Lời giải:

Không mất tổng quát giả sử $m\geq n$. Ta có:

$5^m+5^n=150$

$5^n(5^{m-n}+1)=150=2.3.5^2$

Ta thấy $5^{m-n}+1$ không chia hết cho $5$ với mọi $m,n$ nguyên dương.

Do đó $5^n=5^2\Rightarrow n=2$.

$5^{m-n}+1=2.3.5^2:5^n=2.3.5^2:5^2=6$

$5^{m-2}+1=6$

$5^{m-2}=5=5^1\Rightarrow m-2=1\Rightarrow m=3$

Vậy $(m,n)=(3,2)$ và hoán vị

\(\dfrac{-5}{6}-\left(-\dfrac{7}{6}\right)+4\dfrac{1}{2}\\ =\dfrac{-5}{6}+\dfrac{7}{6}+\dfrac{9}{2}\\ =\dfrac{2}{6}+\dfrac{9}{2}\\ =\dfrac{2}{6}+\dfrac{27}{6}\\ =\dfrac{29}{6}\)

\(\dfrac{1}{4}.5\dfrac{1}{3}:\left(-\dfrac{3}{7}\right)=\dfrac{1}{4}.\dfrac{16}{3}:\left(-\dfrac{3}{7}\right)=\dfrac{4}{3}:\left(-\dfrac{3}{7}\right)=-\dfrac{28}{9}\)

\(\dfrac{1}{4}.\dfrac{2}{3}+2\dfrac{3}{4}=\dfrac{2}{12}+\dfrac{11}{4}=\dfrac{35}{12}\)

\(-\dfrac{5}{6}+\dfrac{7}{6}+4\dfrac{1}{2}=\dfrac{2}{6}+\dfrac{9}{2}=\dfrac{1}{3}+\dfrac{9}{2}=\dfrac{29}{6}\)

\(\dfrac{1}{4}\cdot5\dfrac{1}{3}:\dfrac{-3}{7}\\ =\dfrac{1}{4}\cdot\dfrac{16}{3}\cdot\dfrac{-7}{3}\\ =\dfrac{1\cdot16\cdot-7}{4\cdot3\cdot3}\\ =\dfrac{-28}{9}\)

_____________

\(\dfrac{1}{4}\cdot\dfrac{2}{3}+2\dfrac{3}{4}\\ =\dfrac{1}{4}\cdot\dfrac{2}{3}+\dfrac{11}{4}\\ =\dfrac{2}{12}+\dfrac{11}{4}\\ =\dfrac{2}{12}+\dfrac{33}{12}\\ =\dfrac{35}{12}\)

\(\dfrac{1}{4}.5\dfrac{1}{3}:\left(-\dfrac{3}{7}\right)=\dfrac{1}{4}.\dfrac{16}{3}:\left(-\dfrac{3}{7}\right)=\dfrac{4}{3}:\left(-\dfrac{3}{7}\right)=-\dfrac{28}{9}\)

\(\dfrac{1}{4}.\dfrac{2}{3}+2\dfrac{3}{4}=\dfrac{1}{6}+\dfrac{11}{4}=\dfrac{35}{12}\)

Lời giải:
Ta thấy:

$ab=\frac{3}{4}>0; bc=\frac{3}{10}>0; ca=\frac{1}{10}>0$

$\Rightarrow a,b,c>0$.

$ab=\frac{3}{4}, bc=\frac{3}{10}, ca=\frac{1}{10}$

$\Rightarrow ab.bc.ac=\frac{3}{4}.\frac{3}{10}.\frac{1}{10}$

$\Rightarrow (abc)^2=\frac{9}{400}=(\frac{3}{20})^2=(\frac{-3}{20})^2$

Do $a,b,c>0$ nên $abc>0$

$\Rightarrow abc=\frac{3}{20}$.

