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

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

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

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

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

a

\(5\frac{4}{7}:x+=13\)

\(\frac{39}{7}:x=13\)

\(x=\frac{39}{7}:13\)

\(x=\frac{3}{7}\)

\(\frac{4}{7}x=\frac{9}{8}-0,125\)

\(\frac{4}{7}x=1\)

\(x=1:\frac{4}{7}\)

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

a) \(\left(4,5-2x\right)\cdot1\frac{4}{7}=\frac{11}{7}\)

\(\left(\frac{9}{2}-2x\right)\cdot\frac{11}{7}=\frac{11}{7}\)

\(\left(\frac{9}{2}-2x\right)=\frac{11}{7}\div\frac{11}{7}\)

\(\left(\frac{9}{2}-2x\right)=1\)

\(2x=\frac{9}{2}-1\)

\(x=\frac{7}{2}\div2\)

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

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

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

\(|\frac{3}{4}\cdot x|=\frac{5}{4}+\frac{1}{2}\)

\(x=\frac{7}{4}\div\frac{3}{4}\)

\(x=\frac{7}{3}\)

c) \(\frac{1}{4}-|3-x|=-\frac{3}{4}\)

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

\(|3-x|=1\)

\(x=3-1\)

\(\Rightarrow x=2\)

d) \(4\cdot\left(x-\frac{6}{7}\right)-\frac{3}{5}=1,4\)

\(4\cdot\left(x-\frac{6}{7}\right)-\frac{3}{5}=\frac{7}{5}\)

\(4\cdot\left(x-\frac{6}{7}\right)=\frac{7}{5}+\frac{3}{5}\)

\(4\cdot\left(x-\frac{6}{7}\right)=2\)

\(\left(x-\frac{6}{7}\right)=2\div4\)

\(x=\frac{1}{2}+\frac{6}{7}\)

\(x=\frac{19}{14}\)

\(\)

k cho mình nhé 

chúc bạn học giỏi

Bài 2: 

b: x+25%x=-1,25

=>1,25x=-1,25

hay x=-1

c: x-75%x=1/4

=>1/4x=1/4

hay x=1

Bài 2: 

a: =3/2-11/4=6/4-11/4=-5/4

b: =-49/6-17/2=-49/6-51/6=-100/6=-50/3

|\(\frac32x\) + \(\frac12\)| = |4\(x\) - 1|

\(\left[\begin{array}{l}\frac32x+\frac12=-4x+1\\ \frac32x+\frac12=4x-1\end{array}\right.\)

\(\left[\begin{array}{l}\frac32x+4x=1-\frac12\\ \frac32x-4x=-1-\frac12\end{array}\right.\)

\(\left[\begin{array}{l}\frac{11}{2}x=\frac12\\ -\frac52x=-\frac32\end{array}\right.\)

\(\left[\begin{array}{l}x=\frac12:\frac{11}{2}\\ x=-\frac32:\frac{-5}{2}\end{array}\right.\)

\(\left[\begin{array}{l}x=\frac12\times\frac{2}{11}\\ x=-\frac32\times\frac{-2}{5}\end{array}\right.\)

\(\left[\begin{array}{l}x=\frac{1}{11}\\ x=\frac35\end{array}\right.\)

Vậy \(x\in\) {\(\frac{1}{11};\frac35\)}

|\(\frac54x\) - \(\frac72\)| - |\(\frac58x\) + \(\frac35\)| = 0

|\(\frac54x\) - \(\frac72\)| = |\(\frac58x\) + \(\frac35\)|

\(\left[\begin{array}{l}\frac54x-\frac72=-\frac58x-\frac35\\ \frac54x-\frac72=\frac58x+\frac35\end{array}\right.\)

\(\left[\begin{array}{l}\frac54x+\frac58x=\frac72-\frac35\\ \frac54x-\frac58x=\frac72+\frac35\end{array}\right.\)

\(\left[\begin{array}{l}\frac{15}{8}x=\frac{29}{20}\\ \frac58x=\frac{41}{10}\end{array}\right.\)

\(\left[\begin{array}{l}x=\frac{29}{10}:\frac{15}{8}\\ x=\frac{41}{10}:\frac58\end{array}\right.\)

\(\left[\begin{array}{l}x=\frac{116}{75}\\ x=\frac{164}{25}\end{array}\right.\)

Vậy \(x\in\) {\(\frac{116}{75}\); \(\frac{164}{25}\)}

a) \(\frac{9}{20}\)                                                   c) \(\frac{-55}{4}\)                                                                                                                                 

b) \(\frac{116}{75}\)                                                d) \(\frac{-76}{45}\)

đúng hết đấy nhé mình tính kĩ lắm ko sai đâu

       chúc may mắn

1+³⁺⁴⁺⁹⁼

Ta có : \(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right).......\left(1-\frac{1}{2017}\right)\)

\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}......\frac{2016}{2017}\)

\(=\frac{1.2.3......2016}{2.3.4.......2017}\)

\(=\frac{1}{2017}\)

g)=>x+1/2=0

x=0-1/2

x=-1/2

hoặc 2/3-2x=0

2x=2/3-0

2x=2/3

x=2/3:2

x=1/3

nhìn @_@ hoa cả mắt đăng từng bài thôi bạn

\(a,\frac{3}{4}-\frac{-1}{2}=x+\frac{1}{8}\Leftrightarrow x+\frac{1}{8}=\frac{5}{4}\Leftrightarrow x=\frac{5}{4}-\frac{1}{8}=\frac{9}{8}\)

\(b,\frac{x}{3}+\frac{3}{4}=\frac{2}{24}\Leftrightarrow\frac{x}{3}=\frac{2}{24}-\frac{3}{4}\Leftrightarrow\frac{x}{3}=\frac{-2}{3}\Leftrightarrow x=-2\)

\(c,\frac{-2}{3}-\left(-x\right)=\frac{1}{4}\Leftrightarrow-\left(-x\right)=\frac{1}{4}-\frac{-2}{3}\Leftrightarrow x=\frac{11}{12}\)

\(d,\frac{-5}{7}+x=\frac{-1}{2}+\frac{1}{3}\Leftrightarrow x=\frac{-1}{2}+\frac{1}{3}-\frac{-5}{7}=\frac{23}{42}\)