Giải phương trình sau: \(\frac{x^2+1}{x}+\frac{x}{x^2+1}=\frac{5}{2}\)
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a) \(\frac{x-1}{2}+\frac{x-2}{3}+\frac{x-3}{4}=\frac{x-4}{5}+\frac{x-5}{6}\)
\(\left(\frac{x-1}{2}+1\right)+\left(\frac{x-2}{3}+3\right)+\left(\frac{x-3}{4}+1\right)=\left(\frac{x-4}{5}+1\right)+\left(\frac{x-5}{6}+1\right)\)
\(\frac{x-1}{2}+\frac{x-1}{3}+\frac{x-1}{4}=\frac{x-1}{5}+\frac{x-1}{6}\)
\(\left(x-1\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\right)\)=0
\(x-1=0\)
\(x=1\)
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ĐK: x khác 0
pt (2) \(\Leftrightarrow\left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2=13\)
Đặt \(a=x+\frac{1}{x};b=y+\frac{1}{y}\), hệ pt trở thành:
\(\begin{cases}a+b=5\\a^2+b^2=13\end{cases}\) giải hệ pt đối xứng loại I được
\(\begin{cases}a=2\\b=3\end{cases}\) hoặc \(\begin{cases}a=3\\b=2\end{cases}\)
Thế vào được tập nghiệm của hệ pt đã cho:
\(\left\{\left(1;\frac{3-\sqrt{5}}{2}\right);\left(1;\frac{3+\sqrt{5}}{2}\right);\left(\frac{3-\sqrt{5}}{2};1\right);\left(\frac{3+\sqrt{5}}{2};1\right)\right\}\)
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ĐKXĐ: \(x\ne\left\{0;-1;-2;-3;-4;-5;-6;-7\right\}\)
\(\frac{1}{x}+\frac{1}{x+2}+\frac{1}{x+5}+\frac{1}{x+7}=\frac{1}{x+1}+\frac{1}{x+3}+\frac{1}{x+4}+\frac{1}{x+6}\)
\(\Rightarrow\frac{1}{x}+\frac{1}{x+7}+\frac{1}{x+2}+\frac{1}{x+5}=\frac{1}{x+1}+\frac{1}{x+6}+\frac{1}{x+3}+\frac{1}{x+4}\)
\(\Rightarrow\frac{x+7+x}{x\left(x+7\right)}+\frac{x+5+x+2}{\left(x+2\right)\left(x+5\right)}=\frac{x+6+x+1}{\left(x+1\right)\left(x+6\right)}+\frac{x+4+x+3}{\left(x+3\right)\left(x+4\right)}\)
\(\Rightarrow\frac{2x+7}{x^2+7x}+\frac{2x+7}{x^2+7x+10}=\frac{2x+7}{x^2+7x+6}+\frac{2x+7}{x^2+7x+12}\)
\(\Rightarrow\left(2x+7\right)\left(\frac{1}{x^2+7x}+\frac{1}{x^2+7x+10}-\frac{1}{x^2+7x+6}-\frac{1}{x^2+7x+12}\right)=0\)
mà \(\frac{1}{x^2+7x}+\frac{1}{x^2+7x+10}-\frac{1}{x^2+7x+6}-\frac{1}{x^2+7x+12}\ne0\)
=> 2x + 7 = 0 => x = -7/2
Vậy x = -7/2
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\(a,\Leftrightarrow5\left(x-2\right)-15x\le9+10\left(x+1\right)\)
\(\Leftrightarrow5x-10-15x\le9+10x+10\)
\(\Leftrightarrow-20x\le29\)
\(\Leftrightarrow x\ge-1,45\)
Vậy ...........
\(b,\Rightarrow\left(x+2\right)-3\left(x-3\right)=5\left(x-2\right)\)
\(\Leftrightarrow x+2-3x+9-5x+10=0\)
\(\Leftrightarrow-7x+21=0\)
\(\Leftrightarrow x=3\)
Vậy ..............
