(2\(x\) + 7) ⋮ (\(x\) + 1) (đk \(x\) ≠ -1; \(x\in\)Z)

2\(x\) + 2 + 5 ⋮ \(x\) + 1

2.(\(x\) + 1) + 5 ⋮ \(x\) + 1

                  5 ⋮ \(x\) + 1

\(x\) + 1 \(\in\) Ư(5) =  {-5; -1; 1; 5}

\(x\)        \(\in\) {-6; -2; 0; 4}

a) 2x - 3 = -12

=> 2x = -12 + 3 = -9

=> x = \(-\frac{9}{2}\)

b) \(\frac{1}{2}+2x=-\frac{5}{6}:\frac{2}{3}\)

=> \(\frac{1}{2}+2x=-\frac{5}{6}\cdot\frac{3}{2}\)

=> \(\frac{1}{2}+2x=-\frac{5}{2}\cdot\frac{1}{2}\)

=> \(\frac{1}{2}+2x=-\frac{5}{2}\)

=> \(2x=-\frac{5}{2}-\frac{1}{2}=-3\)

=> \(x=-3:2=-\frac{3}{2}\)

c) \(1< \frac{x}{5}< 2\)

=> \(\frac{5}{5}< \frac{x}{5}< \frac{10}{5}\)

=> 5 < x < 10

=> x \(\in\){6,7,8,9}

Dù bạn có cho âm vào nx thì nó vẫn sai nhá 

d) Đặt \(A=\frac{x+5}{x-2}=\frac{x-2+7}{x-2}=1+\frac{7}{x-2}\)

=> \(x-2\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)

+) x - 2 = 1 => x = 3(T/M)

x - 2 = -1 => x = -1 +2 = 1(t/m)

x - 2  = 7 => x = 9 (t/m)

x - 2 = -7 => x = -7 + 2 = -5(t/m)

e) làm nốt ...

a,\(2x-3=-12\)

\(< =>2x=-12+3=-9\)

\(< =>x=-\frac{9}{2}\)

b,\(\frac{1}{2}+2x=-\frac{5}{6}:\frac{2}{3}\)

\(< =>\frac{1}{2}+\frac{4x}{2}=-\frac{5}{6}.\frac{3}{2}\)\(< =>\frac{4x+1}{2}=-\frac{5}{4}\)

\(< =>\frac{8x+2}{4}=-\frac{5}{4}\)\(< =>8x+2=-5\)

\(< =>8x=-5-2=-7\)\(< =>x=-\frac{7}{8}\)

Ây da,câu b nhầm tận đáy :))

Bạn tham khảo câu b  dcv_new nhé

\(a,\Leftrightarrow\left(5x+1\right)\left(x-4\right)-\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(5x+1-x\right)=0\\ \Leftrightarrow5x\left(x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\\ b,\Leftrightarrow2x^2-10x-2x^2-3x=26\\ \Leftrightarrow-13x=26\\ \Leftrightarrow x=-2\\ c,\Leftrightarrow x^3+1-x^3+3x=15\\ \Leftrightarrow3x=14\\ \Leftrightarrow x=\dfrac{14}{3}\)

\(d,\Leftrightarrow x^3-5x+2x^2-10+5x-2x^2-17=0\\ \Leftrightarrow x^3-27=0\\ \Leftrightarrow x^3=27\\ \Leftrightarrow x=3\)

b) \(\left(x+3\right)^2-5x-15=0\\ \Leftrightarrow\left(x+3\right)^2-5\left(x+3\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x+3-5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+3=0\\x-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)

Vậy tập nghiệm của phương trình là : \(S=\left\{-3;2\right\}\)

c) \(2x^5-4x^3+2x=0\\ \Leftrightarrow2x\left(x^4-2x^2+1\right)=0\\ \Leftrightarrow2x\left(x^2-1\right)^2=0\\ \Rightarrow\left[{}\begin{matrix}2x=0\\\left(x^2-1\right)^2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x^2=1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)

Vậy tập nghiệm của pt là : \(S=\left\{0;1;-1\right\}\)

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

\(=\dfrac{x^2-x^2+10x-25}{\left(x-5\right)\left(x+5\right)}\cdot\dfrac{x\left(x+5\right)}{2x-5}-\dfrac{x^2}{x-5}\)

\(=\dfrac{5\left(2x-5\right)\cdot x}{\left(x-5\right)\left(2x-5\right)}-\dfrac{x^2}{x-5}=\dfrac{5x-x^2}{x-5}=-x\)

b: Để P là số nguyên thì x là số nguyên

2x+5=x-1

x=-6

vậy...

\(2x+5=x-1\)

\(2x-x=-1-5\)

\(x=-6\)

\(2x+5=x-1\)

\(\Leftrightarrow2x-x=-1-5\)

\(\Leftrightarrow x=-6\)

Vậy \(x=-6\)

hok tốt!~!