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\(m-n=3\Leftrightarrow m=n+3\)

Thay vào B ta được :

\(B=\frac{n+3-8}{n-5}=\frac{4\left(n+3\right)-n}{3\left(n+3\right)+3}=\frac{n-5}{n-5}+\frac{3n+12}{3n+12}=1+1=2\)

 m - n = 3 => m = 3+ n 

Thay vào B ta có

\(B=\frac{3+n-8}{n-5}+\frac{4\left(3+n\right)-n}{3\left(3+n\right)+3}=\frac{n-5}{n-5}+\frac{12+4n-n}{9+3n+3}=1+\frac{3n+12}{3n+12}=2\)

Từ m-n=3=>m=n+3

Ta có: \(\frac{m-8}{n-3}=\frac{\left(n+3\right)-8}{n-3}=\frac{n-5}{n-5}=1\)  (1)

\(\frac{4m-n}{3m+3}=\frac{4.\left(n+3\right)-n}{3.\left(n+3\right)+3}=\frac{4n+12-n}{3n+9+3}=\frac{\left(4n-n\right)+12}{3n+12}=\frac{3n+12}{3n+12}=1\)   (2)

Từ (1) và (2) \(\Rightarrow A=1-1=0\)

Vậy A=0

0

Bn có thể làm bằng 2 cahs :

1.Đặt m bằng n là số nào đó thỏa mãn bài

2.bằng cahs suy luận ( mink k thích cahcs này lắm)

a) Với \(\frac{m}{n} = \frac{{ - 5}}{6}\), giá trị của biểu thức là:

\(\begin{array}{l}A = \frac{{ - 2}}{3} - \left( {\frac{{ - 5}}{6} + \frac{{ - 5}}{2}} \right).\frac{{ - 5}}{8}\\A = \frac{{ - 2}}{3} - \frac{{-20}}{6}.\frac{{ - 5}}{8}\\A = \frac{{ - 2}}{3} - \frac{{ 25}}{{12}}\\A = \frac{{ - 33}}{{12}}\end{array}\)

b) Với \(\frac{m}{n} = \frac{5}{2}\) , giá trị của biểu thức là:

\(\begin{array}{l}A = \frac{{ - 2}}{3} - \left( {\frac{5}{2} + \frac{{ - 5}}{2}} \right).\frac{{ - 5}}{8}\\A = \frac{{ - 2}}{3} - 0.\frac{{ - 5}}{8} = \frac{{ - 2}}{3}\end{array}\)

c) Với \(\frac{m}{n} = \frac{2}{{ - 5}}\) , giá trị của biểu thức là:

\(\begin{array}{l}A = \frac{-2}{3} - \left( {\frac{2}{{ - 5}} + \frac{{ - 5}}{2}} \right).\frac{{ - 5}}{8}\\A = \frac{-2}{3} - \left( {\frac{{ - 4}}{{10}} + \frac{{ - 25}}{{10}}} \right).\frac{{ - 5}}{8}\\A = \frac{-2}{3} - \frac{{ - 29}}{{10}}.\frac{{ - 5}}{8}\\A = \frac{-2}{3} - \frac{{29}}{{16}}\\A = \frac{{-32}}{{48}} - \frac{{87}}{{48}}\\A = \frac{{ - 119}}{{48}}\end{array}\).

a) \(\lim \frac{{2{n^2} + 6n + 1}}{{8{n^2} + 5}} = \lim \frac{{{n^2}\left( {2 + \frac{6}{n} + \frac{1}{{{n^2}}}} \right)}}{{{n^2}\left( {8 + \frac{5}{{{n^2}}}} \right)}} = \lim \frac{{2 + \frac{6}{n} + \frac{1}{n}}}{{8 + \frac{5}{n}}} = \frac{2}{8} = \frac{1}{4}\)

b) \(\lim \frac{{4{n^2} - 3n + 1}}{{ - 3{n^3} + 6{n^2} - 2}} = \lim \frac{{{n^3}\left( {\frac{4}{n} - \frac{3}{{{n^2}}} + \frac{1}{{{n^3}}}} \right)}}{{{n^3}\left( { - 3 + \frac{6}{n} - \frac{2}{{{n^3}}}} \right)}} = \lim \frac{{\frac{4}{n} - \frac{3}{{{n^2}}} + \frac{1}{{{n^3}}}}}{{ - 3 + \frac{6}{n} - \frac{2}{{{n^3}}}}} = \frac{{0 - 0 + 0}}{{ - 3 + 0 - 0}} = 0\).

c) \(\lim \frac{{\sqrt {4{n^2} - n + 3} }}{{8n - 5}} = \lim \frac{{n\sqrt {4 - \frac{1}{n} + \frac{3}{{{n^2}}}} }}{{n\left( {8 - \frac{5}{n}} \right)}} = \frac{{\sqrt {4 - 0 + 0} }}{{8 - 0}} = \frac{2}{8} = \frac{1}{4}\).

