\(4^2.120-4^3.\frac{1}{17}+4^2.34\)

\(=4^2\left(120-\frac{4}{17}+34\right)\)

\(=4^2.\frac{2614}{17}\)

\(=\frac{41824}{17}\)

sao lại lôi 1 phần 17  ở đâu bạn giải thích rõ đi

I don't now

mik ko biết 

sorry 

......................

a. 4² x 102 - 4³ x 17 - 4² x 34

= 4² x (102 - 4 x 17 -34)

= 4² x 0

= 0

b. 7³ x 9 + 3² x 7⁴ - 45 x 539

= 3087 + 21609 - 24255

= 24696 - 24255

= 441

\(a.\) \(\frac{1}{8}-\frac{7}{5}-\frac{2}{x}=\frac{-233}{120}\)

\(\frac{1}{8}-\frac{7}{5}+\frac{233}{120}=\frac{2}{x}\)

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

\(=>x=3\)

\(b.\frac{-12}{7}.\left(\frac{3}{4}-x\right).\frac{1}{4}=0\)

\(=>\frac{3}{4}-x=0\)

\(x=0+\frac{3}{4}\)

\(x=\frac{3}{4}\)

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

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

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

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

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

        VẬY: \(x=\frac{389}{210}\)

1. \(x=\frac{61}{42}\)

2. \(x=\frac{-36}{5}\) 

3. \(x=\frac{13}{11}\)

4. \(x=\frac{1}{12}\)

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

a, \(\left(\frac{5}{8}.0,6-5:3\frac{1}{3}\right).\left(\frac{1}{5}-1,4\right).\left(-5\right)^2.x=120\)

\(\left(\frac{5}{8}.\frac{3}{5}-5:\frac{10}{3}\right)\left(\frac{1}{5}-\frac{7}{5}\right).25x=120\)

\(\left(\frac{3}{8}-\frac{3}{2}\right)\left(-\frac{6}{5}\right).25x=120\)

\(\left(-\frac{9}{8}\right).\left(-\frac{6}{5}\right).25x=120\)

\(\frac{27}{20}.25x=120\)\(25x=120:\frac{27}{20}=\frac{800}{9}\)

\(x=\frac{800}{9}:25=\frac{32}{9}\)

đợi tí đi nấu cơm đã 

a)\(\left(\frac{1}{24.25}+\frac{1}{25.26}+...+\frac{1}{29.30}\right).120+x:\frac{1}{3}=-4\)

\(\Rightarrow\left(\frac{1}{24}-\frac{1}{25}+\frac{1}{25}-\frac{1}{26}+...+\frac{1}{29}-\frac{1}{30}\right).120+x:\frac{1}{3}=-4\)

\(\Rightarrow\left(\frac{1}{24}-\frac{1}{30}\right).120+x:\frac{1}{3}=-4\)

\(\Rightarrow\frac{1}{120}.120+x:\frac{1}{3}=-4\)

\(\Rightarrow1+x:\frac{1}{3}=-4\)

\(\Rightarrow x:\frac{1}{3}=-4-1=-5\)

\(\Rightarrow x=-5.\frac{1}{3}=\frac{-5}{3}\)

b)\(1\frac{3}{5}+\left(\frac{\frac{2}{7}+\frac{2}{17}+\frac{2}{37}}{\frac{5}{7}+\frac{5}{17}+\frac{5}{37}}\right).x=\frac{16}{5}\)

\(\Rightarrow\frac{8}{5}+\left[\frac{2.\left(\frac{1}{7}+\frac{1}{17}+\frac{1}{37}\right)}{5.\left(\frac{1}{7}+\frac{1}{17}+\frac{1}{37}\right)}\right].x=\frac{16}{5}\)

\(\Rightarrow\frac{8}{5}+\frac{2}{5}.x=\frac{16}{5}\)

\(\Rightarrow\frac{2}{5}.x=\frac{16}{5}-\frac{8}{5}=\frac{8}{5}\)

\(\Rightarrow x=\frac{8}{5}:\frac{2}{5}=\frac{8}{5}.\frac{5}{2}=\frac{8}{2}=4\)

\(\Rightarrow x=4\)

1) 

a) \(-\frac{8}{15}< \frac{x}{45}< -\frac{2}{5}\)

Lại có: \(-\frac{8}{15}=\frac{-24}{45};-\frac{2}{5}=\frac{-18}{45}\)

=> \(-\frac{24}{45}< \frac{x}{45}< -\frac{18}{45}\)

=> -24 < x < - 18 

=> x \(\in\){ - 23; -22; -21; -20 ; -19 } ( thử lại thỏa mãn )

b) \(x=\frac{-4}{3}+\frac{-7}{5}=-\frac{4.5}{3.5}+\frac{-7.3}{5.3}=-\frac{41}{15}\)

c) \(\frac{83}{x}=\frac{13}{4}+\frac{9}{10}=\frac{83}{20}\)

=> x = 20 ( thử lại thỏa mãn) 

d) \(x=\frac{10}{8}+\frac{-24}{48}+\frac{105}{-120}=-\frac{1}{8}\)

e) \(\left|x-\frac{1}{2}\right|=\left|-\frac{2}{7}\right|+\frac{5}{4}\)

\(\left|x-\frac{1}{2}\right|=\frac{2}{7}+\frac{5}{4}\)

\(\left|x-\frac{1}{2}\right|=\frac{43}{28}\)

TH1: \(x-\frac{1}{2}=\frac{43}{28}\)

         \(x=\frac{57}{28}\)

TH2: \(x-\frac{1}{2}=-\frac{43}{28}\)

      \(x=-\frac{29}{28}\)

A=1/2