a, ( 152 +và 2/4 - 148 và 3/8 ) : 0,2 = x : 0,3

=>  33/8 : 1/5 = x : 3/10

=>  x : 3/10 = 165/8

=>  x = 99/10

b, ( 85 và 7/30 - 83 và 5/18 ) : 2 và 2/3 = 0,01x : 4

=>  88/45 : 8/3 = 0,01x : 4

=> 0,01x : 4 = 11/15

=> 0,01x = 44/15

=> x = 880/3

c, x - 1/ x + 5 = 6/7

=> 7( x - 1 ) = 6( x + 5 )

=> 7x - 7 = 6x + 30

=> 7x - 6x = 7 + 30

=> x = 37

d, x²/6 = 24/25

=> x². 25 = 6 . 24

=> x².25 = 144

=> x² = 144/25

=> x = ( 12/5)² hoặc x = ( -12/5)

g, x - 3/ x + 5 = 5/7

=> 7( x - 3 ) = 5 ( x + 5 )
=> 7x - 21 = 5x + 25

=> 7x - 5x = 21 + 25

=> 2x = 46

=> x = 23

a) \(\frac{x-1}{x+5}=\frac{6}{7}\)

7( x - 1 ) = 6( x + 5 )

7x - 7 = 6x - 30

7x - 6x = -30 + 7

x = -23

b) \(\frac{x^2}{6}=\frac{24}{25}\)

\(x^2.25=6.24\)

\(x^2.25=144\)

\(x^2=\frac{144}{25}\)

\(x=\sqrt{\frac{144}{25}}\)

\(x=\frac{12}{5}\)

c) \(\frac{x-2}{x-4}=\frac{x+4}{x+7}\)

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

\(x^2+7x-2x-14=x^2-4x+4x-16\)

\(x^2+5x-14=x^2^{ }-16\)

\(x^2-x^2+5x=-16+14\)

\(5x=-2\)\(x=\frac{-2}{5}\)

trước tiên tìm điều kiện của x rồi nhân chéo

a) \(x:8=7:4\)

=> \(\frac{x}{8}=\frac{7}{4}\)

=> \(x.4=7.8\)

=> \(x.4=56\)

=> \(x=56:4\)

=> \(x=14\)

Vậy \(x=14.\)

b) \(\left(x+1\right):0,75=1,4:0,25\)

=> \(\left(x+1\right):0,75=5,6\)

=> \(\left(x+1\right)=5,6.0,75\)

=> \(x+1=4,2\)

=> \(x=4,2-1\)

=> \(x=3,2\)

Vậy \(x=3,2.\)

Mình chỉ làm 2 câu thôi nhé, bạn đăng nhiều quá.

Chúc bạn học tốt!

Câu 6 :

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

=> \(\frac{15x}{15}+\frac{5\left(2x+\frac{x-1}{5}\right)}{15}=\frac{15}{15}-\frac{3\left(3x-\frac{1-2x}{3}\right)}{15}\)

=> \(15x+5\left(2x+\frac{x-1}{5}\right)=15-3\left(3x-\frac{1-2x}{3}\right)\)

=> \(15x+10x+\frac{5\left(x-1\right)}{5}=15-9x+\frac{3\left(1-2x\right)}{3}\)

=> \(15x+10x+x-1=15-9x+1-2x\)

=> \(15x+10x+x-1-15+9x-1+2x=0\)

=> \(37x-17=0\)

=> \(x=\frac{17}{37}\)

Vậy phương trình trên có nghiệm là \(S=\left\{\frac{17}{37}\right\}\)

Bài 7 :

a, Ta có : \(\frac{x-23}{24}+\frac{x-23}{25}=\frac{x-23}{26}+\frac{x-23}{27}\)

=> \(\frac{x-23}{24}+\frac{x-23}{25}-\frac{x-23}{26}-\frac{x-23}{27}=0\)

=> \(\left(x-23\right)\left(\frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27}\right)=0\)

=> \(x-23=0\)

=> \(x=23\)

Vậy phương trình trên có nghiệm là \(S=\left\{23\right\}\)

c, Ta có : \(\frac{x+1}{2004}+\frac{x+2}{2003}=\frac{x+3}{2002}+\frac{x+4}{2001}\)

=> \(\frac{x+1}{2004}+1+\frac{x+2}{2003}+1=\frac{x+3}{2002}+1+\frac{x+4}{2001}+1\)

