a) =>

b) =>

a) ${\left(\frac{1}{2}\right)}^m=\frac{1}{32}$

\(\Rightarrow\left(\dfrac{1}{2}\right)^m=\left(\dfrac{1}{2}\right)^5\Rightarrow m=5\)

b)

$\frac{343}{125}={\left(\frac{7}{5}\right)}^n$

\(\Rightarrow\left(\dfrac{7}{5}\right)^3=\left(\dfrac{7}{5}\right)^n\Rightarrow n=3\)

a) =>

b) =>

c)\(7^{2n}+7^{2n+2}=2450\)

⇒\(7^{2n}+7^{2n}.7^2=2450\)

⇒\(7^{2n}.50=2450\)

⇒\(7^{2n}=49\)\(=7^2\)

⇒2n=2

⇒n=1

a)\(\left(-\dfrac{1}{5}\right)^n=-\dfrac{1}{125}\)                   b)\(\left(-\dfrac{2}{11}\right)^m=\dfrac{4}{121}\)

\(\left(-\dfrac{1}{5}\right)^n=\left(-\dfrac{1}{5}\right)^3\)                    \(=\left(-\dfrac{2}{11}\right)^m=\left(-\dfrac{2}{11}\right)^2\)

⇒n=3                                          ⇒m=2

a) \(\left(\frac{1}{2}\right)^m=\frac{1}{32}\)

\(\Rightarrow\left(\frac{1}{2}\right)^m=\left(\frac{1}{2}\right)^5\)

=> m = 5

Vậy m = 5

b) \(\frac{343}{125}=\left(\frac{7}{5}\right)^n\)

\(\Rightarrow\left(\frac{7}{5}\right)^3=\left(\frac{7}{5}\right)^n\)

=> n = 3

Vậy n = 3

ĐKXĐ: \(x\ne\pm1;x\ne0\)

a)\(\left(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}\right):\dfrac{2x}{5x-5}-\dfrac{x^2-1}{x^2+2x+1}\)

\(=\left(\dfrac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}\right):\dfrac{2x}{5x-5}-\dfrac{x^2-1}{x^2+2x+1}\)

\(=\dfrac{x^2+2x+1-\left(x^2-2x+1\right)}{\left(x-1\right)\left(x+1\right)}:\dfrac{2x}{5x-5}-\dfrac{x^2-1}{x^2+2x+1}\)

\(=\dfrac{4x}{\left(x-1\right)\left(x+1\right)}:\dfrac{2x}{5x-5}-\dfrac{x^2-1}{x^2+2x+1}\)

\(=\dfrac{4x}{\left(x-1\right)\left(x+1\right)}.\dfrac{5\left(x-1\right)}{2x}-\dfrac{x^2-1}{x^2+2x+1}\)

\(=\dfrac{10}{x+1}-\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x+1\right)^2}\)

\(=\dfrac{10}{x+1}-\dfrac{x-1}{x+1}\)

\(=\dfrac{11-x}{x+1}\)

b) \(A=\dfrac{11-x}{x+1}=2\)

\(\Leftrightarrow11-x=2\left(x+1\right)\)

\(\Leftrightarrow11-x=2x+2\)

\(\Leftrightarrow-x-2x=2-11\)

\(\Leftrightarrow-3x=-9\)

\(\Leftrightarrow x=3\left(nhận\right)\)

c) -Để \(A=\dfrac{11-x}{x+1}\in Z\) thì:

\(\left(11-x\right)⋮\left(x+1\right)\)

\(\Rightarrow\left(12-x-1\right)⋮\left(x+1\right)\)

\(\Rightarrow12⋮\left(x+1\right)\)

\(\Rightarrow\left(x+1\right)\inƯ\left(12\right)\)

\(\Rightarrow\left(x+1\right)\in\left\{1;2;3;4;6;12;-1;-2;-3;-4;-6;-12\right\}\)

\(\Rightarrow x\in\left\{2;3;5;11;-2;-3;-4;-5;-7;-13\right\}\)

a) \(\left(\dfrac{1}{2}\right)^{2n-1}=\dfrac{1}{8}\)

\(\Rightarrow\left(\dfrac{1}{2}\right)^{2n-1}=\left(\dfrac{1}{2}\right)^3\)

\(\Rightarrow2n-1=3\)

\(\Rightarrow2n=4\)

\(\Rightarrow n=2\)

a) \(\left(\dfrac{1}{2}\right)^{2n-1}=\dfrac{1}{8}\)

\(\Rightarrow2^{-\left(2n-1\right)}=2^{-3}\)

\(\Rightarrow2^{-2n+1}=2^{-3}\)

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

\(\Rightarrow-2n=-4\)

\(\Rightarrow n=-2\)

Vậy ...

b) \(\left(\dfrac{7}{5}\right)^n=\dfrac{343}{125}\)

\(\Rightarrow\left(\dfrac{7}{5}\right)^n=\left(\dfrac{7}{5}\right)^3\)

\(\Rightarrow n=3\)

Vậy ....

a. )\(\left(\dfrac{1}{2}\right)^{2n-1}\)=\(\dfrac{1}{8}\)

