x + y) - x(x + y)

= (2 - x)(x + y)

2/ 5x² - 5xy -10x + 10y

= 5x(x - y) - 10(x - y)

= (5x - 10(x - y)

3/ 4x² + 8xy - 3x - 6y

= 4x(x + 2y) - 3(x + 2y)

= (4x - 3)(x + 2y)

1) 2x + 2y - x(x + y) 

= 2(x + y) - x(x + y)

= (2 - x)(x + y)

2) 5x² - 5xy - 10x + 10y 

= 5x(x - y) - 10(x - y)

= (5x - 10)(x - y)

= 5(x - 2)(x - y)

3) 4x² + 8xy - 3x - 6y  

= 4x(x + 2y) - 3(x + 2y)

= (4x - 3)(x + 2y)

4) 2x² + 2y² - x²z + z - y²z - 2 

= 2(x² + y² - z(x² + y²) - (2 - z)

= (2 - z)(x² + y²) - (2 - z)

= (2 - z)(x² + y²)

5) x² + xy - 5x - 5y

= x(x + y) - 5(x + y)

= (x - 5)(x + y)

6) x(2x - 7) - 4x + 14 

= x(2x - 7) - 2(2x - 7) 

= (x - 2)(2x - 7)

7)x² - 3x + xy - 3y  

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

= (x - 3)(x + y)

5/ x² + xy - 5x - 5y 

= x(x + y) - 5(x + y)

= (x - 5)(x + y)

6/ x(2x - 7) - 4x + 14

= 2x² - 7x - 4x + 14

= (2x² - 4x) - (7x - 14)

= 2x(x - 2) -7(x - 2)

= (2x - 7)(x - 2)

7/ x² - 3x + xy - 3y

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

= (x + y)(x - 3) 

\(A=\dfrac{4x^2\left(x+y\right)-9y^2\left(x+y\right)}{4x^2\left(x-y\right)-9y^2\left(x-y\right)}=\dfrac{\left(4x^2-9y^2\right)\left(x+y\right)}{\left(4x^2-9y^2\right)\left(x-y\right)}=\dfrac{x+y}{x-y}\)

Với x=2014, y=14:

\(A=\dfrac{2014+14}{2014-14}=\dfrac{2028}{2000}=\dfrac{507}{500}\)

thèm ăn cục đường phèn.

x² - 5x - 14

= x² + 2x - 7x - 14

= x(x + 2) - 7(x + 2)

= (x - 7)(x + 2)

a) \(x^2-5x-14\)

\(=x^2-7x+2x-14\)

\(=\left(x^2+2x\right)-\left(7x-14\right)\)

\(=x\left(x+2\right)-7\left(x+2\right)\)

\(=\left(x+2\right)\left(x-7\right)\)

b) \(x^2y+xy^2+xz^2+yz^2+x^2z+y^2z+2xyz\)

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

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

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

a)   5(x-1)=x-1

     5x-5=x-1

     5x-x=5-1

     4x=4=>x=1

b)x(x-2)+(x-2)=0

(x-2)(x+1)=0

=>x=2 hay x=-1

c)5x(x-3)-x+3=0

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

(5x-1)(x-3)=0

=>x=\(\frac{1}{5}\)hay x=3

d)x(2x-7)-4x+17=0

x(2x-7-4)+17=0

x(2x-11)+17=0

=> đa thức này không có nghiệm

3.x=2.y , 4.x=2.z  và x² + 

Làm giống như bạn Should A Person

\(a)\) Ta có : 

\(3x=2y\)\(\Leftrightarrow\)\(\frac{x}{2}=\frac{y}{3}\)

\(4x=2z\)\(\Leftrightarrow\)\(\frac{x}{2}=\frac{z}{4}\)

\(\Rightarrow\)\(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}\)\(\Leftrightarrow\)\(\frac{x^2}{4}=\frac{y^2}{9}=\frac{2z^2}{32}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có : 

\(\frac{x^2}{4}=\frac{y^2}{9}=\frac{2z^2}{32}=\frac{x^2+y^2+2z^2}{4+9+32}=\frac{108}{45}=\frac{12}{5}\)

Do đó : 

\(x^2=\frac{12}{5}.4=\frac{48}{5}\)\(\Rightarrow\)\(x=\sqrt{\frac{48}{5}}\) hoặc \(-\sqrt{\frac{48}{5}}\)

\(y^2=\frac{12}{5}.9=\frac{108}{5}\)\(\Rightarrow\)\(y=\sqrt{\frac{108}{5}}\) hoặc \(y=-\sqrt{\frac{108}{5}}\)

\(2z^2=\frac{12}{5}.32=\frac{384}{5}\)\(\Rightarrow\)\(z=\sqrt{\frac{192}{5}}\) hoặc \(z=-\sqrt{\frac{192}{5}}\)

Vì \(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}\) nên x, y, z cùng dấu 

Vậy \(x=\sqrt{\frac{48}{5}}\)\(;\)\(y=\sqrt{\frac{108}{5}}\)\(;\)\(z=\sqrt{\frac{192}{5}}\) hoặc \(x=-\sqrt{\frac{48}{5}}\)\(;\)\(y=-\sqrt{\frac{108}{5}}\)\(;\)\(z=-\sqrt{\frac{192}{5}}\)

Chúc bạn học tốt ~