(\(x\) + 2)ⁿ⁺¹ = ( \(x\) + 2)ⁿ⁺¹¹

(\(x+2\))ⁿ⁺¹ -  ( \(x\) + 2)ⁿ⁺¹¹ = 0

(\(x\) + 2)ⁿ⁺¹.(  1 + (\(x\) + 2)¹⁰) = 0

(\(x\) + 2)¹⁰ + 1 > 0 ∀ \(x\)

=> (\(x\) + 2)ⁿ⁺¹ = 0 ⇒ \(x\) + 2  = 0 ⇒ \(x\) = -2

vậy \(x\) = -2

x = 0

x = 1000

x = 1000

x = k có số nào hoặc là mk chịu

x = 16 ; 17 ; 18 ; 19 ; 20 ; ... 99

x = 100 ; 101 ; 102 ; 103 ; 104 

x = 2 ; 3 ; 4 ; 5 ; 6 ; 7 ; 8 ; 9 ; 10

x = chịu

x = nhiều hoặc mk k biết

\(1,A=5^{n+2}+26\cdot5^n+8^{2n+1}\\ A=5^n\cdot25+26\cdot5^n+8\cdot8^{2n+1}\\ A=51\cdot5^n+8\cdot64^n\)

Ta có \(64:59R5\Rightarrow64^n:59R5\)

Vì vậy \(51\cdot5^n+8\cdot64^n:59R=5^n\cdot51+8\cdot5^n=5^n\left(51+8\right)=5^n\cdot59⋮59\)

Vậy \(A⋮59\)

(\(R\) là dư)

\(2,\\ a,2x\ge0;\left(x+2\right)^2\ge0,\forall x\\ \Leftrightarrow P=\dfrac{\left(x+2\right)^2}{2x}\ge0\\ P_{min}=0\Leftrightarrow x+2=0\Leftrightarrow x=-2\)

ta có : 13/20 x 20/17 < X < 11/6 x 12

                   13/17 <   X    <   33/4 

                    X = 14/17,15/17,16/17,......... 

                     k đúng cho mình nhé

Ta có : 

\(\frac{\text{13}}{\text{20}}\text{x}\frac{\text{20}}{\text{17}}< \text{x}< \frac{\text{11}}{\text{6}}\text{x}\frac{\text{12}}{\text{1}}\) ( x ∈ N )

=> \(\frac{\text{13}}{\text{17}}< x< 22\)

=> 0,764.... < x < 22

=> x ∈ { 1 ; 2 ; 3 ; ... ; 21 }

Vậy x ∈ { 1 ; 2 ; 3 ; ... ; 21 }

\(\left(1\dfrac{3}{4}-\dfrac{4}{6}\right):\left(1\dfrac{1}{5}+2\dfrac{2}{5}+\dfrac{1}{5}\right)< x< 1\dfrac{1}{5}.1\dfrac{1}{4}+3\dfrac{2}{11}:2\dfrac{3}{121}\)

x = 0

x = 1000

x = 1000

x = k có số nào hoặc là mk chịu

x = 16 ; 17 ; 18 ; 19 ; 20 ; ... 99

x = 100 ; 101 ; 102 ; 103 ; 104 

x = 2 ; 3 ; 4 ; 5 ; 6 ; 7 ; 8 ; 9 ; 10

x = chịu

x = nhiều hoặc mk k biết

cho mình hỏi

giai ho cau

a, ĐK: \(x\ge0;x\ne1\)

\(P=\left(1+\dfrac{2}{\sqrt{x}+1}+\dfrac{3}{\sqrt{x}-1}\right).\left(1-\dfrac{6}{\sqrt{x}+5}\right)\)

\(=\left[\dfrac{x-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\dfrac{2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\dfrac{3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right].\dfrac{\sqrt{x}+5-6}{\sqrt{x}+5}\)

\(=\dfrac{x+5\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}.\dfrac{\sqrt{x}-1}{\sqrt{x}+5}\)

\(=\dfrac{\sqrt{x}}{\sqrt{x}+1}\)

b, \(P=\dfrac{\sqrt{x}}{\sqrt{x}+1}=1-\dfrac{1}{\sqrt{x}+1}\in Z\)

\(\Leftrightarrow\sqrt{x}+1\in\left\{-1;1\right\}\)

\(\Leftrightarrow\sqrt{x}=0\)

\(\Leftrightarrow x=0\left(tm\right)\)

Vậy ta có điều phải chứng minh.

a: Ta có: \(P=\left(1+\dfrac{2}{\sqrt{x}+1}+\dfrac{3}{\sqrt{x}-1}\right)\cdot\left(1-\dfrac{6}{\sqrt{x}+5}\right)\)

\(=\dfrac{x-1+2\sqrt{x}-2+3\sqrt{x}+3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+5-6}{\sqrt{x}+5}\)

\(=\dfrac{x+5\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}+5}\)

\(=\dfrac{\sqrt{x}}{\sqrt{x}+1}\)