\(\frac{x+14}{86}+\frac{x+15}{85}+\frac{x+16}{84}+\frac{x+17}{83}+\frac{x+116}{4}=0\)

\(\Leftrightarrow\frac{x+14}{86}+\frac{x+15}{85}+\frac{x+16}{84}+\frac{x+17}{83}+\frac{x+100}{4}+4=0\)

\(\Leftrightarrow\left(\frac{x+14}{86}+1\right)+\left(\frac{x+15}{85}+1\right)+\left(\frac{x+14}{86}+1\right)+\left(\frac{x+13}{87}+1\right)+\frac{x+100}{4}=0\)

\(\Leftrightarrow\frac{x+100}{86}+\frac{x+100}{85}+\frac{x+100}{84}+\frac{x+100}{83}+\frac{x+100}{4}=0\)

\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{86}+\frac{1}{85}+\frac{1}{84}+\frac{1}{83}+\frac{1}{4}\right)=0\)

\(\Leftrightarrow x+100=0\left(vì\frac{1}{86}+\frac{1}{85}+\frac{1}{84}+\frac{1}{83}+\frac{1}{4}\ne0\right)\)

\(\Leftrightarrow x=-100\)

vậy.............................

Kham khảo 

\(\left[\frac{\left(x+14\right)}{86}+1\right]+\left[\frac{\left(x+15\right)}{85}+1\right]+\left[\frac{\left(x+16\right)}{84}+1\right]+\left[\frac{\left(x+17\right)}{83}+1\right]\)\(+\left[\frac{\left(x+116\right)}{4}-4\right]=0\)

\(\frac{\Rightarrow\left(x+100\right)}{86}+\frac{\left(x+100\right)}{85}+\frac{\left(x+100\right)}{84}+\frac{\left(x+100\right)}{83}+\frac{\left(x+100\right)}{16}=0\)

\(\Rightarrow\left(x+100\right)\left(\frac{1}{86}+\frac{1}{85}+\frac{1}{84}+\frac{1}{83}+\frac{1}{16}\right)=0\)

\(\Leftrightarrow x+100=0\Rightarrow x=-100\)

Mặt khác :\(\frac{1}{86}+\frac{1}{85}+\frac{1}{84}+\frac{1}{83}+\frac{1}{16}\ne0\)

Vậy x=-100

_Chúc bn hc tốt

28 và 12 vì 28+12=40

28-12=16

Ta có x + y = 40

x - y = 16

=> x + y + x - y = 40 + 16

=> 2x = 56

=> x = 28

Lại có x + y - (x - y) = 40 - 16

=> x + y - x + y = 24

=> 2y = 24

=> y = 12

Vậy x = 28 ; y = 12

Ta có

\(\hept{\begin{cases}x+y=40\\x-y=16\end{cases}\Rightarrow\left(x+y\right)-\left(x-y\right)=40-16}\)

\(\Rightarrow x+y-x+y=24\)

\(\Rightarrow2y=24\)

\(\Rightarrow y=12\)

\(\Rightarrow x=40-12=28\)

Vậy (x;y)=(28;12)

Hok Tốt !!!!!!!!!!!!!!!!!!

\(\left(x^2-16\right)-\left(x-4\right)^2=0\)

\(\Rightarrow x^2-16-\left(x^2-8x+16\right)=0\)

\(\Rightarrow x^2-16-x^2+8x-16=0\)

\(\Rightarrow8x-32=0\)

\(\Rightarrow8x=0+32=32\)

\(\Rightarrow x=32:8=4\)

Để \(\left(n+8\right)⋮\left(n+5\right)\) thì

\(\left(n+8\right)-\left(n+5\right)⋮\left(n+5\right)\)

\(\Rightarrow\)\(3⋮\left(n+5\right)\)

\(\Rightarrow\)​​\(\left(n+5\right)\inƯ\left(3\right)\)

\(\Rightarrow\)\(\left(n+5\right)\in\left(1;-1;3;-3\right)\)

\(\Rightarrow\)\(n\in\left(-4;-6;-2;-8\right)\)

Để \(\left(16-3n\right)⋮\left(n+4\right)\) thì

\(\left(16-3n\right)+\left(n+4\right)⋮\left(n+4\right)\)

\(\Rightarrow\)\(\left(16-3n\right)+3\left(n+4\right)⋮\left(n+4\right)\)

\(\Rightarrow\)\(16-3n+3n+12⋮\left(n+4\right)\)

\(\Rightarrow\)\(28⋮\left(n+4\right)\)

\(\Rightarrow\)\(\left(n+4\right)\inƯ\left(28\right)\)

\(\Rightarrow\)\(\left(n+4\right)\in\left\{\pm1;\pm2;\pm4;\pm7;\pm14;\pm28\right\}\)

\(\Rightarrow\)\(n\in\left\{-3;-4;-2;-6;0;-8;3;-11;10;-18;24;-32\right\}\)

Phần a, 2 dòng cuối lỗi tí 

\(\left(n+5\right)\in\left\{1;-1;3;-3\right\}\)

\(n\in\left\{-4;-6;-2;-8\right\}\)

\(< =>-2\sqrt{x}=-4\\ < =>\sqrt{x}=2\\ < =>x=4\)

ĐKXĐ: x>=0

\(3-2\sqrt{x}=-1\)

=>\(2\sqrt{x}=3+1=4\)

=>\(\sqrt{x}=2\)

=>x=4(nhận)

a) Ta có: \(B=\left(\dfrac{x+3\sqrt{x}-3}{x-16}-\dfrac{1}{\sqrt{x}+4}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}-4}\)

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

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

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+4}\)