a. \(x\ne5\) là ĐKXĐ của biểu thức P

b. P =\(\dfrac{\left(x-5\right)^2}{x-5}\)=\(x-5\)

c. P = -1 <=> x-5 =-1 <=> x=4

x: 5 dư 3 ; x : 6 dư 3 ; 55 < x < 65

Theo bài ra ta có : \(\left\{{}\begin{matrix}x-3⋮5\\x-3⋮6\end{matrix}\right.\)  

                              => x - 3 \(\in\) BC(5;6)

                            5 = 5;         6 = 2.3 => BCNN(5; 6) = 2.3.5 = 30

=> BC(5;6) = {30; 60; 90; ...;}

=> x - 3 \(\in\) { 30; 60; 90} => x \(\in\) { 33; 63; 93;...}

vì 55 < x < 65 mà x \(\in\) { 33; 63; 93} nên x = 63

Kết luận x = 63 

\(\dfrac{x}{x-5}+\dfrac{4x}{x+5}+\dfrac{x\left(x-15\right)}{x^2-25}\)

  = \(\dfrac{x\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}+\dfrac{4x\left(x-5\right)}{\left(x+5\right)\left(x-5\right)}+\dfrac{x\left(x-15\right)}{\left(x+5\right)\left(x-5\right)}\)

  = \(\dfrac{x^2+5x+4x^2-20x+x^2-15x}{\left(x-5\right)\left(x+5\right)}\)

  = \(\dfrac{6x^2-30x}{\left(x-5\right)\left(x+5\right)}\)

  = \(\dfrac{6x\left(x-5\right)}{\left(x-5\right)\left(x+5\right)}\)

  = \(\dfrac{6x}{x+5}\)

23

230000m2

bạn ơi có hình đâu

Chờ mình sẽ cho hình

\(đk:x\ne1\\ \dfrac{x^2+5x}{3x^2-6x+3}:\dfrac{7x+35}{6x-6}\\ =\dfrac{x\left(x+5\right)}{3\left(x^2-2x+1\right)}:\dfrac{7\left(x+5\right)}{6\left(x-1\right)}\\ =\dfrac{x\left(x+5\right)}{3\left(x-1\right)^2}\times\dfrac{6\left(x-1\right)}{7\left(x+5\right)}\\ =\dfrac{2x}{7\left(x-1\right)}\)

\(đk:x\ne1\)

\(\dfrac{x^2+5}{3x^2-6x+3}.\dfrac{7x+35}{6x-6}\\ =\dfrac{x^2+5}{3\left(x^2-2x+1\right)}.\dfrac{7\left(x+5\right)}{6\left(x-1\right)}\\ =\dfrac{x^2+5}{3\left(x-1\right)^2}.\dfrac{7\left(x+5\right)}{6\left(x-1\right)}\\ =\dfrac{7\left(x^2+5\right)\left(x+5\right)}{18.\left(x-1\right)^3}\)