=> \(\frac{x+2005}{2004}+\frac{x+2005}{2003}=\frac{x+2005}{2002}+\frac{x+2005}{2001}\)

=> \(\frac{x+2005}{2004}+\frac{x+2005}{2003}-\frac{x+2005}{2002}-\frac{x+2005}{2001}=0\)

=> \(\left(x+2005\right)\left(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\right)=0\)

=> \(x+2005=0\)

=> \(x=-2005\)

Vậy phương trình trên có nghiệm là \(S=\left\{-2005\right\}\)

e, Ta có : \(\frac{x-45}{55}+\frac{x-47}{53}=\frac{x-55}{45}+\frac{x-53}{47}\)

=> \(\frac{x-45}{55}-1+\frac{x-47}{53}-1=\frac{x-55}{45}-1+\frac{x-53}{47}-1\)

=> \(\frac{x-100}{55}+\frac{x-100}{53}=\frac{x-100}{45}+\frac{x-100}{47}\)

=> \(\frac{x-100}{55}+\frac{x-100}{53}-\frac{x-100}{45}-\frac{x-100}{47}=0\)

=> \(\left(x-100\right)\left(\frac{1}{55}+\frac{1}{53}-\frac{1}{45}-\frac{1}{47}\right)=0\)

=> \(x-100=0\)

Vậy phương trình trên có nghiệm là \(S=\left\{100\right\}\)

a, \(\left|x+\frac{1}{3}\right|=0\Leftrightarrow x=-\frac{1}{3}\)

b, \(\left|\frac{5}{18}-x\right|-\frac{7}{24}=0\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{5}{18}-x=\frac{7}{24}\\\frac{5}{18}-x=-\frac{7}{24}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{72}\\x=\frac{41}{72}\end{cases}}\)

c, \(\frac{2}{5}-\left|\frac{1}{2}-x\right|=6\Leftrightarrow\left|\frac{1}{2}-x\right|=-\frac{28}{5}\)vô lí 

Vì \(\left|\frac{1}{2}-x\right|\ge0\forall x\)*luôn dương* Mà \(-\frac{28}{5}< 0\)

=> Ko có x thỏa mãn 

\(|x+\frac{1}{3}|=0\)

\(< =>x+\frac{1}{3}=0< =>x=-\frac{1}{3}\)

\(|x+\frac{3}{4}|=\frac{1}{2}\)

\(< =>\orbr{\begin{cases}x+\frac{3}{4}=\frac{1}{2}\\x+\frac{3}{4}=-\frac{1}{2}\end{cases}}\)

\(< =>\orbr{\begin{cases}x=-\frac{1}{4}\\x=-\frac{5}{4}\end{cases}}\)

\(|\frac{5}{18}-x|-\frac{7}{24}=0\)

\(< =>|\frac{5}{18}-x|=\frac{7}{24}\)

\(< =>\orbr{\begin{cases}\frac{5}{18}-x=\frac{7}{24}\\\frac{5}{18}-x=-\frac{7}{24}\end{cases}}\)

\(< =>\orbr{\begin{cases}x=-\frac{1}{72}\\x=\frac{41}{72}\end{cases}}\)

\(\frac{2}{5}-|\frac{1}{2}-x|=6\)

\(< =>\frac{2}{5}-6=|\frac{1}{2}-x|\)

\(< =>\orbr{\begin{cases}\frac{1}{2}-x=-\frac{28}{5}\\\frac{1}{2}-x=\frac{28}{5}\end{cases}}\)

\(< =>\orbr{\begin{cases}x=\frac{61}{10}\\x=-\frac{51}{10}\end{cases}}\)

a.

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

\(\left(x-1\right)\times7=6\times\left(x-5\right)\)

\(7x-7=6x-30\)

\(7x-6x=-30+7\)

\(x=-23\)

b.

\(\frac{x^2}{6}=\frac{24}{25}\)

\(x^2=\frac{24}{25}\times6\)

\(x^2=\frac{144}{25}\)

\(x^2=\left(\pm\frac{12}{5}\right)^2\)

\(x=\pm\frac{12}{5}\)

Vậy \(x=\frac{12}{5}\) hoặc \(x=-\frac{12}{5}\)

CÓ VẤN ĐỀ ._.

b) \(\frac{x^2}{6}=\frac{24}{25}\)

\(\Rightarrow x.25=24.6\)

\(\Rightarrow x.25=144\)

\(\Rightarrow x=\frac{144}{25}\)

\(\Rightarrow x=5,76\)

Vậy x = 5,76