\(\frac{2x+1}{2x^2+5x+2}-\frac{3}{x^2-4}=2\)

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

\(\frac{2x+1}{x\left(2x+1\right)+2\left(2x+1\right)}-\frac{3}{\left(x-2\right)\left(x+2\right)}=2\)

\(\frac{2x+1}{\left(2x+1\right)\left(x+2\right)}-\frac{3}{\left(x-2\right)\left(x+2\right)}=2\)

\(\frac{\left(2x+1\right)\left(x-2\right)-3\left(2x+1\right)}{\left(2x+1\right)\left(x+2\right)\left(x-2\right)}=2\)

\(\frac{\left(2x+1\right)\left(x-2-3\right)}{\left(2x+1\right)\left(x+2\right)\left(x-2\right)}=2\)

\(x-5=2\left(x+2\right)\left(x-2\right)\)

\(x-5=2x^2-8\)

\(2x^2-x-3=0\)

\(2x^2-3x+2x-3=0\)

\(2x\left(x+1\right)-3\left(x+1\right)=0\)

\(\left(x+1\right)\left(2x-3\right)=0\)

\(\orbr{\begin{cases}x+1=0\\2x-3=0\end{cases}< =>\orbr{\begin{cases}x=-1\\x=\frac{3}{2}\end{cases}}}\)

KL ............

\(\frac{2x+1}{2x^2+5x+2}-\frac{3}{x^2-4}=0\)

\(\Leftrightarrow\frac{2x+1}{2x^2+x+4x+2}-\frac{3}{x^2-4}=2\)

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

\(\Leftrightarrow\frac{2x+1}{\left(x+2\right)\left(2x+1\right)}-\frac{3}{\left(x+2\right)\left(x-2\right)}=2\)

\(\Leftrightarrow\frac{1}{x+2}-\frac{3}{\left(x+2\right)\left(x-2\right)}=2\)

\(\Leftrightarrow\frac{x-2}{\left(x+2\right)\left(x-2\right)}-\frac{3}{\left(x+2\right)\left(x-2\right)}=2\)

\(\Leftrightarrow\frac{x-2-3}{\left(x+2\right)\left(x-2\right)}=2\)

\(\Leftrightarrow\frac{x-5}{\left(x+2\right)\left(x-2\right)}-2=0\)

\(\Leftrightarrow\frac{x-5}{\left(x+2\right)\left(x-2\right)}-\frac{2\left(x+2\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=0\)

\(\Leftrightarrow\frac{x-5}{\left(x+2\right)\left(x-2\right)}-\frac{2\left(x^2-4\right)}{\left(x+2\right)\left(x-2\right)}=0\)

\(\Leftrightarrow\frac{x-5-2x^2+8}{\left(x-2\right)\left(x+2\right)}=0\)

\(\Leftrightarrow\frac{-2x^2+x+3}{\left(x-2\right)\left(x+2\right)}=0\)

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

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

\(\Leftrightarrow-2x\left(x+1\right)+3\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(-2x+3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\-2x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=\frac{3}{2}\end{cases}}}\)

đk: \(x\ne\left\{\pm2;-\frac{1}{2}\right\}\)

Ta có: \(\frac{2x+1}{2x^2+5x+2}-\frac{3}{x^2-4}=2\)

\(\Leftrightarrow\frac{\left(2x+1\right)\left(x-2\right)-3\left(2x+1\right)}{\left(x+2\right)\left(x-2\right)\left(2x+1\right)}=2\)

\(\Leftrightarrow2x^2-3x-2-6x-3=2\left(x^2-4\right)\left(2x+1\right)\)

\(\Leftrightarrow2x^2-9x-5=4x^3+2x^2-16x-8\)

\(\Leftrightarrow4x^3-7x-3=0\)

\(\Leftrightarrow\left(2x-3\right)\left(2x+1\right)\left(x+1\right)=0\)

