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10 tháng 6 2017

Bài 1:

\(\left(\dfrac{x}{x^2-49}-\dfrac{x-7}{x^2+7x}\right):\dfrac{2x-7}{x^2+7x}+\dfrac{x}{7-x}\)

\(=\left(\dfrac{x}{\left(x-7\right)\left(x+7\right)}-\dfrac{x-7}{x\cdot\left(x+7\right)}\right)\cdot\dfrac{x^2+7x}{2x-7}+\dfrac{x}{-\left(x-7\right)}\)

\(=\dfrac{x^2-\left(x-7\right)^2}{x\cdot\left(x-7\right)\left(x+7\right)}\cdot\dfrac{x\cdot\left(x+7\right)}{2x-7}-\dfrac{x}{x-7}\)

\(=\dfrac{\left(x-\left(x-7\right)\right)\cdot\left(x+x-7\right)}{x-7}\cdot\dfrac{1}{2x-7}-\dfrac{x}{x-7}\)

\(=\dfrac{\left(x-x+7\right)\cdot\left(2x-7\right)}{x-7}\cdot\dfrac{1}{2x-7}-\dfrac{x}{x-7}\)

\(=\dfrac{7}{x-7}-\dfrac{x}{x-7}\)

\(=\dfrac{7-x}{x-7}\)

\(=\dfrac{-\left(x-7\right)}{x-7}\)

\(=-1\)

10 tháng 6 2017

A = \(\left(\dfrac{x}{x^2-49}-\dfrac{x-7}{x^2+7x}\right):\dfrac{2x-7}{x^2+7x}+\dfrac{x}{7-x}\)

A = \(\left(\dfrac{x}{\left(x+7\right)\left(x-7\right)}-\dfrac{x-7}{x\left(x+7\right)}\right):\dfrac{2x-7}{x\left(x+7\right)}+\dfrac{x}{7-x}\)

A = \(\left(\dfrac{x^2-\left(x-7\right)^2}{\left(x+7\right)\left(x-7\right)x}\right):\dfrac{2x-7}{x\left(x+7\right)}-\dfrac{x}{x-7}\)

A = \(\left(\dfrac{x^2-\left(x^2-14x+49\right)}{\left(x+7\right)\left(x-7\right)x}\right):\dfrac{\left(2x-7\right)\left(x-7\right)-\left(x^3+7x^2\right)}{\left(x+7\right)\left(x-7\right)x}\)

A = \(\dfrac{14x-49}{\left(x+7\right)\left(x-7\right)x}:\dfrac{-x^3-5x^2-21x+49}{\left(x+7\right)\left(x-7\right)x}\)

A = \(\dfrac{14x-49}{\left(x+7\right)\left(x-7\right)x}.\dfrac{\left(x+7\right)\left(x-7\right)x}{-x^3-5x^2-21x+49}\)

A = \(\dfrac{14x-49}{-x^3-5x^2-21x+49}\)

25 tháng 2 2021

`a,(25xy^3(2x-y)^2)/(75xy^2(y-2x))(x,y ne 0)(y ne 2x)`

`=(25xy^3(y-2x)^2)/(75xy^2(y-2x))`

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

`b,(x^2-y^2)/(x^2-y^2+xz-yz)`

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

`=(x+y)/(x+y+z)`

`c,((2x+3)-x^2)/(x^2-1)(x ne +-1)`

`=(-(x^2-3x+x-3))/((x-1)(x+1))`

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

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

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

`d,(3x^3-7x^2+5x-1)/(2x^3-x^2-4x+3)`

`=(3x^3-3x^2-4x^2+4x+x-1)/(2x^3-2x^2+x^2-x-3x+3)`

`=(3x^2(x-1)-4x(x-1)+x-1)/(2x^2(x-1)+x(x-1)-3(x-1))`

`=(3x^2-4x+1)/(2x^2+x-3)`

`=(3x^2-3x-x+1)/(2x^2-2x+3x-3)`

`=(3x(x-1)-(x-1))/(2x(x-1)+3(x-1))`

`=(3x-1)/(2x+3)`

a) Ta có: \(\dfrac{25xy^3\cdot\left(2x-y\right)^2}{75xy^2\cdot\left(y-2x\right)}\)

