a Đ

b S

c S

d Đ

a ) S 

b ) Đ

c ) S

d ) Đ

k cho mk nhé 

Mai Hồng Ngọc? Vũ Thị Thu Hằng? Ai đúng dzậy -_-*

a) Đk: x \(\ne\)-2

Ta có: \(\frac{2}{x+2}-\frac{2x^2+16}{x^2+8}=\frac{5}{x^2-2x+4}\)

<=> \(\frac{2\left(x^2-2x+4\right)-\left(2x^2+16\right)}{\left(x+2\right)\left(x^2-2x+4\right)}=\frac{5\left(x+2\right)}{\left(x+2\right)\left(x^2-2x+4\right)}\)

<=> 2x² - 4x + 8 - 2x² - 16 = 5x + 10

<=> -4x - 8 = 5x + 10

<=> -4x - 5x = 10 + 8

<=> -9x = 18

<=> x = -2 (ktm)

=> pt vô nghiệm

b) Đk: x \(\ne\)2; x \(\ne\)-3

Ta có: \(\frac{1}{x-2}-\frac{6}{x+3}=\frac{5}{6-x^2-x}\)

<=> \(\frac{x+3}{\left(x-2\right)\left(x+3\right)}-\frac{6\left(x-2\right)}{\left(x-2\right)\left(x+3\right)}=-\frac{5}{\left(x-2\right)\left(x+3\right)}\)

<=> x + 3 - 6x + 12 = -5

<=> -5x = -5 - 15

<=> -5x = -20

<=> x = 4 

vậy S = {4}

c) Đk: x \(\ne\)8; x \(\ne\)9; x \(\ne\)10; x \(\ne\)11

Ta có: \(\frac{8}{x-8}+\frac{11}{x-11}=\frac{9}{x-9}+\frac{10}{x-10}\)

<=> \(\left(\frac{8}{x-8}+1\right)+\left(\frac{11}{x-11}+1\right)=\left(\frac{9}{x-9}+1\right)+\left(\frac{10}{x-10}+1\right)\)

<=> \(\frac{x}{x-8}+\frac{x}{x-11}-\frac{x}{x-9}-\frac{x}{x-10}=0\)

<=> \(x\left(\frac{1}{x-8}+\frac{1}{x-11}-\frac{1}{x-9}-\frac{1}{x-10}\right)=0\)

<=> x = 0 (vì \(\frac{1}{x-8}+\frac{1}{x-11}-\frac{1}{x-9}-\frac{1}{x-10}\ne0\)

Vậy S = {0}

\(x:\left(9\frac{1}{2}-\frac{3}{2}\right)=\frac{\frac{2}{9}+\frac{2}{5}-\frac{2}{11}}{\frac{8}{5}+\frac{8}{9}-\frac{8}{11}}\)

\(x:8=\frac{2\left(\frac{1}{9}+\frac{1}{5}-\frac{1}{11}\right)}{8\left(\frac{1}{5}+\frac{1}{9}-\frac{1}{11}\right)}\)

\(x:8=\frac{1}{4}\)

\(x=2\)

Vậy..........

\(\frac{5-\frac{5}{3}+\frac{5}{9}-\frac{5}{27}}{8-\frac{8}{3}+\frac{8}{9}-\frac{8}{27}}:\frac{15-\frac{15}{11}+\frac{15}{121}}{16-\frac{16}{11}+\frac{16}{121}}\)

\(=\frac{5\left(1-\frac{1}{3}+\frac{1}{9}-\frac{1}{27}\right)}{8\left(1-\frac{1}{3}+\frac{1}{9}-\frac{1}{27}\right)}:\frac{15\left(1-\frac{1}{11}+\frac{1}{121}\right)}{16\left(1-\frac{1}{11}+\frac{1}{121}\right)}\)

\(=\frac{5}{8}:\frac{15}{16}\)

\(=\frac{2}{3}\)

2/3

Tk mk nha

\(1,\)\(x-\frac{3}{5}=\frac{3}{35}-\frac{-7}{6}\)

        \(x-\frac{3}{5}=\frac{3}{35}+\frac{7}{6}\)

        \(x-\frac{3}{5}=\frac{263}{210}\)

                    \(x=\frac{263}{210}+\frac{3}{5}\)

                    \(x=\frac{389}{210}\)

        VẬY: \(x=\frac{389}{210}\)

\(\frac{5\times\left(1-\frac{1}{3}+\frac{1}{9}-\frac{1}{27}\right)}{8\times\left(1-\frac{1}{3}+\frac{1}{9}-\frac{1}{27}\right)}\div\frac{15\times\left(1-\frac{1}{11}+\frac{1}{121}\right)}{16\times\left(1-\frac{1}{11}+\frac{1}{121}\right)}=\frac{5}{8}\div\frac{15}{16}=\frac{2}{3}\)

F=5-5x(1/3+1/9-1/27) /8-8x(1/3+1/9-1/27)

: 15-15x(1/11+1/121) /16-16x(1/11+1/121)

=5-5x1/8-8x1

: 15-15x1/16-16x1

=0:0=0

chắc vậy!