\(\Rightarrow x\in\left\{\frac{3}{2};-1\right\}\) (đkxđ)

\(\frac{2x+1}{2x^2+5x+2}-\frac{3}{x^2-4}=2\left(x\ne-\frac{1}{2};\pm2\right)\)

\(\Leftrightarrow\frac{2x+1}{2x\left(x+2\right)+\left(x+2\right)}-\frac{3}{\left(x-2\right)\left(x+2\right)}=2\)

\(\Leftrightarrow\frac{2x+1}{\left(2x+1\right)\left(x+2\right)}-\frac{3}{\left(x-2\right)\left(x+2\right)}=\frac{2\left(x-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)

\(\Leftrightarrow\frac{x-2}{\left(x-2\right)\left(x+2\right)}-\frac{3}{\left(x-2\right)\left(x+2\right)}=\frac{2x^2-8}{\left(x-2\right)\left(x+2\right)}\)

\(\Rightarrow x-2-3=2x^2-8\)

\(\Leftrightarrow x-2x^2=-8+2+3\)

\(\Leftrightarrow x-2x^2=-3\)

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

\(\Leftrightarrow2x\left(x+1\right)-3\left(x+1\right)=0\)

\(\Leftrightarrow\left(2x-3\right)\left(x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x-3=0\\x+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{2}\left(tm\right)\\x=-1\left(tm\right)\end{cases}}\)

Vậy nghiệm của phương trình là: \(S=\left\{\frac{3}{2};-1\right\}\)

a) \(A=\frac{\frac{x+1}{x-1}-\frac{x-1}{x+1}}{\frac{2x}{5x-5}}=\frac{\left(x+1\right)^2-\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)}\cdot\frac{5\left(x-1\right)}{2x}\)  \(\left(x\ne\pm1\right)\)

\(=\frac{4x}{\left(x+1\right)\left(x-1\right)}\cdot\frac{5\left(x-1\right)}{2x}=\frac{10}{x+1}\)

b) \(A=\frac{10}{-3+1}=-5\)

c) \(\left|x-2\right|=4-2x\Leftrightarrow\orbr{\begin{cases}x-2=4-2x\\x-2=2x-4\end{cases}}\Leftrightarrow x=2\)

\(\Rightarrow A=\frac{10}{2+1}=\frac{10}{3}\)

d) \(A=2\Rightarrow x+1=5\Rightarrow x=4\)

e) \(A< 0\Rightarrow x+1< 0\Rightarrow x< -1\)

f) \(\left(x+1\right)\in\left\{\pm1;\pm2;\pm5;\pm10\right\}\)

\(\Rightarrow x\in\left\{0;-2;-3;4;-6;9;-11\right\}\)

\(A=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}\right)\div\left(\frac{2x}{5x-5}\right)\)

\(=\left(\frac{\left(x+1\right)\left(x-1\right)-\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\right)\)

\(=\frac{\left(x^2-1\right)-\left(x^2-1\right)}{\left(x-1\right)\left(x+1\right)}\div\frac{2x}{5\left(x-1\right)}\)

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

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

* Sửa '' 5A '' => '' 5a ''

\(\text{Theo đề ra:}\)\(3AB=2BC\)

\(\Rightarrow BC=\frac{3}{2}AB\)

\(\text{Áp dụng định lý Py-ta-go, ta có: }\)

\(AB^2+AC^2=BC^2\)

\(\Rightarrow AB^2+25a^2=\frac{9}{4}AB^2\)

\(\Rightarrow25a^2=\frac{5}{4}AB^2\)

\(\Rightarrow AB^2=20a^2\)

\(\Rightarrow AB=2\sqrt{5a}\)

\(Q=-x^2+6x+1=-\left(x^2-6x+9\right)+10=-\left(x-3\right)^2+10\le10\)

Dấu \(=\)khi \(x-3=0\Leftrightarrow x=3\). 

Vậy \(maxQ=10\).