\(=\dfrac{25xy^2\cdot y\cdot\left(y-2x\right)^2}{25xy\cdot y\cdot\left(y-2x\right)\cdot3}\)

\(=\dfrac{y\left(y-2x\right)}{3}\)

 

b: \(\Leftrightarrow\dfrac{7x+10}{x+1}\left(x^2-x-2-2x^2+3x+5\right)=0\)

\(\Leftrightarrow\left(7x+10\right)\left(-x^2+2x+3\right)=0\)

\(\Leftrightarrow\left(7x+10\right)\left(x^2-2x-3\right)=0\)

=>(7x+10)(x-3)=0

hay \(x\in\left\{-\dfrac{10}{7};3\right\}\)

d: \(\Leftrightarrow\dfrac{13}{2x^2+7x-6x-21}+\dfrac{1}{2x+7}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)

\(\Leftrightarrow\dfrac{13}{\left(2x+7\right)\left(x-3\right)}+\dfrac{1}{\left(2x+7\right)}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)

\(\Leftrightarrow26x+91+x^2-9-12x-14=0\)

\(\Leftrightarrow x^2+14x+68=0\)

hay \(x\in\varnothing\)

29 tháng 6 2017

a, \(\left(\dfrac{x}{x^2-49}-\dfrac{x-7}{x^2+7x}\right):\dfrac{2x-7}{x^2+7x}+\dfrac{x}{7-x}\)

\(=\left[\dfrac{x^2}{x\left(x-7\right)\left(x+7\right)}-\dfrac{\left(x-7\right)^2}{x\left(x+7\right)\left(x-7\right)}\right].\dfrac{x\left(x+7\right)}{2x-7}+\dfrac{x}{7-x}\)

\(=\left[\dfrac{x^2-\left(x-7\right)^2}{x\left(x-7\right)\left(x+7\right)}\right].\dfrac{x\left(x+7\right)}{2x-7}+\dfrac{x}{7-x}\)

\(=\left[\dfrac{\left(x-x+7\right)\left(x+x-7\right)}{x\left(x-7\right)\left(x+7\right)}\right].\dfrac{x\left(x+7\right)}{2x-7}+\dfrac{x}{7-x}\)

\(=\dfrac{7\left(2x-7\right)}{x\left(x-7\right)\left(x+7\right)}.\dfrac{x\left(x+7\right)}{2x-7}+\dfrac{x}{7-x}\)

\(=\dfrac{7}{x-7}+\dfrac{x}{7-x}\)

\(=\dfrac{7\left(7-x\right)+x\left(x-7\right)}{\left(x-7\right)\left(7-x\right)}=\dfrac{49-7x+x^2-7x}{7x-x^2-49+7x}\)

\(=\dfrac{49-14x+x^2}{14x-x^2-49}=\dfrac{-\left(14-x^2-49\right)}{14x-x^2-49}=-1\)

\(\Rightarrow\)Giá trị biểu thức trên không phụ thuộc vào biến

\(\Rightarrowđpcm\)

b, tương tự

29 tháng 6 2017

Bạn chỉ cần tính ra thôi .

+ Nếu kết quả rút gọn có x thì kết luận giá trị của biểu thức phụ thuộc vào giá trị của biến .

+ Nếu kết quả rút gọn không có x thì kết luận giá trị của biểu thức không phụ thuộc vào giá trị của biến .

Chúc bạn làm bài tốt nha .

7 tháng 6 2023

\(\dfrac{x}{x+2}+\dfrac{7x-16}{\left(x+2\right)\left(4x-7\right)}\left(dkxd:x\ne-2;x\ne\dfrac{7}{4}\right)\)

\(=\dfrac{x\left(4x-7\right)+7x-16}{\left(x+2\right)\left(4x-7\right)}\)

\(=\dfrac{4x^2-7x+7x-16}{\left(x+2\right)\left(4x-7\right)}\)

\(=\dfrac{4x^2-16}{\left(x+2\right)\left(4x-7\right)}\)

\(=\dfrac{4\left(x^2-4\right)}{\left(x+2\right)\left(4x-7\right)}\)

\(=\dfrac{4\left(x-2\right)\left(x+2\right)}{\left(x+2\right)\left(4x-7\right)}\)

\(=\dfrac{4\left(x-2\right)}{4x-7}\)