$a=abc:(bc)=\frac{3}{20}: \frac{3}{10}=\frac{1}{2}$

$b=abc:(ac)=\frac{3}{20}:\frac{1}{10}=\frac{3}{2}$

$c=abc:(ab)=\frac{3}{20}: \frac{3}{4}=\frac{1}{5}$

\(\dfrac{ac}{bc}=\dfrac{1}{10}:\dfrac{3}{10}=\dfrac{1}{3}\Rightarrow\dfrac{a}{b}=\dfrac{1}{3}\Rightarrow a=\dfrac{1}{3}b\)

Mà: \(ab=\dfrac{3}{4}=>\dfrac{1}{3}b\cdot b=\dfrac{3}{4}=>b^2=\dfrac{3}{4}:\dfrac{1}{3}=\dfrac{9}{4}\) 

\(\Rightarrow\left[{}\begin{matrix}b=\dfrac{3}{2}\\b=-\dfrac{3}{2}\end{matrix}\right.\)

TH1: \(b=\dfrac{3}{2}=>a=\dfrac{1}{3}\cdot\dfrac{3}{2}=\dfrac{1}{2}\)

\(\Rightarrow\dfrac{3}{2}c=\dfrac{3}{10}=>c=\dfrac{3}{10}:\dfrac{3}{2}=\dfrac{1}{5}\)

TH2: \(b=-\dfrac{3}{2}=>a=\dfrac{1}{3}\cdot\left(-\dfrac{3}{2}\right)=-\dfrac{1}{2}\)

\(\Rightarrow-\dfrac{3}{2}c=\dfrac{3}{10}=>c=\dfrac{3}{10}:-\dfrac{3}{2}=\dfrac{-1}{5}\)

\(\dfrac{-15}{20}+\left(5-\dfrac{7}{8}\right)=\dfrac{-3}{4}+5-\dfrac{7}{8}=\dfrac{-6}{8}+\dfrac{40}{8}-\dfrac{7}{8}=\dfrac{27}{8}\)

\(\dfrac{4}{-5}\cdot\left(\dfrac{2}{3}-\dfrac{7}{8}\right)=\dfrac{4}{-5}\cdot\left(\dfrac{16}{24}-\dfrac{21}{24}\right)=\dfrac{4}{-5}\cdot\dfrac{-5}{24}=\dfrac{1}{6}\)

\(\dfrac{13}{17}\cdot\dfrac{4}{5}+\dfrac{13}{17}\cdot\dfrac{-3}{4}=\dfrac{13}{17}\cdot\left(\dfrac{4}{5}+\dfrac{-3}{4}\right)=\dfrac{13}{17}\cdot\left(\dfrac{16}{20}-\dfrac{15}{20}\right)=\dfrac{13}{17}\cdot\dfrac{1}{20}=\dfrac{13}{340}\)

\(-\dfrac{15}{20}+\left(5-\dfrac{7}{8}\right)=-\dfrac{3}{4}+5-\dfrac{7}{8}=\dfrac{-6}{8}-\dfrac{7}{8}+5=-\dfrac{13}{8}+5=\dfrac{27}{8}\)

\(-\dfrac{4}{5}\left(\dfrac{2}{3}-\dfrac{7}{8}\right)=-\dfrac{4}{5}\left(-\dfrac{5}{24}\right)=\dfrac{1}{6}\)

\(\dfrac{13}{17}.\dfrac{4}{5}+\dfrac{13}{17}.\left(-\dfrac{3}{4}\right)=\dfrac{13}{17}\left(\dfrac{4}{5}-\dfrac{3}{4}\right)=\dfrac{13}{17}.\dfrac{1}{20}=\dfrac{13}{340}\)

\(\left(x-\dfrac{3}{2}\right)^2=\dfrac{9}{16}=\left(\dfrac{3}{4}\right)^2\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{2}=\dfrac{3}{4}\\x-\dfrac{3}{2}=-\dfrac{3}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{4}\\x=\dfrac{3}{4}\end{matrix}\right.\)

