Bài 1 :

a) \(3x^2+4x-7\)

\(=3x^2-3x+7x-7\)

\(=3x\left(x-1\right)+7\left(x-1\right)\)

\(\left(x-1\right)\left(3x+7\right)\)

b) \(3x^2+48+24x-12y^2\)

\(=3\left(x^2+16+8x-4y^2\right)\)

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

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

Bài 2 :

a) Phân thức xác định \(\Leftrightarrow\hept{\begin{cases}x-3y\ne0\\2xy-1\ne0\\x+2\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne3y\\2xy\ne1\\x\ne-2\end{cases}}}\)

b) \(A=\left(\frac{x+2y}{x-3y}+\frac{5y}{3y-x}-2xy\right)\cdot\frac{x+2}{2xy-1}+\frac{x^2-3}{x+2}\)

\(A=\left(\frac{x+2y}{x-3y}-\frac{5y}{x-3y}-\frac{2xy\left(x-3y\right)}{x-3y}\right)\cdot\frac{x+2}{2xy-1}+\frac{x^2-3}{x+2}\)

\(A=\left(\frac{x+2y-5y-2x^2y+6xy^2}{x-3y}\right)\cdot\frac{x+2}{2xy-1}+\frac{x^2-3}{x+2}\)

\(A=\left(\frac{x-3y-2x^2y+6xy^2}{x-3y}\right)\cdot\frac{x+2}{2xy-1}+\frac{x^2-3}{x+2}\)

\(A=\frac{\left(x-3y\right)-2xy\left(x-3y\right)}{x-3y}\cdot\frac{x+2}{2xy-1}+\frac{x^2-3}{x+2}\)

\(A=\frac{-\left(x-3y\right)\left(2xy-1\right)\left(x+2\right)}{\left(x-3y\right)\left(2xy-1\right)}+\frac{x^2-3}{x+2}\)

\(A=\frac{-\left(x+2\right)\left(x+2\right)}{\left(x+2\right)}+\frac{x^2-3}{x+2}\)

\(A=\frac{-x^2-4x-4+x^2-3}{x+2}\)

\(A=\frac{-4x-7}{x+2}\)

c) Thay x = 3 ( vì y bị triệt tiêu hết nên ko xét đến đỡ mệt ng :) )

\(A=\frac{-4\cdot3-7}{3+2}=\frac{-19}{5}\)

Lm hộ luôn bài hình nha :33

a) ta có: \(|4x^2-1|\ge0\forall x\)

\(|2x-1|\ge0\forall x\Leftrightarrow3x|2x-1|\ge0\forall x\)

Mà \(|4x^2-1|+3x|2x-1|=0\)

=> I4x^2-1I và 3xI2x-1I=0

=> 4x^2-1=0 và 3x=0 hoặc 2x-1=0

=> 4x^2=1 và x=0 hoặc 2x=1

=> x^2=1/4 và x=0 hoặc x=1/2

=> x=\(\pm\frac{1}{2}\)và x=0 hoặc x=1/2

Vậy x=\(\pm\frac{1}{2}\); x=0

Phạm Nhật Quỳnh

Bạn xem lại nhé x chưa chắc đã dương nha 

bài 2

a,6xz+9yz/4y^2

Ta có: \(\frac{x^2y+2xy^2+y^3}{2x^2+xy-y^2}\)

\(=\frac{x^2y+xy^2+xy^2+y^3}{2x^2+2xy-xy-y^2}\)

\(=\frac{xy\left(x+y\right)+y^2\left(x+y\right)}{2x\left(x+y\right)-y\left(x+y\right)}\)

\(=\frac{\left(x+y\right)\left(xy+y^2\right)}{\left(2x-y\right)\left(x+y\right)}=\frac{xy+y^2}{2x-y}\left(đpcm\right)\)

Ta có: \(\frac{x^2+3xy+2y^2}{x^3+2x^2y-xy^2-2y^3}\)

\(=\frac{x^2+xy+2xy+2y^2}{x^2\left(x+2y\right)-y^2\left(x+2y\right)}\)

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

\(=\frac{\left(x+2y\right)\left(x+y\right)}{\left(x+y\right)\left(x-y\right)\left(x+2y\right)}=\frac{1}{x-y}\left(đpcm\right)\)

dài lắm bạn ạ mk đang đau tay

lam giup minh di ma huheo

1) a) \(\frac{x}{x+1}+\frac{x^3-2x^2}{x^3+1}=\frac{x}{x+1}+\frac{x^3-2x^2}{\left(x+1\right)\left(x^2-x+1\right)}\)

\(=\frac{x\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}+\frac{x^3-2x^2}{\left(x+1\right)\left(x^2-x+1\right)}=\frac{x^3-x^2+x+x^3-2x^2}{\left(x+1\right)\left(x^2-x+1\right)}\)

\(=\frac{2x^3-3x^2+x}{\left(x+1\right)\left(x^2-x+1\right)}=\frac{x\left(x-1\right)\left(2x-1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)

b) \(\frac{x+1}{2x-2}+\frac{3}{x^2-1}+\frac{x+3}{2x+2}=\frac{x+1}{2\left(x-1\right)}+\frac{3}{\left(x-1\right)\left(x+1\right)}+\frac{x+3}{2\left(x+1\right)}\)

