b: \(\Leftrightarrow\dfrac{20}{x}-\dfrac{20}{x+20}=\dfrac{1}{6}\)

=>\(\dfrac{20x+400-20x}{x\left(x+20\right)}=\dfrac{1}{6}\)

=>x*(x+20)=400*6=2400

=>x^2+20x-2400=0

=>(x+60)(x-40)=0

=>x=-60 hoặc x=40

c: \(\dfrac{2x+1}{2x-1}-\dfrac{2x-1}{2x+1}=\dfrac{8}{4x^2-1}\)

=>(2x+1)^2-(2x-1)^2=8

=>4x^2+4x+1-4x^2+4x-1=8

=>8x=8

=>x=1(nhận)

giup m nha minh s tick

X =  150

150 nhá bạn tick cho mình nhá

\(B=\left(1+\dfrac{2}{3}-\dfrac{1}{4}\right).\left(\dfrac{4}{5}-\dfrac{3}{4}\right)^2:\dfrac{17}{4800}\)

\(=\left(\dfrac{12}{12}+\dfrac{8}{12}-\dfrac{3}{12}\right).\left(\dfrac{16}{20}-\dfrac{15}{20}\right)^2.\dfrac{4800}{17}\)

\(=\dfrac{17}{12}.\left(\dfrac{1}{20}\right)^2.\dfrac{4800}{17}\)

\(=\dfrac{17}{12}.\dfrac{1}{400}.\dfrac{4800}{17}\)

\(=1\)

\(B=\left(1+\dfrac{2}{3}-\dfrac{1}{4}\right).\left(\dfrac{4}{5}-\dfrac{3}{4}\right)^2:\dfrac{17}{4800}\\ =\dfrac{17}{12}.\dfrac{1}{400}:\dfrac{17}{4800}\\ =\dfrac{17}{4800}:\dfrac{17}{4800}\\ =1\)

4800 : 8 x 4 = 2400

`100/x-100/(x+10)=1/2`
`<=>(100x+1000-100x)/(x^2+10x)=1/2`
`<=>1000/(x^2+10x)=1/2`
`<=>x^2+10x=2000`
`<=>x^2+10x-2000=0`
`Delta'=25+2000=2025`
`<=>x_1=40,x_2=-50`
Vậy `S={40,-50}`

\(\dfrac{100}{x}=\dfrac{1}{2}+\dfrac{100}{x+10}\Rightarrow\dfrac{100}{x}=\dfrac{x+210}{2x+20}\Rightarrow200x+2000=x^2+210x\)

\(x^2+10x-2000=0\Leftrightarrow\left(x+50\right)\left(x-40\right)=0\Rightarrow x=40\)

=) đây là bài giải bằng cách lập pt mà nãy bạn đã đăng nè:v mà giải thì ra vô nghiệm á bạn nên mik ko có làm:v

3000 x 2 : 3 = 2000

4800 : 8 x 4 = 2400

4000 : 5 : 2 = 400

- Đề lỗi phải khum ạ ?

Lời giải:

ĐKXĐ: $x\neq \pm 4$

PT $\Leftrightarrow \frac{8(x-4)+8(x+4)}{x^2-16}=\frac{25}{3}$

$\Leftrightarrow \frac{16x}{x^2-16}=\frac{25}{3}$

$\Rightarrow 48x=25x^2-400$

$\Leftrightarrow 25x^2-48x-400=0$

$\Leftrightarrow (5x-\frac{24}{5})^2=\frac{10576}{25}$

$\Rightarrow x=\frac{24\pm 4\sqrt{661}}{25}$ (đều thỏa mãn)

ĐKXĐ \(x\ne8;x\ne11;x\ne9;x\ne10\)

\(\dfrac{8}{x-8}+\dfrac{11}{x-11}=\dfrac{9}{x-9}+\dfrac{10}{x-10}\)

\(\Leftrightarrow\left(\dfrac{8}{x-8}+1\right)+\left(\dfrac{11}{x-11}+1\right)=\left(\dfrac{9}{x-9}+1\right)+\left(\dfrac{10}{x-10}+1\right)\)

\(\Leftrightarrow\dfrac{x}{x-8}+\dfrac{x}{x-11}=\dfrac{x}{x-9}+\dfrac{x}{x-10}\)

\(\Leftrightarrow\dfrac{x}{x-8}+\dfrac{x}{x-11}-\dfrac{x}{x-9}-\dfrac{x}{x-10}=0\)

\(\Leftrightarrow x\left(\dfrac{1}{x-8}+\dfrac{1}{x-11}-\dfrac{1}{x-9}-\dfrac{1}{x-10}\right)=0\)

\(\Leftrightarrow x=0\) hoặc \(\dfrac{1}{x-8}+\dfrac{1}{x-11}-\dfrac{1}{x-9}-\dfrac{1}{x-10}=0\)

1) x=0

2) \(\dfrac{1}{x-8}+\dfrac{1}{x-11}-\dfrac{1}{x-9}-\dfrac{1}{x-10}=0\)

\(\Leftrightarrow\dfrac{x-11+x-8}{\left(x-8\right)\left(x-11\right)}-\dfrac{x-10+x-9}{\left(x-9\right)\left(x-10\right)}=0\)

\(\Leftrightarrow\dfrac{2x-19}{\left(x-8\right)\left(x-11\right)}=\dfrac{2x-19}{\left(x-9\right)\left(x-10\right)}\)

\(\Leftrightarrow\dfrac{2x-19}{x^2-19x+88}=\dfrac{2x-19}{x^2-19x+90}\)

do \(x^2-19x+88\ne x^2-19x+90\)

\(\Rightarrow2x-19=0\)

=> x=\(\dfrac{19}{2}\)

Vậy x=\(0\); x=\(\dfrac{19}{2}\)

Tik