\(Q=-x^2+6x+1=-\left(x-3\right)^2+10\le10\)

dấu "=" xảy ra \(< =>x=3\)

Bài này không có GTLN, chỉ có GTNN nha. 

\(A=3x^2+y^2+4x-y=3x^2+4x+y^2-y=\frac{1}{3}\left(9x^2+12x+4\right)+\left(y^2-y+\frac{1}{4}\right)-\frac{19}{12}\)

\(=\frac{1}{3}\left(3x+2\right)^2+\left(y-\frac{1}{2}\right)^2-\frac{19}{12}\ge-\frac{19}{12}\)

Dấu \(=\)khi \(\hept{\begin{cases}3x+2=0\\y-\frac{1}{2}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-\frac{2}{3}\\y=\frac{1}{2}\end{cases}}\)

Ta có:\(3x^2-18y^2+2z^2+3y^2z^2-18x=27\)

\(\Leftrightarrow3x^2-18y^2+2z^2+3y^2z^2-18x-27=0\)

\(\Leftrightarrow3\left(x^2-6x+9\right)-18y^2+2z^2+3y^2z^2-54=0\)

\(\Leftrightarrow3\left(x-3\right)^2-18y^2+2z^2+3y^2z^2=54\)

Để pt có nghiệm nguyên thì:\(z^2⋮3\) \(\Rightarrow z⋮3\)\(\Rightarrow z^2⋮9\)\(\Rightarrow z^2\ge9\)

\(\Leftrightarrow3\left(x-3\right)^2+3y^2\left(z^2-6\right)+2z^2=54\)

\(\Rightarrow54=3\left(x-3\right)^2+3y^2\left(z^2-6\right)+2z^2\ge3\left(x-3\right)^2\le12\)

\(\Rightarrow y^2\le4\Rightarrow\hept{\begin{cases}y^2=1\\y^2=4\end{cases}}\)

Với \(y^2=1\Rightarrow y=1\)pt có dạng :

\(3\left(x-3\right)^2+5z^2=72\)

\(\Leftrightarrow5z^2\le72\)

\(\Leftrightarrow z^2=9\Leftrightarrow z=3\)

\(\Rightarrow x=6\)

Với \(y^2=4\Rightarrow y=2\)pt có dạng:

\(3\left(x-3\right)^2+14z^2=126\)

\(\Leftrightarrow14z^2\le126\)

\(\Leftrightarrow z^2\le9\Rightarrow z=3\)

\(\Rightarrow x=3\)

Vậy ......

 3X- 2 =2X -3

<=> X=-1

học tốt

\(3x-2x=2-3\)

\(\Leftrightarrow x=-1\)

Tìm \(x\)

\(\left|4x+2\right|=x+5\)   (Điều kiện: \(x\ge-5\)

\(\Rightarrow\orbr{\begin{cases}4x+2=x+5\\4x+2=-x-5\end{cases}}\Rightarrow\orbr{\begin{cases}4x-x=5-2\\4x+x=-7\end{cases}}\Rightarrow\orbr{\begin{cases}3x=3\\5x=-7\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\left(tm\right)\\x=-1,4\left(tm\right)\end{cases}}\)

Vậy \(x\in\left\{-1,4;1\right\}\).

\(\left|4x+2\right|=x+5\left(1\right)\)

\(đk:x+5\ge0\)

\(\Leftrightarrow x\ge-5\)

\(pt\left(1\right)\Leftrightarrow\orbr{\begin{cases}4x+2=x+5\\4x+2=-\left(x+5\right)\end{cases}}\Leftrightarrow\orbr{\begin{cases}4x-x=5-2\\4x+x=-5-2\end{cases}}\Leftrightarrow\orbr{\begin{cases}3x=3\\5x=-7\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\left(tm\right)\\x=\frac{-7}{5}\left(tm\right)\end{cases}}\)

\(\Leftrightarrow x^4+2x^2+2x^2+4-4x^2\)

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

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

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