A=\(\frac{\frac{3}{7}-\frac{3}{17}+\frac{3}{37}}{\frac{5}{7}-\frac{5}{17}+\frac{5}{37}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}}{\frac{7}{5}-\frac{7}{4}+\frac{7}{3}-\frac{7}{2}}\)

\(=\frac{3\left(\frac{1}{7}-\frac{1}{17}+\frac{1}{37}\right)}{5\left(\frac{1}{7}-\frac{1}{17}+\frac{1}{37}\right)}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}}{-7\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}\right)}\)

\(=\frac{3}{5}+\frac{1}{-7}=\frac{3}{5}-\frac{1}{7}\)

\(=\frac{21}{35}-\frac{5}{35}=\frac{16}{35}\)

\(\frac{x_1-1}{5}=\frac{x_2-2}{4}=\frac{x_3-3}{3}=\frac{x_4-4}{2}=\frac{x_5-5}{1}=\frac{x_1+x_2+x_3+x_4+x_5-\left(1+2+3+4+5\right)}{5+4+3+2+1}\)

\(=\frac{30-\left(1+2+3+4+5\right)}{15}=1\)

Vậy \(\frac{x_{1-1}}{5}=1\)

 phần sau nữa bạn tự làm nhé

ta có \(\frac{\frac{3}{7}-\frac{3}{11}+\frac{3}{13}}{\frac{5}{7}-\frac{5}{11}+\frac{5}{13}}=\frac{3\left(\frac{1}{7}-\frac{1}{11}+\frac{1}{13}\right)}{5\left(\frac{1}{7}-\frac{1}{11}+\frac{1}{13}\right)}=\frac{3}{5}\)

và \(\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{4}+\frac{5}{8}-\frac{5}{6}}=\frac{2\left(\frac{1}{2.2}-\frac{1}{3.2}+\frac{1}{4.2}\right)}{5\left(\frac{1}{4}+\frac{1}{8}-\frac{1}{6}\right)}=\frac{2\left(\frac{1}{4}+\frac{1}{8}-\frac{1}{6}\right)}{5\left(\frac{1}{4}+\frac{1}{8}-\frac{1}{6}\right)}=\frac{2}{5}\)

Vậy \(\frac{\frac{3}{7}-\frac{3}{11}+\frac{3}{13}}{\frac{5}{7}-\frac{5}{11}+\frac{5}{13}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{4}+\frac{5}{8}-\frac{5}{6}}=\frac{3}{5}+\frac{2}{5}=\frac{5}{5}=1\)

ĐS: 1

Mệt hết cả hơi

Câu 1:

d: =>4x+6=15x-12

=>4x-15x=-12-6=-18

=>-11x=-18

hay x=18/11

e: =>\(45x+27=12+24x\)

=>21x=-15

hay x=-5/7

f: =>35x-5=96-6x

=>41x=101

hay x=101/41

g: =>3(x-3)=90-5(1-2x)

=>3x-9=90-5+10x

=>3x-9=10x+85

=>-7x=94

hay x=-94/7

làm rõ ra giúp với ạ, ghi v k hỉu j hết ;-;

\(B=4\cdot\left(-\frac{1}{2}\right)^3:\left(\frac{4}{5}\right)^0\cdot\frac{1}{2}-\frac{\frac{3}{5}-\frac{3}{9}+\frac{3}{13}}{\frac{7}{5}-\frac{7}{9}+\frac{7}{13}}\)

\(=4\cdot\frac{-1}{8}:1\cdot\frac{1}{2}-\frac{3\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{13}\right)}{7\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{13}\right)}\)

\(=-\frac{1}{4}-\frac{3}{7}=-\frac{19}{28}\)

khó thí

Làm được nhưng ngại làm lắm .

