`1/8 xx 16^n =2^n`

`1/(2^3) xx (2^4)^n =2^n`

` 2^(-3) xx 2^(4n) =2^n`

` 2^(4n-3) =2^n`

`4n-3=n`

`3n=3`

`n=1` 

Mk chỉ lm mẫu cho bn 2 câu thôi , các câu khác tương tự nhóa ~~~

a, 10 chia hết cho n - 1

=> n - 1 thuộc Ư(10)

Mà : Ư(10) = { 1;2;5;10 }

+) n - 1 = 1 => n = 1 + 1 => n = 2

+) n - 1 = 2 => n = 2 + 1 => n = 3

+) n - 1 = 5 => n = 5 + 1 => n = 6

+) n - 1 = 10 => n = 10 + 1 => n = 11

Vậy n thuộc { 2;3;6;11 }

b, n + 9 chia hết cho n - 1

Mà : n - 1 chia hết cho n - 1

Nên : ( n + 9 ) - ( n - 1 ) chia hết cho n - 1

=> n + 9 - n + 1 chia hết cho n - 1

=> 10 chia hết cho n - 1 

=> n - 1 thuộc Ư(10)

Mà : Ư(10) = { 1;2;5;10 }

+) n - 1 = 1 => n = 1 + 1 =>n = 2

+) n - 1 = 2 =>n = 2 + 1 => n = 3

+) n - 1 = 5 => n = 5 + 1 => n = 6

+) n - 1 = 10 => n = 10 + 1 => n = 11

Vậy n thuộc { 2;3;6;11 }

a. \(N=\left(\dfrac{x+2}{x\sqrt{x}+1}-\dfrac{1}{\sqrt{x}+1}\right).\dfrac{4\sqrt{x}}{3}\)  \(\left(ĐKXĐ:x\ge0\right)\)

\(N=\left(\dfrac{x+2}{x\sqrt{x}+1}-\dfrac{x-\sqrt{x}+1}{x\sqrt{x}+1}\right).\dfrac{4\sqrt{x}}{3}\)

\(\text{​​}\text{​​}N=\dfrac{\sqrt{x}+1}{x\sqrt{x}+1}.\dfrac{4\sqrt{x}}{3}\)

\(N=\dfrac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}\)

b.\(N=\dfrac{8}{9}\Leftrightarrow\dfrac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}=\dfrac{8}{9}\)

\(\Leftrightarrow3\sqrt{x}=2x-2\sqrt{x}+2\)

\(\Leftrightarrow\left(2\sqrt{x}-1\right)\left(\sqrt{x}-2\right)=0\)

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

c.\(\dfrac{1}{N}>\dfrac{3\sqrt{x}}{4}\Leftrightarrow\dfrac{3\left(x-\sqrt{x}+1\right)}{4\sqrt{x}}>\dfrac{3\sqrt{x}}{4}\)

\(\Leftrightarrow x-\sqrt{x}+1>x\)

\(\Leftrightarrow x< 1\)

a: ĐKXĐ: \(x\ge0\)

Ta có: \(N=\left(\dfrac{x+2}{x\sqrt{x}+1}-\dfrac{1}{\sqrt{x}+1}\right)\cdot\dfrac{4\sqrt{x}}{3}\)

\(=\dfrac{x+2-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\dfrac{4\sqrt{x}}{3}\)

\(=\dfrac{4\sqrt{x}}{3x-3\sqrt{x}+3}\)

Bài 1: 

\(\Leftrightarrow2^n\cdot\dfrac{9}{2}=9\cdot5^n\)

\(\Leftrightarrow2^n=2\cdot5^n\)

\(\Leftrightarrow2^{n-1}=5^n\)

Bài 2: 

a: \(A=\dfrac{2^8+3^8}{2^8}=1+\dfrac{3^8}{2^8}\)

b: \(B=\left(2^{17}+17^2\right)\cdot\left(9^{15}-15^9\right)\cdot\left(16-16\right)=0\)

Ta có : N = x³ + 8 = x³ + 2³ = (x + 2)(x² - 2x + 4)

Mà : M = N 

Nên :  (x + 2)(x² - 2x + 4) = (x + a)(x² - 2x + b)

Vậy a = 2 ; b = 4

a, Theo bài ra ta có : M = N 

hay \(\frac{2}{3}x-\frac{1}{3}=3x-2\left(x-1\right)\)

\(\Leftrightarrow\frac{2x-1}{3}=3x-2x+2\)

\(\Leftrightarrow\frac{2x-1}{3}=x+2\Leftrightarrow\frac{2x-1}{3}=\frac{3x+6}{3}\)

Khử mẫu : \(\Rightarrow2x-1=3x+6\Leftrightarrow-x-7=0\Leftrightarrow x=-7\)

b, Theo bài ra ta có : M + N = 8 

hay \(\frac{2x}{3}-\frac{1}{3}+2x-2\left(x-1\right)=8\)

\(\Leftrightarrow\frac{2x-1}{3}+2x-2x+2=8\)

\(\Leftrightarrow\frac{2x-1}{3}-6=0\Leftrightarrow\frac{2x-1-18}{3}=0\Leftrightarrow2x-19=0\Leftrightarrow x=\frac{19}{2}\)

ta có\(\frac{2\sqrt{x-8}}{\sqrt{x+1}}\in z\)

\(\Leftrightarrow\frac{\sqrt{4x-32}}{\sqrt{x+1}}\in z\)

\(\Leftrightarrow\sqrt{\frac{4x-32}{x+1}}\in z\)

\(\Leftrightarrow\frac{4x-32}{x+1}\)là số chính phương

\(\Rightarrow4x-32⋮x+1\)

\(\Leftrightarrow\left(4x+4\right)-36⋮x+1\)

mà \(\left(4x+4\right)⋮x+1\)

\(\Leftrightarrow36⋮x+1\)

\(\Rightarrow x+1\inƯ\left(36\right)\)

\(\Rightarrow x+1\in\left\{\pm1;\pm2;\pm3;\pm4;\pm6;\pm9;\pm13;\pm36\right\}\)

đến đây lập bảng tự làm tiếp sao cho thỏa mãn nhé
 

`a,`

`P(x)=M(x)+N(x)`

`P(x)=`\(\left(5x^4+8x^2-9x^3-12x-6\right)+\left(-5x^2+9x^3-5x^4+12x-8\right)\)

`P(x)= 5x^4+8x^2-9x^3-12x-6-5x^2+9x^3-5x^4+12x-8`

`P(x)=(5x^4-5x^4)+(-9x^3+9x^3)+(8x^2-5x^2)+(-12x+12x)+(-6-8)`

`P(x)=3x^2-14`

`b,`

`M(x)=N(x)+Q(x)`

`-> Q(x)=M(x)-N(x)`

`-> Q(x)=(5x^4+8x^2-9x^3-12x-6)-(-5x^2+9x^3-5x^4+12x-8)`

`Q(x)=5x^4+8x^2-9x^3-12x-6+5x^2-9x^3+5x^4-12x+8`

`Q(x)=(5x^4+5x^4)+(-9x^3-9x^3)+(8x^2+5x^2)+(-12x-12x)+(-6+8)`

`Q(x)=10x^4-18x^3+13x^2-24x+2`