\(=\frac{\left(x+1\right)^2}{2\left(x-1\right)\left(x+1\right)}+\frac{6}{2\left(x-1\right)\left(x+1\right)}+\frac{\left(x+3\right)\left(x-1\right)}{2\left(x+1\right)\left(x-1\right)}\)

\(=\frac{\left(x+1\right)^2+6+\left(x+3\right)\left(x-1\right)}{2\left(x-1\right)\left(x+1\right)}=\frac{x^2+2x+1+6+x^2+2x-3}{2\left(x-1\right)\left(x+1\right)}\)

\(=\frac{2x^2+4x+2}{2\left(x-1\right)\left(x+1\right)}=\frac{2\left(x+1\right)^2}{2\left(x-1\right)\left(x+1\right)}=\frac{x+1}{x-1}\)

2) Ta có A = \(\left(\frac{x^2+y^2}{x^2-y^2}-1\right).\frac{x-y}{4y}=\frac{2y^2}{x^2-y^2}.\frac{x-y}{4y}=\frac{2y^2\left(x-y\right)}{\left(x-y\right)\left(x+y\right).4y}=\frac{y}{2\left(x+y\right)}\)

Thay x = 14 ; y = -15 vào biểu thức ta được 

\(A=\frac{y}{2\left(x+y\right)}=\frac{-15}{2\left(14-15\right)}=\frac{-15}{-2}=7,5\)

.các bác giúp em với ạ,em cảm ơn trc ạ

bài 1 :

1, 

a, x^2 - xy = x(x - y)

b, x^2 + 2xy + y^2 - 4

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

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

2,

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

= 4x^2 - 1 + 4x - 4x^2

= 4x - 1

3,  x^2 - 6x + 5 = 0

<=> x^2 - x - 5x + 5 = 0

<=> x(x - 1) - 5(x - 1) = 0

<=> (x - 5)(x - 1) = 0

<=> x = 5 hoặc x = 1

.giúp em bài 2 với ạ

\(\frac{3x}{5x+5y}-\frac{x}{10x-10y}\)

\(=\frac{3x}{5\left(x+y\right)}-\frac{x}{10\left(x+y\right)}\)

\(=\frac{30x\left(x-y\right)-5x\left(x+y\right)}{5\left(x+y\right).10\left(x+y\right)}\)

\(=\frac{5x\left(5x-7y\right)}{50\left(x+y\right)\left(x-y\right)}\)

\(=\frac{x\left(5x-7y\right)}{\left(x+y\right)\left(x-y\right)}\)

chỗ cuối tớ sai 

\(=\frac{x\left(5x-7y\right)}{10\left(x+y\right)\left(x-y\right)}\)

đây nha , e xin lỗi

a) \(\frac{3}{2x^2+2x}+\frac{2x-1}{x^2-1}-\frac{2}{x}=\frac{3}{2x\left(x+1\right)}+\frac{2x-1}{\left(x-1\right)\left(x+1\right)}-\frac{2}{x}\)

                                                          \(=\frac{3\left(x-1\right)+\left(2x-1\right)-2.2\left(x-1\right)\left(x+1\right)}{2x\left(x-1\right)\left(x+1\right)}\)

                                                          \(=\frac{3x-2x+4x^2-2x-4x^2+4x-4x+4}{2x\left(x-1\right)\left(x+1\right)}\)

                                                          \(=\frac{x+1}{2x\left(x-1\right)\left(x+1\right)}\)

                                                          \(=\frac{1}{2x\left(x-1\right)}\)

b) \(\frac{3x}{5x+5y}-\frac{x}{10x-10y}=\frac{3x}{5\left(x+y\right)}-\frac{x}{10\left(x-y\right)}\)

                                                   \(=\frac{3x.10\left(x-y\right)-x.5\left(x+y\right)}{50\left(x-y\right)\left(x+y\right)}\)

                                                   \(=\frac{30x\left(x-y\right)+5x\left(x+y\right)}{50\left(x-y\right)\left(x+y\right)}\)

                                                   \(=\frac{5x\left[6\left(x-y\right)-\left(x+y\right)\right]}{50\left(x-y\right)\left(x+y\right)}\)

                                                   \(=\frac{5x\left(5x-7y\right)}{50\left(x-y\right)\left(x+y\right)}\)

                                                   \(=\frac{x\left(5x-7y\right)}{10\left(x-y\right)\left(x+y\right)}\)

c) \(\frac{5x^2-y^2}{xy}-\frac{3x-2y}{y}=\frac{5x^2-y-x\left(3x-2y\right)}{xy}\)

                                                \(=\frac{5x^2-y-3x^2+2xy}{xy}\)

                                               \(=\frac{2x^2-y+2xy}{xy}\)

d) \(\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}=\frac{3}{2\left(x+3\right)}-\frac{x-6}{2x\left(x+3\right)}\)

                                            \(=\frac{3x-x+6}{2x\left(x+3\right)}\)

                                            \(=\frac{2x+6}{2x\left(x+3\right)}\)

                                            \(=\frac{2\left(x+3\right)}{2x\left(x+3\right)}\)

                                            \(=\frac{2}{2x}=\frac{1}{x}\)