Bài 1:

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

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

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

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

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

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

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

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

\(=-1\)

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

Bài 2:

\(B=\left[\dfrac{3}{x+1}+\left(\dfrac{3}{x}-\dfrac{x}{x^2+2x+1}\right):\dfrac{2x^2+3x}{x+1}\right]:\dfrac{1+3x}{x^2+x}\)

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

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

\(=\left(\dfrac{3}{x+1}+\dfrac{2x^2+6x+3}{x\left(x+1\right)}\cdot\dfrac{1}{2x^2+3x}\right)\cdot\dfrac{x\left(x+1\right)}{1+3x}\)

\(=\left(\dfrac{3}{x+1}+\dfrac{2x^2+6x+3}{x\left(x+1\right)\left(2x^2+3x\right)}\right)\cdot\dfrac{x\left(x+1\right)}{1+3x}\)

\(=\dfrac{3x\cdot\left(2x^2+3x\right)+2x^2+6x+3}{x\left(x+1\right)\left(2x^2+3x\right)}\cdot\dfrac{x\left(x+1\right)}{1+3x}\)

\(=\dfrac{6x^3+9x^2+2x^2+6x+3}{2x^2+3x}\cdot\dfrac{1}{1+3x}\)

\(=\dfrac{6x^3+11x^2+6x+3}{2x^2+3x}\cdot\dfrac{1}{1+3x}\)

\(=\dfrac{6x^3+11x^2+6x+3}{\left(2x^2+3x\right)\left(1+3x\right)}\)

\(=\dfrac{6x^3+11x^2+6x+3}{2x^2+6x^3+3x+9x^2}\)

\(=\dfrac{6x^3+11x^2+6x+3}{11x^2+6x^3+3x}\)

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

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

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

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

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

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

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

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

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

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

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

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

\(=1\)

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

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 .

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

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

\(\Leftrightarrow\dfrac{x^2-x^2+14x-49}{\left(x-7\right).\left(2x-7\right)}-\dfrac{x}{x-7}\)

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

\(\Leftrightarrow\dfrac{7-x}{x-7}=-1\)

Vậy bt không phụ thuộc vào biến

b) Tương tự

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 \(x\in\varnothing\)

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

Lời giải:

ĐKXĐ:.....

Ta có: \(\frac{x+5}{x(2x-5)+2}+\frac{x+2}{x(2x-7)+3}=\frac{2x+7}{2x^2-7x+3}\)

\(\Leftrightarrow \frac{x+5}{x(2x-5)+2}=\frac{2x+7}{2x^2-7x+3}-\frac{x+2}{2x^2-7x+3}\)

\(\Leftrightarrow \frac{x+5}{x(2x-5)+2}=\frac{2x+7-(x+2)}{2x^2-7x+3}\)

\(\Leftrightarrow \frac{x+5}{x(2x-5)+2}=\frac{x+5}{2x^2-7x+3}\)

\(\Leftrightarrow (x+5)\left(\frac{1}{x(2x-5)+2}-\frac{1}{2x^2-7x+3}\right)=0\)

TH1: \(x+5=0\Leftrightarrow x=-5\) (thỏa mãn)

TH2: \(\frac{1}{x(2x-5)+2}-\frac{1}{2x^2-7x+3}=0\)

\(\Leftrightarrow x(2x-5)+2=2x^2-7x+3\)

\(\Leftrightarrow -5x+2=-7x+3\)

\(\Leftrightarrow 2x=1\Leftrightarrow x=\frac{1}{2}\). Thử lại thây mẫu số bằng 0 (không thỏa mãn ĐKXĐ)
Vậy PT có nghiệm duy nhất \(x=-5\)

ĐỀ BÀI LÀ GÌ VẬY BẠN.

\(1\text{)}\left(x-5\right)^2+3\left(x-5\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(x-5+3\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(x-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)

\(2\text{)}\dfrac{2x-1}{3}-\dfrac{5x+2}{7}=x+13\)

\(\Leftrightarrow\dfrac{7\left(2x-1\right)-3\left(5x+2\right)}{21}=\dfrac{21\left(x+13\right)}{21}\)

\(\Leftrightarrow14x-7-15x-6=21x+273\)

\(\Leftrightarrow-x-13=21x+273\)

\(\Leftrightarrow-22x=286\)

\(\Rightarrow x=-\dfrac{286}{22}=-\dfrac{143}{11}\)

\(3\text{)}\dfrac{x-1}{x+2}-\dfrac{x}{x-2}=\dfrac{7x-6}{4-x^2}\left(đk:x\ne\pm2\right)\)

\(\Leftrightarrow\dfrac{x-1}{2+x}+\dfrac{x}{2-x}=\dfrac{7x-6}{4-x^2}\)

\(\Leftrightarrow\dfrac{\left(x-1\right)\left(2-x\right)+x\left(2+x\right)}{4-x^2}=\dfrac{7x-6}{4-x^2}\)

\(\Leftrightarrow2x-x^2-2+x+2x+x^2=7x-6\)

\(\Leftrightarrow x-2=7x-6\)

\(\Leftrightarrow-6x=-4\)

\(\Rightarrow x=\dfrac{2}{3}\)

1.(x−5)²+3(x−5)=0

=>(x-5)(x-5)+3.(x-5)=0

=>(x-5).(x-5+3)=0

=>x-5=0 hoặc x-2=0

=>x=5 hoặc x=2

2)\(\dfrac{2x-1}{3}\)-\(\dfrac{5x+2}{7}\)=x+13

=>\(\dfrac{7.\left(2x-1\right)}{7.3}\)-\(\dfrac{3.\left(5x+2\right)}{3.7}\)=\(\dfrac{21.\left(x+13\right)}{21}\)

=>khử mẫu:

=>7.(2x-1)-3.(5x+2)=21.(x+13)

=>14x-7-15x-6=21x+273

=>14-7-15x-6-21x-273=0

=>-36x-272=0

=>-36x=272

=>x=...