\(=\dfrac{4x-8}{4x-7}\)

===============================

\(\dfrac{x}{x-2}+\dfrac{2}{2-x}\left(dkxd:x\ne2\right)\)

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

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

\(=1\)

 

7 tháng 6 2023

cái này có phải tìm x đâu mà đkxđ nee

9 tháng 6 2017

\(A=\left(\dfrac{x}{x^2-49}-\dfrac{x-7}{x^2+7x}\right):\dfrac{2x-7}{x^2+7x}+\dfrac{x}{7-x}\)

\(A=\left(\dfrac{x}{\left(x-7\right)\left(x+7\right)}-\dfrac{x-7}{x\left(x+7\right)}\right):\dfrac{2x-7}{x^2+7x}+\dfrac{x}{7-x}\)

\(A=\dfrac{x^2-\left(x-7\right)^2}{x\left(x-7\right)\left(x+7\right)}\times\dfrac{x^2+7x}{2x-7}+\dfrac{x}{7-x}\)

\(A=\dfrac{x^2-x^2+14x-49}{x\left(x-7\right)\left(x+7\right)}\times\dfrac{x\left(x+7\right)}{2x-7}+\dfrac{x}{7-x}\)

\(A=\dfrac{7\left(2x-7\right)}{x\left(x-7\right)\left(x+7\right)}\times\dfrac{x\left(x+7\right)}{2x-7}+\dfrac{x}{7-x}\)

\(A=\dfrac{7}{x-7}-\dfrac{x}{x-7}\)

\(A=\dfrac{7-x}{x-7}\)

\(A=-\dfrac{7-x}{7-x}\)

\(A=-1\)

10 tháng 6 2017

\(A=\left(\dfrac{x}{x^2-49}-\dfrac{x-7}{x^2+7x}\right):\dfrac{2x-7}{x^2+7x}+\dfrac{x}{7-x}\)

\(A=\left(\dfrac{x}{\left(x-7\right)\left(x+7\right)}-\dfrac{x-7}{x\left(x+7\right)}\right).\dfrac{x^2+7x}{2x-7}-\dfrac{x}{x-7}\)

\(A=\dfrac{x^2-\left(x-7\right)^2}{x\left(x-7\right)\left(x+7\right)}.\dfrac{x\left(x+7\right)}{2x-7}-\dfrac{x}{x-7}\)

\(A=\dfrac{\left(x-x+7\right)\left(x+x-7\right)}{x-7}.\dfrac{1}{2x-7}-\dfrac{x}{x-7}\)

\(A=\dfrac{\left(x-x+7\right)\left(2x-7\right)}{x-7}.\dfrac{1}{2x-7}-\dfrac{x}{x-7}\)

\(A=\dfrac{7}{x-7}-\dfrac{x}{x-7}\)

\(A=\dfrac{7-x}{x-7}=\dfrac{-\left(x-7\right)}{x-7}=-1\)

Bài 2:

a: \(=2x^4-x^3-10x^2-2x^3+x^2+10x=2x^3-3x^3-9x^2+10x\)

b: \(=\left(x^2-15x\right)\left(x^2-7x+3\right)\)

\(=x^4-7x^3+3x^2-15x^3+105x^2-45x\)

\(=x^4-22x^3+108x^2-45x\)

c: \(=12x^5-18x^4+30x^3-24x^2\)

d: \(=-3x^6+2.4x^5-1.2x^4+1.8x^2\)

a: \(\Leftrightarrow\dfrac{7x+10}{x+1}\left(x^2-x-2-2x^2+3x+5\right)=0\)

\(\Leftrightarrow\left(7x+10\right)\left(-x^2+2x+3\right)=0\)

\(\Leftrightarrow\left(7x+10\right)\cdot\left(x^2-2x-3\right)=0\)

=>(7x+10)(x-3)=0

=>x=3 hoặc x=-10/7

b: \(\Leftrightarrow\dfrac{13}{\left(2x+7\right)\left(x-3\right)}+\dfrac{1}{2x+7}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)

\(\Leftrightarrow13\left(x+3\right)+x^2-9-12x-42=0\)

\(\Leftrightarrow x^2-12x-51+13x+39=0\)

\(\Leftrightarrow x^2+x-12=0\)

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

=>x=-4