SORRY ...

d,

\(|x-\frac{1}{3}|=\frac{5}{6}\Rightarrow \left[\begin{matrix} x-\frac{1}{3}=\frac{5}{6}\\ x-\frac{1}{3}=-\frac{5}{6}\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{7}{6}\\ x=\frac{-1}{2}\end{matrix}\right.\)

e,

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

\(\Leftrightarrow 2|2x-\frac{2}{3}|=\frac{3}{4}-2=\frac{-5}{4}\)

\(\Leftrightarrow |2x-\frac{2}{3}|=-\frac{5}{8}<0\) (vô lý vì trị tuyệt đối của 1 số luôn không âm)

Vậy không tồn tại $x$ thỏa mãn đề bài.

f, 

\(\frac{2x-1}{2}=\frac{5+3x}{3}\Leftrightarrow 3(2x-1)=2(5+3x)\)

\(\Leftrightarrow 6x-3=10+6x\)

\(\Leftrightarrow 13=0\) (vô lý)

Vậy không tồn tại $x$ thỏa mãn đề bài.

a,

$0-|x+1|=5$

$|x+1|=0-5=-5<0$ (vô lý do trị tuyệt đối của một số luôn không âm)

Do đó không tồn tại $x$ thỏa mãn điều kiện đề.

b,

\(2-|\frac{3}{4}-x|=\frac{7}{12}\)

\(|\frac{3}{4}-x|=2-\frac{7}{12}=\frac{17}{12}\)

\(\Rightarrow \left[\begin{matrix} \frac{3}{4}-x=\frac{17}{12}\\ \frac{3}{4}-x=\frac{-17}{12}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{-2}{3}\\ x=\frac{13}{6}\end{matrix}\right.\)

c, 

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

\(2|\frac{1}{2}x-\frac{1}{3}|=\frac{7}{4}\)

\(|\frac{1}{2}x-\frac{1}{3}|=\frac{7}{8}\)

\(\Rightarrow \left[\begin{matrix} \frac{1}{2}x-\frac{1}{3}=\frac{7}{8}\\ \frac{1}{2}x-\frac{1}{3}=-\frac{7}{8}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{29}{12}\\ x=\frac{-13}{12}\end{matrix}\right.\)

a: 0-|x+1|=5

=>|x+1|=0-5=-5<0(vô lý)

=>x∈∅

b: \(2-\left|\frac34-x\right|=\frac{7}{12}\)

=>\(\left|x-\frac34\right|=2-\frac{7}{12}=\frac{17}{12}\)

=>\(\left[\begin{array}{l}x-\frac34=\frac{17}{12}\\ x-\frac34=-\frac{17}{12}\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{17}{12}+\frac{9}{12}=\frac{26}{12}=\frac{13}{6}\\ x=-\frac{17}{12}+\frac34=-\frac{17}{12}+\frac{9}{12}=-\frac{8}{12}=-\frac23\end{array}\right.\)

c: \(2\left|\frac12x-\frac13\right|-\frac32=\frac14\)

=>\(\left|2\left(\frac12x-\frac13\right)\right|=\frac32+\frac14=\frac74\)

=>\(\left|x-\frac23\right|=\frac74\)

=>\(\left[\begin{array}{l}x-\frac23=\frac74\\ x-\frac23=-\frac74\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac74+\frac23=\frac{21+8}{12}=\frac{29}{12}\\ x=-\frac74+\frac23=\frac{-21+8}{12}=-\frac{13}{12}\end{array}\right.\)

d: \(\left|x-\frac13\right|=\frac56\)

=>\(\left[\begin{array}{l}x-\frac13=\frac56\\ x-\frac13=-\frac56\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac56+\frac13=\frac76\\ x=-\frac56+\frac13=-\frac36=-\frac12\end{array}\right.\)

e: \(\frac34-2\left|2x-\frac23\right|=2\)

=>\(2\left|2x-\frac23\right|=\frac34-2=-\frac54<0\) (vô lý)

=>x∈∅

f: \(\frac{2x-1}{2}=\frac{5+3x}{3}\)

=>3(2x-1)=2(3x+5)

=>6x-3=6x+10

=>-3=10(vô lý)

=>x∈∅