\(\left(x-\dfrac{3}{2}\right)^2=\dfrac{9}{16}\\ \left(x-\dfrac{3}{2}\right)^2=\left(\dfrac{3}{4}\right)^2\)

TH1: \(x-\dfrac{3}{2}=\dfrac{3}{4}\Rightarrow x=\dfrac{3}{4}+\dfrac{3}{2}\Rightarrow x=\dfrac{3}{4}+\dfrac{6}{4}\Rightarrow x=\dfrac{9}{4}\)

TH2: \(x-\dfrac{3}{2}=-\dfrac{3}{4}\Rightarrow x=-\dfrac{3}{4}+\dfrac{3}{2}\Rightarrow x=-\dfrac{3}{4}+\dfrac{6}{4}\Rightarrow x=\dfrac{3}{4}\)

\(10\cdot10^2\cdot10^3\cdot...\cdot10^x=10^{12}\\ 10^{1+2+3+...+x}=10^{12}\\ 1+2+3+...+x=12\\ \dfrac{x\left(x+1\right)}{2}=12\\ x\left(x+1\right)=24\\ x^2+x-24=0\)

=> Không có x thuộc N thỏa 

anh giải thích cho em phần không có x thuộc N thỏa là sao

Đặt \(x^2+3x=t\)

\(\left(t+1\right)\left(t-3\right)-5=t^2-2t-8=\left(t-1\right)^2-9=\left(t-4\right)\left(t+2\right)\)

\(\Rightarrow\left(x^2+3x-4\right)\left(x^2+3x+2\right)=\left(x-1\right)\left(x+4\right)\left(x+1\right)\left(x+2\right)\)

\(\left(x^2+3x+1\right)\left(x^2+3x-3\right)-5\\ =\left(x^2+3x-1+2\right)\left(x^2+3x-1-2\right)-5\\ =\left(x^2+3x-1\right)^2-2^2-5\\ =\left(x^2+3x-1\right)^2-3^2\\ =\left(x^2+3x-1-3\right)\left(x^2+3x-1+3\right)\\ =\left(x^2+3x-4\right)\left(x^2+3x+2\right)\\ =\left(x-1\right)\left(x+4\right)\left(x+1\right)\left(x+2\right)\)

\(x-\dfrac{8}{15}=\dfrac{11}{3}\Leftrightarrow x=\dfrac{11}{3}+\dfrac{8}{15}=\dfrac{21}{5}\)

\(\dfrac{3x}{2}+\dfrac{7}{3}=\dfrac{5}{4}\Leftrightarrow\dfrac{3x}{2}=\dfrac{5}{4}-\dfrac{7}{3}=-\dfrac{13}{12}\Leftrightarrow x=-\dfrac{13}{12}:\dfrac{3}{2}=-\dfrac{13}{18}\)

\(\dfrac{7}{8}-x=-\dfrac{4}{5}:\dfrac{3}{10}=-\dfrac{8}{3}\Leftrightarrow x=\dfrac{7}{8}+\dfrac{8}{3}=\dfrac{85}{24}\)

\(x-\dfrac{8}{15}=\dfrac{11}{3}\\ x=\dfrac{8}{15}+\dfrac{11}{3}\\ x=\dfrac{8}{15}+\dfrac{55}{15}\\ x=\dfrac{63}{15}\)

___________

\(\dfrac{3}{2}x+\dfrac{7}{3}=\dfrac{5}{4}\\ \dfrac{3}{2}x=\dfrac{5}{4}-\dfrac{7}{3}\\ \dfrac{3}{2}x=\dfrac{-13}{12}\\ x=\dfrac{-13}{12}:\dfrac{3}{2}\\ x=\dfrac{-13}{18}\)

____________

\(\dfrac{7}{8}-x=-\dfrac{4}{5}:\dfrac{3}{10}\\ \dfrac{7}{8}-x=\dfrac{-8}{3}\\ x=\dfrac{7}{8}+\dfrac{8}{3}\\ x=\dfrac{85}{24}\)

Mẹ là cái cung tên, con là cái mũi tên.