d) \(\lim \left( {4 - \frac{{{2^{{\rm{n}} + 1}}}}{{{3^{\rm{n}}}}}} \right) = \lim \left( {4 - 2 \cdot {{\left( {\frac{2}{3}} \right)}^{\rm{n}}}} \right) = 4 - 2.0 = 4\).

e) \(\lim \frac{{{{4.5}^{\rm{n}}} + {2^{{\rm{n}} + 2}}}}{{{{6.5}^{\rm{n}}}}} = \lim \frac{{{{4.5}^{\rm{n}}} + {2^2}{{.2}^{\rm{n}}}}}{{{{6.5}^{\rm{n}}}}} = \lim \frac{{{5^n}.\left[ {4 + 4.{{\left( {\frac{2}{5}} \right)}^{\rm{n}}}} \right]}}{{{{6.5}^n}}} = \lim \frac{{4 + 4.{{\left( {\frac{2}{5}} \right)}^{\rm{n}}}}}{6} = \frac{{4 + 4.0}}{6} = \frac{2}{3}\).

g) \(\lim \frac{{2 + \frac{4}{{{n^3}}}}}{{{6^{\rm{n}}}}} = \lim \left( {2 + \frac{4}{{{{\rm{n}}^3}}}} \right).\lim {\left( {\frac{1}{6}} \right)^{\rm{n}}} = \left( {2 + 0} \right).0 = 0\).

a) Câu này thiếu đề nhé bạn.

b) \(\frac{25}{5^n}=5\)

\(\Rightarrow5^n=25:5\)

\(\Rightarrow5^n=5\)

\(\Rightarrow5^n=5^1\)

\(\Rightarrow n=1\)

Vậy \(n=1.\)

c) \(\frac{81}{\left(-3\right)^n}=-243\)

\(\Rightarrow\left(-3\right)^n=81:\left(-243\right)\)

\(\Rightarrow\left(-3\right)^n=-\frac{1}{3}\)

\(\Rightarrow\left(-3\right)^n=\left(-3\right)^{-1}\)

\(\Rightarrow n=-1\)

Vậy \(n=-1.\)

e) \(\left(\frac{1}{3}\right)^n=\frac{1}{81}\)

\(\Rightarrow\left(\frac{1}{3}\right)^n=\left(\frac{1}{3}\right)^4\)

\(\Rightarrow n=4\)

Vậy \(n=4.\)

f) \(\left(-\frac{3}{4}\right)^n=\frac{81}{256}\)

\(\Rightarrow\left(-\frac{3}{4}\right)^n=\left(-\frac{3}{4}\right)^4\)

\(\Rightarrow n=4\)

Vậy \(n=4.\)

Chúc bạn học tốt!

d) \(\frac{1}{2}.2^n+4.2^n=9.2^5\)

\(\Rightarrow2^n.\left(\frac{1}{2}+4\right)=288\)

\(\Rightarrow2^n.\frac{9}{2}=288\)

\(\Rightarrow2^n=288:\frac{9}{2}\)

\(\Rightarrow2^n=64\)

\(\Rightarrow2^n=2^6\)

\(\Rightarrow n=6\)

Vậy \(n=6.\)

g) \(-\frac{512}{343}=\left(-\frac{8}{7}\right)^n\)

\(\Rightarrow\left(-\frac{8}{7}\right)^n=\left(-\frac{8}{7}\right)^3\)

\(\Rightarrow n=3\)

Vậy \(n=3.\)

h) \(5^{-1}.25^n=125\)

\(\Rightarrow5^{-1}.5^{2n}=5^3\)

\(\Rightarrow5^{-1+2n}=5^3\)

\(\Rightarrow-1+2n=3\)

\(\Rightarrow2n=3+1\)

\(\Rightarrow2n=4\)

\(\Rightarrow n=4:2\)

\(\Rightarrow n=2\)

Vậy \(n=2.\)

k) \(3^{-1}.3^n+6.3^{n-1}=7.3^6\)

\(\Rightarrow3^{n-1}+6.3^{n-1}=7.3^6\)

\(\Rightarrow3^{n-1}.\left(1+6\right)=7.3^6\)

\(\Rightarrow3^{n-1}.7=7.3^6\)

\(\Rightarrow n-1=6\)

\(\Rightarrow n=6+1\)

\(\Rightarrow n=7\)

Vậy \(n=7.\)

Chúc bạn học tốt!

b)\(\frac{25}{5^n}\)=5

\(5^n\)=25/5

\(5^n\)=\(5^1\)

⇒ n = 1

c) \(\frac{81}{\left(-3\right)^n}\)=-243

\(\left(-3\right)^n\)=81/-243

\(\left(-3\right)^n\)=\(\frac{-1}{3}\)

\(\left(-3\right)^n\)=\(\left(-3\right)^{-1}\)

⇒n=-1