cau a : (3x^2y-6xy+9x)(-4/3xy)

           =-4/3xy.3x^2y+4/3xy.6xy-4/3xy.9x

           =-4x+8-8y

cau b : (1/3x+2y)(1/9x^2-2/3xy+4y^2)

            =(1/3)^3-2/9x^2y+8y^3+4/3xy^2+2/9x^2y-4/3xy^2+8y^3

             =(1/3)^3 + (2y)^3x-2

cau c :  (x-2)(x^2-5x+1)+x(x^2+11)

            =x^3-5x^2+x-2x^2+10x-2+x^3+11x

            =2x^3-7x^2+22x-2

cau d := x^3 + 6xy^2 -27y^3

cau e := x^3 + 3x^2 -5x - 3x^2y - 9xy = 15y

cau f := x^2-2x+2x -4-2x-1

          = x(x-2)-5

cau e la + 15y ko phai =15y

Make questions to the underlines works: 

We went to school by bus.

-->  Are they going to school by car?

Where does he live?

-> What!You don't know where I live!

Câu cuối chế

1.how did we go to school

Bài 1.

a) -2x( -3x + 2 ) - ( x + 2 )²

= 6x² - 4x - ( x² + 4x + 4 )

= 6x² - 4x - x² - 4x - 4

= 5x² - 8x - 4

b) ( x + 2 )( x² - 2x + 4 ) - 2( x + 1 )( 1 - x )

= x³ + 8 + 2( x + 1 )( x - 1 )

= x³ + 8 + 2( x² - 1 )

= x³ + 8 + 2x² - 2

= x³ + 2x² + 6

c) ( 2x - 1 )² - 2( 4x² - 1 ) + ( 2x + 1 )²

= 4x² - 4x + 1 - 8x² + 2 + 4x² + 4x + 1

= 4

d) x² - 3x + xy - 3y

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

= ( x - 3 )( x + y )

Bài 2.

a) 4x² - 4xy + y² = ( 2x - y )²

b) 9x³ - 9x²y - 4x + 4y

= 9x²( x - y ) - 4( x - y )

= ( x - y )( 9x² - 4 )

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

c) x³ + 2 + 3( x³ - 2 )

= x³ + 2 + 3x³ - 6

= 4x³ - 4

= 4( x³ - 1 )

= 4( x - 1 )( x² + x + 1 )

Bài 3.

2( x - 2 ) = x² - 4x + 4

⇔ ( x - 2 )² - 2( x - 2 ) = 0

⇔ ( x - 2 )( x - 2 - 2 ) = 0

⇔ ( x - 2 )( x - 4 ) = 0

⇔ x = 2 hoặc x = 4

Muốn rút gọn một phân thức đại số ta phải:

- Phân tích tử và mẫu thành nhân tử (nếu cần) để tìm nhân tử chung

- Chia cả tử và mẫu cho nhân tử chung giống nhau

1)x=1;2)x=1;3)x=3;4)x=2;5)x=2;6)x=4.

moi hoc lop 6 thoi

1) \(x^2+x-2=0\)

\(\Leftrightarrow x^2+2x-x-2=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x=-2\\x=1\end{cases}}\)

2) \(x^2+2x-3=0\)

\(\Leftrightarrow x^2+3x-x-3=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x=-3\\x=1\end{cases}}\)

3) \(x^2-x-6=0\)

\(\Leftrightarrow x^2-3x+2x-6=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)

4) \(x^2+x-6=0\)

\(\Leftrightarrow x^2+3x-2x-6=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x=-3\\x=2\end{cases}}\)

5) \(2x^2-x-6=0\)

\(\Leftrightarrow2x^2-4x+3x-6=0\)

\(\Leftrightarrow2x\left(x-2\right)+3\left(x-2\right)=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-\frac{3}{2}\end{cases}}\)

\(x^2+5x+6\)

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

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

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

Bài này bạn dùng phương pháp tách số nhé.

Đề:.......

<=> x² + 6x - x + 6

<=> (x² + 6x) - (x + 6)

<=> x(x + 6) - (x + 6)

<=> (x + 6)(x - 1)

Biểu thức hay đa thức z?

biểu thức lớp 7,8 á 

\(\left(x+1\right)\left(x+1\right)^2=0\)

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

\(\Leftrightarrow x+1=0\)

\(\Leftrightarrow x=-1\)

Vậy \(x=-1\)

\(\left(x+1\right)\left(x+1\right)^2=0\)

\(\left(x+1\right)\left(x+1+1\right)=0\)

\(\left(x+1\right)\left(x+2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+1=0\\x+2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-1\\x=-2\end{cases}}}\)

2x( 3 - x ) + 5x - 15

= 2x( 3 - x ) - ( 15 - 5x )

= 2x( 3 - x ) - 5( 3 - x )

= ( 3 - x )( 2x - 5 )

Đề:.......

<=> 6x - 2x² + 5x - 15

<=> 2x(3 - x) + 5(x - 3)

<=> 2x(3 - x) - 5(3 - x)

<=> (3 - x)(2x - 5)

a) ( 2x - 1 )( 2x + 1 ) - 4( x² + x ) = 16

⇔ 4x² - 1 - 4x² - 4x = 16

⇔ -4x - 1 = 16

⇔ -4x = 17

⇔ x = -17/4

b) 5x( x - 2013 ) - x + 2013 = 0

⇔ 5x( x - 2013 )  - ( x - 2013 ) = 0

⇔ ( x - 2013 )( 5x - 1 ) = 0

⇔ \(\orbr{\begin{cases}x-2013=0\\5x-1=0\end{cases}}\)

⇔ \(\orbr{\begin{cases}x=2013\\x=\frac{1}{5}\end{cases}}\)

a) \(\left(2x-1\right)\left(2x+1\right)-4.\left(x^2+x\right)=16\)

\(4x^2-1-4x^2-4x=16\)

\(-1-4x=16\)

\(-4x=17\)

\(x=-\frac{17}{4}\)

b) \(5x\left(x-2013\right)-x+2013=0\)

\(\left(x-2013\right)\left(5x-1\right)=0\)

\(\Rightarrow\hept{\begin{cases}x-2013=0\\5x-1=0\end{cases}\Rightarrow\hept{\begin{cases}x=2013\\x=\frac{1}{5}\end{cases}}}\)