a) 5xy ( x - y ) - 2x + 2y

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

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

b) 6x-2y-x(y-3x)

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

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

c)  x² + 4x - xy-4y

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

( x + 4 ) ( x - y )

d) 3xy + 2z - 6y - xz 

= ( 3xy - 6y ) + ( 2z - xz )

= 3y ( x - 2 ) + z ( x - 2 )

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

a,5xy(x-y)-2x+2y=5xy(x-y)-2(x-y)=(x-y)(5xy-2)

b,6x-2y-x(y-3x)=-2(y-3x)-x(y-3x)=(y-3x)(-2-x)

c,x^2+4x-xy-4y=x(x+4)-y(x+4)=(x+4)(x-y)

d,3xy+2z-6y-xz=(3xy-6y)+(2z-xz)=3y(x-2)+z(2-x)=3y(x-2)-z(x-2)=(x-2)(3y-z)

11)

a,4-9x^2=0

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

2-3x=0=>x=2/3 hoặc 2+3x=0=>x=-2/3

b,x^2 +x+1/4=0

(x+1/2)^2 =0

x+1/2=0

x=-1/2

c,2x(x-3)+(x-3)=0

(x-3)(2x+1)=0

x-3=0=>x=3 hoặc 2x+1=0=>x=-1/2

d,3x(x-4)-x+4=0

3x(x-4)-(x-4)=0

(x-4)(3x-1)=0

x-4=0=>x=4 hoặc 3x-1=0=>x=1/3

e,x^3-1/9x=0

x(x^2-1/9)=0

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

x=0 hoặc x+1/3=0=>x=-1/3 hoặc x-1/3=0=>x=1/3

f,(3x-y)^2-(x-y)^2 =0

(3x-y-x+y)(3x-y+x-y)=0

2x(4x-2y)=0

4x(2x-y)=0

x=0hoặc 2x-y=0=>x=y/2

thu thủy học lớp 9 chưa có long lone

a: (x-1)(2y-4)=0

=>x-1=0 và 2y-4=0

=>x=1 và y=2

b: (3x-2)(y-3)=6

mà x,y là số nguyên

nên \(\left(3x-2;y-3\right)\in\left\{\left(1;6\right);\left(-2;-3\right)\right\}\)

\(\Leftrightarrow\left(x,y\right)\in\left\{\left(1;9\right);\left(0;0\right)\right\}\)

d: \(\left(3x-4\right)\left(2y-1\right)=2\)

\(\Leftrightarrow\left(3x-4;2y-1\right)\in\left\{\left(2;1\right);\left(-2;-1\right)\right\}\)

\(\Leftrightarrow\left(x,y\right)=\left(2;1\right)\)

a) Theo đề bài, ta có: 

\(x:y:z=2:4:6\Rightarrow\frac{x}{2}=\frac{y}{4}=\frac{z}{6}\)và \(3x-y+z=24\)

Theo tính chất của dãy tỉ số bằng nhau, ta có:

   \(\frac{x}{2}=\frac{y}{4}=\frac{z}{6}=\frac{3x-y+z}{2.3-4+6}=\frac{24}{8}=3\)

\(.\frac{x}{2}=3\Rightarrow x=3.2=6\)

\(.\frac{y}{4}=3\Rightarrow y=3.4=12\)

\(.\frac{z}{6}=3\Rightarrow z=3.6=18\)

Vậy\(x,y,z\) lần lượt là: \(6,12,18\)

b) Vì x, y, z tỉ lệ nghịch với 6, 10, 4 nên ta có:

       \(6x=10y=4z\Rightarrow\frac{x}{\frac{1}{6}}=\frac{y}{\frac{1}{10}}=\frac{z}{\frac{1}{4}}\)

Theo tính chất của dãy tỉ số bằng nhua, ta có: 

       \(\frac{x}{\frac{1}{6}}=\frac{y}{\frac{1}{10}}=\frac{z}{\frac{1}{4}}=\frac{x+2y-3z}{\frac{1}{6}+2.\frac{1}{10}-3.\frac{1}{4}}=\frac{115}{\frac{-23}{60}}=-300\)

\(.\frac{x}{\frac{1}{6}}=-300\Rightarrow x=-300.\frac{1}{6}=-50\)

\(.\frac{y}{\frac{1}{10}}=-300\Rightarrow y=-300.\frac{1}{10}=-30\)

\(.\frac{z}{\frac{1}{4}}=-300\Rightarrow z=-300.\frac{1}{4}=-75\)

Vậy x, y, z lần lượt là: -50; -30; -75

Minh cảm ơn nha!

Ta có:

EQ\F(x,6)=EQ\F(y,10)=EQ\F(z,4)

x+2y-3z=115

Áp dụng tính chất của dãy tỉ số bằng nhau

EQ\F(x,6)=EQ\F(y,10)=EQ\F(z,4)=EQ\F(x+2y-3z,6+20-12)=EQ\F(115,14)

EQ\F(x,6)=EQ\F(115,14)=>x=EQ\F(345,7)

EQ\F(y,10)=EQ\F(115,14)=>y=EQ\F(575,7)

EQ\F(z,4)=EQ\F(115,14)=>z=EQ\F(230,7)

Vậy x=EQ\F(345,7)

       y=EQ\F(575,7)

       z=EQ\F(230,7)

Bài 10 :

Câu a :

\(5xy\left(x-y\right)-2x+2y\)

\(=5xy\left(x-y\right)-2\left(x-y\right)\)

\(=\left(x-y\right)\left(5xy-2\right)\)

Câu b :

\(6x-2y-x\left(y-3x\right)\)

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

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

Câu c :

\(x^2+4x-xy-4y\)

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

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

Câu d :

\(3xy+2z-6y-xz\)

\(=\left(3xy-6y\right)-\left(xz-2z\right)\)

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

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

Bài 11 :

Câu a :

\(4-9x^2=0\)

\(\Leftrightarrow\left(2-3x\right)\left(2+3x\right)=0\)

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

Vậy ........................

Câu b :

\(x^2+x+\dfrac{1}{4}=0\)

\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=0\)

\(\Leftrightarrow x+\dfrac{1}{2}=0\)

\(\Leftrightarrow x=-\dfrac{1}{2}\)

Vậy........................

Câu c :

\(2x\left(x-3\right)+\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(2x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{1}{2}\end{matrix}\right.\)

Vậy..................

Câu d :

\(3x\left(x-4\right)-x+4=0\)

\(\Leftrightarrow3x\left(x-4\right)-\left(x-4\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(3x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{1}{3}\end{matrix}\right.\)

Vậy................................

Câu e :

\(x^3-\dfrac{1}{9}x=0\)

\(\Leftrightarrow x\left(x^2-\dfrac{1}{9}\right)=0\)

\(\Leftrightarrow x\left(x-\dfrac{1}{3}\right)\left(x+\dfrac{1}{3}\right)=0\)

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

Vậy........................

Câu f :

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

\(\Leftrightarrow\left(3x-y-x+y\right)\left(3x-y+x-y\right)=0\)

\(\Leftrightarrow2x\left(4x-2y\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=0\\4x-2y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)

Vậy..........................

Ta có: 3x = 4y = 5z

\(\Rightarrow\frac{x}{\frac{1}{3}}=\frac{y}{\frac{1}{4}}=\frac{z}{\frac{1}{5}}=\frac{x-2y+3z}{\frac{1}{3}-\frac{1}{2}+\frac{3}{5}}=\frac{0}{\frac{13}{30}}=0\)

=> x = 0 x 1/3 = 0

      y = 0 x 1/4 = 0

      z = 0 x 1/5 = 0