thiếu dấu

a) x - 1/2 = 3/5

x = 3/5 + 1/2

x = 11/10

b) x - 1/2 = -2/3

x = -2/3 + 1/2

x = -1/6

c) 2/5 - x = 0,25

x = 2/5 - 0,25

x = 2/5 - 1/4

x = 3/20

a, ĐKXĐ:\(x\ne-3\)

\(x+1+\dfrac{2}{x+3}=\dfrac{x+5}{x+3}\\ \Leftrightarrow x+1=\dfrac{x+5}{x+3}-\dfrac{2}{x+3}\\ \Leftrightarrow x+1=\dfrac{x+3}{x+3}\\ \Leftrightarrow x+1=1\\ \Leftrightarrow x=0\left(tm\right)\)

b, ĐKXĐ:\(x>2\)

\(\dfrac{x^2-4x-2}{\sqrt{x-2}}=\sqrt{x-2}\\ \Leftrightarrow x^2-4x-2=x-2\\ \Leftrightarrow x^2-5x=0\\ \Leftrightarrow x\left(x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=5\left(tm\right)\end{matrix}\right.\)

\(x^2 + 10x + 25 + 4x^2 - 12x + 9 - 5(x^2 - 4) = 0\\ ⇔ -2x + 54 = 0\\ ⇔ x = 27\)

Ta có: \(\left(x+5\right)^2+\left(2x-3\right)^2-5\left(x-2\right)\left(x+2\right)=0\)

\(\Leftrightarrow x^2+10x+25+4x^2-12x+9-5\left(x^2-4\right)=0\)

\(\Leftrightarrow5x^2-2x+34-5x^2+20=0\)

\(\Leftrightarrow-2x+54=0\)

hay x=27

Vậy: S={27}

x(x-5).(x+5)-(x+2).(x^2-2x+4)=17

\(\Leftrightarrow x\left(x^2-25\right)-\left(x^3-2x^2+4x+2x^2-4x+8\right)=17\)

\(\Leftrightarrow x^3-25x-x^3+2x^2-4x-2x^2+4x-8=17\)

\(\Leftrightarrow-25x=17+8\)

\(\Leftrightarrow-25x=25\)

\(\Leftrightarrow x=-1\)

#)Giải :

\(x\left(x-5\right)\left(x+5\right)-\left(x+2\right)\left(x^2-2x+4\right)=17\)

\(\Rightarrow x\left(x^2-25\right)-\left(x^3-2x^2+4x+2x^2-4x+8\right)=17\)

\(\Rightarrow x^3-25-\left(x^3+8\right)=17\)

\(\Rightarrow x^3-25x-x^3-8=17\)

\(\Rightarrow-25x=25\Rightarrow x=-1\)

Vậy x = -1

x.(x - 5).(x + 5) - (x + 2).(x² - 2x + 4) = 17

<=> -25x - 8 = 17

<=> -17x = 17

<=> x = -1

=> x = -1

a: \(\dfrac{x}{6}=\dfrac{8}{3}\)

=>\(x=6\cdot\dfrac{8}{3}=\dfrac{6}{3}\cdot8=8\cdot2=16\)

b: \(\dfrac{5}{x}=\dfrac{4}{9}\)

=>\(x=\dfrac{5\cdot9}{4}=\dfrac{45}{4}\)

c: \(\dfrac{x+3}{-4}=\dfrac{5}{20}\)

=>\(x+3=\dfrac{-4\cdot5}{20}=-1\)

=>x=-1-3=-4

d: \(\dfrac{7}{3+4x}=\dfrac{-2}{9}\)

=>\(4x+3=\dfrac{9\cdot7}{-2}=-\dfrac{63}{2}\)

=>\(4x=-\dfrac{63}{2}-3=-\dfrac{69}{2}\)

=>\(x=-\dfrac{69}{8}\)

f: ĐKXĐ: x<>1

\(\dfrac{3}{x-1}=\dfrac{x-1}{27}\)

=>\(\left(x-1\right)^2=3\cdot27=81\)

=>\(\left[{}\begin{matrix}x-1=9\\x-1=-9\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=10\left(nhận\right)\\x=-8\left(nhận\right)\end{matrix}\right.\)

a: \(B=\dfrac{x^2+5x+5x+25}{x\left(x+5\right)}=\dfrac{x+5}{x}\)

= \(x^2+5x+6-\left(x^2+3x-10\right)\)

= \(x^2+5x+6-x^2-3x+10\)

= \(2x+16\)

= \(2\left(x+8\right)\)

\(\left(x+2\right)\left(x+3\right)-\left(x-2\right)\left(x+5\right)\)

\(=x^2+5x+6-\left(x^2+3x-10\right)\)

\(=2x+16\)

   (x+2).(x+3)-(x-2).(x+5)

=2x+16

bạn cảm ơn ai vay có bn ấy có giup bn làm đau

mk chua hok den nen ko co bit lam

cảm ơn b nhé