\(a,25\%x=42,6-3,28\\ \Leftrightarrow\dfrac{1}{4}x=39,32\\ \Leftrightarrow x=157,28\)

\(b,x:40\%=7,25+6,75\\ \Leftrightarrow x:40\%=14\\ \Leftrightarrow x=5,6\)

\(c,\dfrac{7011}{25}:\left(x-42,6\right)=117,14-47,03\\ \Leftrightarrow\dfrac{7011}{25}:\left(x-42,6\right)=70,11\\ \Leftrightarrow x-42,6=4\\ \Leftrightarrow x=46,6\)

\(d,0,3\left(x+\dfrac{1}{3}\right)=\dfrac{1}{2}\cdot\left(0,81:27\right)\\ \Leftrightarrow0,3\left(x+\dfrac{1}{3}\right)=0,015\\ \Leftrightarrow x+\dfrac{1}{3}=0,05\\ \Leftrightarrow x=-\dfrac{17}{60}\)

\(a,\frac{7011}{25}:\left(x-42,6\right)=117,14-47,03\)

\(=\frac{7011}{25}:\left(x-42,6\right)=70,11\)

\(\Rightarrow\left(x-42,6\right)=\frac{7011}{25}:70,11\)

\(\Rightarrow x=4+42,6\)

\(\Rightarrow x=46,6\)

b ko co bai b a

Phần a thì có bạn giải rồi nên mình sẽ giải phần b :

0,3 x ( X+ (1)/(3) ) =(1)/(2) x (0,81 : 2,7)

(x+0,2)+(x+0.7)+...+(x+4,7)=65,5

có số x là:(4,7-0,2):0,5+1=10(số)

=> x.10+(0,2+0,7+...+4,7)=65,5

=>x.10+24,5=65,5

=>x.10=65,5-24,5

=>x.10=41

=>x     =41:10

x=4,1

vậy x=1 tí nữa mik làm tiếp cho

(x+0,2)+(x+0.7)+...+(x+4,7)=65,5

có số x là:(4,7-0,2):0,5+1=10(số)

=> x.10+(0,2+0,7+...+4,7)=65,5

=>x.10+24,5=65,5

=>x.10=65,5-24,5

=>x.10=41

=>x     =41:10

x=4,1

vậy x=1

`-(x-42,6+47,03): 27/4=0`

`-[x-(42,6-47,03)]=0`

`-(x+4,43)=0`

`x+4,43=0`

`x=-4,43`

\(\dfrac{7011}{25}:\left(x-42.6\right).15=13\)

\(\Rightarrow\dfrac{7011}{\dfrac{25}{\left(x-42,6\right)}}.0,15=13\)

\(\Rightarrow\dfrac{280,44}{\left(x-42,6\right)}.0,15=13\)

\(x-\left(280,44-42,6\right)\)

\(x=237.8+86,7\)

\(x=324,54\)

\(\)Vậy x=324,54

b.

\(7,2-\left(x+3,75:2,1\right):1,9=5,2\)

\(\Rightarrow7,2+\dfrac{-\left(x-\dfrac{3,75}{2,1}\right)}{1,9}=5,2\)

\(7,2+\dfrac{-\left(x-1,8\right)}{1,9}=5,2\)

b) Ta có: \(-5+\left|3x-1\right|+6=\left|-4\right|\)

\(\Leftrightarrow\left|3x+1\right|+1=4\)

\(\Leftrightarrow\left|3x+1\right|=3\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+1=3\\3x+1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=2\\3x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{4}{3}\end{matrix}\right.\)

Vậy: \(x\in\left\{\dfrac{2}{3};-\dfrac{4}{3}\right\}\)

c) Ta có: \(\left(x-1\right)^2=\left(x-1\right)^4\)

\(\Leftrightarrow\left(x-1\right)^2-\left(x-1\right)^4=0\)

\(\Leftrightarrow\left(x-1\right)^4-\left(x-1\right)^2=0\)

\(\Leftrightarrow\left(x-1\right)^2\cdot\left[\left(x-1\right)^2-1\right]=0\)

\(\Leftrightarrow\left(x-1\right)^2\cdot\left(x-1-1\right)\left(x-1+1\right)=0\)

\(\Leftrightarrow x\cdot\left(x-1\right)^2\cdot\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\left(x-1\right)^2=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=2\end{matrix}\right.\)

Vậy: \(x\in\left\{0;1;2\right\}\)

d) Ta có: \(5^{-1}\cdot25^x=125\)

\(\Leftrightarrow5^{-1}\cdot5^{2x}=5^3\)

\(\Leftrightarrow5^{2x-1}=5^3\)

\(\Leftrightarrow2x-1=3\)

\(\Leftrightarrow2x=4\)

hay x=2

Vậy: x=2

Đi đâu cx gặp A.R.M.Y. Hihi. Kb nha

\(\left(\frac{1}{4}.x-\frac{1}{8}\right).\frac{3}{4}=\frac{1}{4}\)

<=> \(\frac{1}{4}x-\frac{1}{8}=\frac{1}{3}\)

<=> \(\frac{1}{4}x=\frac{11}{24}\)

<=> \(x=\frac{11}{6}\)

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

<=> \(\frac{1}{2}:x+\frac{3}{4}=\frac{5}{3}\)

<=> \(\frac{x}{2}=\frac{11}{12}\)

=> \(12x=22\)

<=> \(x=\frac{11}{6}\)

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

<=> \(2x-\frac{3}{7}=\frac{29}{28}\)

<=> \(2x=\frac{41}{28}\)

<=> \(x=\frac{41}{56}\)

\(25\%.x=42,6-3,28\)

<=> \(0,25x=39,32\)

<=> \(x=157,28\)

\(\frac{15}{12}.x+\frac{3}{7}.x=\frac{6}{5}-\frac{1}{2}\)

<=> \(\left(\frac{15}{12}+\frac{3}{7}\right).x=\frac{7}{10}\)

<=> \(\frac{47}{28}.x=\frac{7}{10}\)

<=> \(x=\frac{98}{235}\)

\(\left(x-\frac{1}{2}\right).\frac{5}{3}=\frac{7}{4}-\frac{1}{2}\)

<=> \(\left(x-\frac{1}{2}\right).\frac{5}{3}=\frac{5}{4}\)

<=> \(x-\frac{1}{2}=\frac{3}{4}\)

<=> \(x=\frac{5}{4}\)

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