=> 2n-1=3

=>2n=3+1

=>2n=4

=>n=4:2=2

\(A=\left(\dfrac{1}{x^2-1}+\dfrac{1}{x+1}\right):\left(\dfrac{1}{x-1}-\dfrac{1}{x}\right)\)

\(\Rightarrow A=\left(\dfrac{1}{\left(x-1\right)\left(x+1\right)}+\dfrac{x-1}{\left(x-1\right)\left(x+1\right)}\right):\left(\dfrac{x}{x\left(x-1\right)}-\dfrac{x-1}{x\left(x-1\right)}\right)\)

\(\Rightarrow A=\dfrac{1+x-1}{\left(x-1\right)\left(x+1\right)}:\dfrac{x-x+1}{x\left(x-1\right)}\)

\(\Rightarrow A=\dfrac{x}{\left(x-1\right)\left(x+1\right)}:\dfrac{1}{x\left(x-1\right)}\)

\(\Rightarrow A=\dfrac{x}{\left(x-1\right)\left(x+1\right)}.x\left(x-1\right)\)

\(\Rightarrow A=\dfrac{x^2}{x+1}\)

đk : xkhác -1 ; 1 

\(A=\left(\dfrac{1+x-1}{\left(x+1\right)\left(x-1\right)}\right):\left(\dfrac{x-x+1}{x\left(x-1\right)}\right)=\dfrac{x}{\left(x+1\right)\left(x-1\right)}:\dfrac{1}{x\left(x-1\right)}=\dfrac{x^2}{x+1}\)

\(A=\dfrac{x-1+1}{\left(x+1\right)\left(x-1\right)}:\dfrac{x-x+1}{x\left(x-1\right)}\)

\(=\dfrac{x}{\left(x+1\right)\left(x-1\right)}\cdot\dfrac{x\left(x-1\right)}{1}\)

\(=\dfrac{x^2}{x+1}\)

Câu 1:

Ta có: \(\left[\dfrac{1}{2.5}+\dfrac{1}{5.8}+...+\dfrac{1}{65.68}\right]x-\dfrac{7}{34}=\dfrac{19}{68}\)

\(\Rightarrow\left[\dfrac{1}{3}\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+...+\dfrac{3}{65.68}\right)\right]x=\dfrac{33}{68}\)

\(\Rightarrow\left[\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{65}-\dfrac{1}{68}\right)\right]x=\dfrac{33}{68}\)

\(\Rightarrow\left[\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{68}\right)\right]x=\dfrac{33}{68}\)

\(\Rightarrow\dfrac{11}{68}x=\dfrac{33}{68}\)

\(\Rightarrow x=3\)

Vậy \(x=3.\)

câu 4:B=8

a)\(\left(\dfrac{1}{2}\right)^n=\dfrac{1}{32}\)

=>\(\left(\dfrac{1}{2}\right)^n=\left(\dfrac{1}{2}\right)^5\)

=>n=5

b)\(\left(\dfrac{343}{125}\right)=\left(\dfrac{7}{5}\right)^n\)

=>\(\left(\dfrac{7}{5}\right)^3=\left(\dfrac{7}{5}\right)^n\)

=>n=3

c)\(\dfrac{16}{2^n}=2\)

=>2ⁿ=\(\dfrac{16}{2}\)

=>2ⁿ=8

=>2ⁿ=2³

=>n=3

d)\(\dfrac{\left(-3\right)^n}{81}=-27\)

=>(-3)ⁿ=-27.81

=>(-3)ⁿ=-2187

=>(-3)ⁿ=(-3)⁷

=>n=7

e)8ⁿ:2ⁿ=4

=>(2³)ⁿ:2ⁿ=4

=>2³ⁿ:2ⁿ=4

=>2^3n-n=4

=>2²ⁿ=4

=>2²ⁿ=2²

=>2n=2

=>n=1

f)3².3ⁿ=3⁵

=>3ⁿ=3⁵:3²

=>3ⁿ=3^5-2

=>3ⁿ=3³

=>n=3

g) (2²:4).2ⁿ=4

=>1.2ⁿ=2²

=>n=2

h)3^-2.3⁴.3ⁿ=3⁷

=>\(\left(\dfrac{1}{3}\right)^2\).3⁴.3ⁿ=3⁷

=>3².3ⁿ=3⁷

=>3²⁺ⁿ=3⁷

=>2+n=7

=>n=5

a) ĐKXĐ: \(x\ne0;x\ne\pm1\)

Ta có: \(A=\left(\dfrac{1}{x^2+x}-\dfrac{2-x}{x+1}\right).\dfrac{x}{x+1}\)

\(A=\left[\dfrac{1}{x\left(x+1\right)}-\dfrac{2-x}{x+1}\right].\dfrac{x}{x+1}\)

\(A=\dfrac{1-x\left(2-x\right)}{x\left(x+1\right)}.\dfrac{x}{x+1}\)

\(A=\dfrac{1-2x+x^2}{x\left(x+1\right)}.\dfrac{x}{x+1}\)

\(A=\dfrac{\left(1-x\right)^2}{\left(x+1\right)^2}\)