\(A=11\sqrt{x+5}\left(đk:x\ge-5\right)\)

\(A=6\Leftrightarrow11\sqrt{x+5}=6\)

\(\Leftrightarrow\sqrt{x+5}=\dfrac{6}{11}\Leftrightarrow x+5=\dfrac{36}{121}\)

\(\Leftrightarrow x=-\dfrac{569}{121}\left(tm\right)\)

ĐKXĐ: \(x\ge0\)

\(A=11\sqrt{x}+5=6\)

\(\Leftrightarrow11\sqrt{x}=1\Leftrightarrow x=\dfrac{1}{11}\)

\(\Leftrightarrow x=\dfrac{1}{121}\left(tm\right)\)

\(\Rightarrow3xy=12-11y\Leftrightarrow3xy+11y=12\)

\(\Leftrightarrow y\left(3x+11\right)=12\Rightarrow y;3x+11\inƯ\left(12\right)=12\left\{\pm1;\pm2;\pm3;\pm4;\pm6;\pm\right\}\)

-> bạn tự lập bảng 

$\frac{1}{}$

$\frac{n+3}{2n-2}=\frac{n-4+7}{2n-2}=\frac{2(n-2)+7}{2n-2}=2+\frac{7}{2n-2}$

Để $\frac{n+3}{2n-2}$ thì $7⋮2n-2$

$\Rightarrow 2n-2\in Ư\left(7\right)\in \{\pm 1;\pm 7\}$

$2n-2=1\Rightarrow n=1,5$

$2n-2=-1\Rightarrow n=0,5$

$2n-2=7\Rightarrow n=4,5$

$2n-2=-7\Rightarrow n=-2,5$

Vì $n\in Z\Rightarrow$ Không có giá trị n thõa mãn

x+y=11

y=11-x

thay pt tren ta co

\(\frac{x-5}{11-x-7}=\frac{-12}{15}\)

\(\frac{x-5}{4-x}=\frac{-12}{15}\)

-12*(4-x)=15*(x-5)

12x-48=15x-75

3x=27

x=9

suy ra y=2

a = 2

b = 8

c = 1

d = 7

e = 3

h = 2

trả lời như v k ai hiểu đc

(x-1)(x+5) / (x-1)(2x+6) = 1

=> (x-1)(x+5) = (x-1)(2x+6)

=> \(x^2+5x-x-5=2x^2+6x-2x-6\)

=> \(x^2-2x^2+4x-4x-5+6=0\)

=> \(-x^2+1=0\)

=> \(-x^2=-1\)

=> \(x^2=1\)

=> x thuộc {-1; 1}

\(\frac{\left(x-1\right)\left(x+5\right)}{\left(x-1\right)\left(2x+6\right)}\)

đề vầy à

\(\frac{\left(x-1\right)\left(x+5\right)}{\left(x-1\right)\left(2x+6\right)}\) = 1 => (x-1)(x+5)=(x-1)(2x+6)

Mà x-1=x-1 => x+5=2x+6

=> 5-6= 2x-x

=> -1=x. Vậy x=-1