a, \(\dfrac{x-2}{x+1}\ge\dfrac{x+1}{x-2}\) 

⇔ \(\dfrac{\left(x-2\right)^2-\left(x+1\right)^2}{\left(x-1\right)\left(x+2\right)}\ge0\)

⇔ \(\dfrac{3-6x}{\left(x+1\right)\left(x-2\right)}\) ≥ 0

⇔ \(\dfrac{2x-1}{\left(x+1\right)\left(x-2\right)}\) ≤ 0

⇔ \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge\dfrac{1}{2}\\-1< x< 2\end{matrix}\right.\\\left\{{}\begin{matrix}x\le\dfrac{1}{2}\\\left[{}\begin{matrix}x< -1\\x>2\end{matrix}\right.\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}\le x< 2\\x< -1\end{matrix}\right.\)

Vậy tập nghiệm là \(\left(-\infty;-1\right)\cup\) \(\left[\dfrac{1}{2};2\right]\)\ {2}

Bạn có thể biến cái ngoặc vuông kia (ở chỗ số 2) thành ngoặc tròn

Còn vì sao mình không biến cái ngoặc vuông kia (ở chỗ số 2) thành ngoặc tròn thì đó là một câu chuyện dài

b, tương tự, chuyển vế đổi dấu 

hoc gioi the hihiihihihhhihihihihiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii

,mnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnn

Mình không biết sin lỗi vạn

Câu 1 : D

Câu 2 : A

Câu 3 : B

Câu 4 : A

Câu 5 : C

lớp 8 thì mấy bài này dễ thôi

1.\(\dfrac{x+2}{x-3}+\dfrac{x}{x+2}=\dfrac{x^2+6}{x^2-x-6}\)

\(\Leftrightarrow\dfrac{x+2}{x-3}+\dfrac{x}{x+2}=\dfrac{x^2+6}{\left(x+2\right)\left(x-3\right)}\)

\(ĐK:x\ne3;-2\)

\(\Leftrightarrow\dfrac{\left(x+2\right)\left(x+2\right)+x\left(x-3\right)}{\left(x+2\right)\left(x-3\right)}=\dfrac{x^2+6}{\left(x+2\right)\left(x-3\right)}\)

\(\Leftrightarrow\left(x+2\right)\left(x+2\right)+x\left(x-3\right)=x^2+6\)

\(\Leftrightarrow x^2+4x+4+x^2-3x-x^2-6=0\)

\(\Leftrightarrow x^2+x-2=0\)

\(\Leftrightarrow\left(x^2-x\right)+\left(2x-2\right)=0\)

\(\Leftrightarrow x\left(x-1\right)+2\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=-2\left(ktm\right)\end{matrix}\right.\)

Vậy \(S=\left\{1\right\}\)

b.\(\left(x+1\right)^2+\left|x-1\right|=x^2+4\)

\(\Leftrightarrow\)    \(\left(x+1\right)^2+x-1=x^2+4\) hoặc   \(\left(x+1\right)^2+1-x=x^2+4\)

Xét \(\left(x+1\right)^2+x-1=x^2+4\)

\(\Leftrightarrow x^2+2x+1+x-1-x^2-4=0\)

\(\Leftrightarrow3x-4=0\)

\(\Leftrightarrow x=\dfrac{4}{3}\)

Xét \(\left(x+1\right)^2+1-x=x^2+4\)

\(\Leftrightarrow x^2+2x+1+1-x-x^2-4=0\)

\(\Leftrightarrow x-2=0\)

\(\Leftrightarrow x=2\)

Vậy \(S=\left\{\dfrac{4}{3};2\right\}\)

2.\(1-\dfrac{x-1}{3}< \dfrac{x+3}{3}-\dfrac{x-2}{2}\)

\(\Leftrightarrow\dfrac{6-2\left(x-1\right)}{6}< \dfrac{2\left(x+3\right)-3\left(x-2\right)}{6}\)

\(\Leftrightarrow6-2\left(x-1\right)< 2\left(x+3\right)-3\left(x-2\right)\)

\(\Leftrightarrow6-2x+2< 2x+6-3x+6\)

\(\Leftrightarrow-x< 4\)

\(\Leftrightarrow x>4\)

Vậy \(S=\left\{x|x>4\right\}\)

1.

\(\Leftrightarrow2.4^x-5.2^x+2\le0\)

Đặt \(2^x=t>0\Rightarrow2.t^2-5t+2\le0\)

\(\Rightarrow\frac{1}{2}\le t\le2\Rightarrow\frac{1}{2}\le2^x\le2\)

\(\Rightarrow-1\le x\le1\)

2.

\(\Leftrightarrow\left\{{}\begin{matrix}x\left(5-x\right)>0\\x\left(5-x\right)< 6\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}0< x< 5\\\left[{}\begin{matrix}x< 2\\x>3\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}0< x< 2\\3< x< 5\end{matrix}\right.\)

3.

\(\Leftrightarrow1\ge\left(\sqrt{6}+\sqrt{5}\right)^{x-1}.\left(\sqrt{6}+\sqrt{5}\right)^{2x-5}\)

\(\Leftrightarrow\left(\sqrt{6}+\sqrt{5}\right)^{3x-6}\le1\)

\(\Leftrightarrow3x-6\le0\Rightarrow x\le2\)

4.

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-2\\1>3^{-x}.3^{\sqrt{x+2}}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-2\\3^{\sqrt{x+2}-x}< 1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-2\\\sqrt{x+2}-x< 0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-2\\\sqrt{x+2}\le x\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x+2< x^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x^2-x-2>0\end{matrix}\right.\) \(\Rightarrow x>2\)

5.

\(\Leftrightarrow\left(\frac{4}{3}\right)^{2x-1}.\left(\frac{4}{3}\right)^{-2x^2+x}\ge1\)

\(\Leftrightarrow\left(\frac{4}{3}\right)^{-2x^2+3x-1}\ge1\)

\(\Leftrightarrow-2x^2+3x-1\ge0\)

\(\Rightarrow\frac{1}{2}\le x\le1\)

6.

\(\Leftrightarrow\left\{{}\begin{matrix}x>-1\\\frac{1}{2}log_2\left(x+7\right)>log_2\left(x+1\right)\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>-1\\\sqrt{x+7}>x+1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>-1\\x+7>x^2+2x+1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>-1\\x^2+x-6< 0\end{matrix}\right.\) \(\Rightarrow-1< x< 2\)

7.

\(\left(\frac{1}{5}\right)^{x^2-2x}\ge\left(\frac{1}{5}\right)^3\)

\(\Leftrightarrow x^2-2x\le3\)

\(\Leftrightarrow x^2-2x-3\le0\Rightarrow-1\le x\le3\)

\(\Rightarrow x=\left\{1;2;3\right\}\Rightarrow\) có 3 nghiệm nguyên dương