\(\frac{x-2}{6}-\frac{x}{2}\le\frac{3}{10}+\frac{x+1}{3}\Leftrightarrow\frac{5\left(x-2\right)}{30}-\frac{15x}{30}\le\frac{9}{30}+\frac{10\left(x+1\right)}{30}\)
\(\Leftrightarrow5x-10-15x-9-10x-10\le0\)
\(\Leftrightarrow-20x-29\le0\Leftrightarrow\left(-20x\right)\cdot\frac{-1}{20}\ge29\cdot-\frac{1}{20}\)
\(\Leftrightarrow x\ge-\frac{29}{20}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài làm:
PT:
đkxđ: \(x\ne0;x\ne2\)
Ta có: \(\frac{x+2}{x-2}=\frac{2}{x^2-2x}+\frac{1}{x}\)
\(\Leftrightarrow\frac{x\left(x+2\right)}{x\left(x-2\right)}=\frac{2}{x\left(x-2\right)}+\frac{x-2}{x\left(x-2\right)}\)
\(\Rightarrow x^2+2x=2+x-2\)
\(\Leftrightarrow x^2+x=0\)
\(\Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\left(vl\right)\\x+1=0\end{cases}}\Rightarrow x=-1\)
BPT:
Ta có: \(\frac{x+1}{2}-x\le\frac{1}{2}\)
\(\Leftrightarrow\frac{x+1}{2}-x-\frac{1}{2}\le0\)
\(\Leftrightarrow\frac{x+1-2x-1}{2}\le0\)
\(\Leftrightarrow\frac{-x}{2}\le0\)
\(\Rightarrow-x\le0\)
\(\Rightarrow x\ge0\)
a) \(ĐKXĐ:\hept{\begin{cases}x\ne0\\x\ne2\end{cases}}\)
\(\frac{x+2}{x-2}=\frac{2}{x^2-2x}+\frac{1}{x}\)
\(\Leftrightarrow\frac{2}{x\left(x-2\right)}+\frac{1}{x}-\frac{x+2}{x-2}=0\)
\(\Leftrightarrow\frac{2+x-2-x^2-2x}{x\left(x-2\right)}=0\)
\(\Leftrightarrow-x^2-x=0\)
\(\Leftrightarrow-x\left(x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\left(ktm\right)\\x=-1\left(tm\right)\end{cases}}}\)
Vậy \(S=\left\{-1\right\}\)
b) \(\frac{x+1}{2}-x\le\frac{1}{2}\)
\(\Leftrightarrow x+1-2x-1\le0\)
\(\Leftrightarrow-x\le0\)
\(\Leftrightarrow x\ge0\)
Vậy \(x\ge0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
- Điều kiện \(\hept{\begin{cases}x\ne5\\x\ne-5\end{cases}}\)\(\Leftrightarrow\frac{x+5}{x\left(x-5\right)}-\frac{\left(x-5\right)}{2x\left(x+5\right)}=\frac{x+25}{2\left(x+5\right)\left(x-5\right)}\)\(\Leftrightarrow\frac{2\left(x+5\right)^2-\left(x-5\right)^2}{2x\left(x-5\right)\left(x+5\right)}=\frac{x\left(x+25\right)}{2x\left(x+5\right)\left(x-5\right)}\)\(\Leftrightarrow x^2+30x+25=x^2+25\Leftrightarrow x=0\)
- Điều Kiện : \(x\ne1\)\(\Leftrightarrow\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{3x}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)\(\Leftrightarrow x^2+x+1-3x=2x^2-2x\Leftrightarrow x^2=1\Leftrightarrow\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)so sánh điều kiện có nghiệm phương trình là : \(x=-1\)
\(\frac{x+5}{x\left(x-5\right)}-\frac{x-5}{2x\left(x+5\right)}=\frac{x+25}{2\left(x-5\right)\left(x+5\right)}\)
\(\Leftrightarrow\)tu giai ra de ma
Điều kiện: x khác 0
Đặt \(\frac{x^2+1}{x}=t\Rightarrow\frac{x}{x^2+1}=\frac{1}{t}\)
Khi đó: \(\frac{x^2+1}{x}+\frac{x}{x^2+1}=\frac{5}{2}\)
\(\Leftrightarrow t+\frac{1}{t}=\frac{5}{2}\)
\(\Leftrightarrow\frac{t^2+1}{t}=\frac{5}{2}\Rightarrow2t^2+2=5t\)
\(\Leftrightarrow2t^2-5t+2=0\Leftrightarrow\left(2t-1\right)\left(t-2\right)=0\Leftrightarrow\orbr{\begin{cases}t=\frac{1}{2}\\t=2\end{cases}}\)
Nếu \(t=\frac{1}{2}\Rightarrow\frac{x^2+1}{x}=\frac{1}{2}\Rightarrow2x^2+2=x\)
\(\Leftrightarrow2x^2-x+2=0\)
Mà \(2x^2-x+2=2\left(x-\frac{1}{4}\right)^2+\frac{15}{8}>0\forall x\)
Nên \(x\in\varnothing\)
Nếu \(t=2\Rightarrow\frac{x^2+1}{x}=2\Rightarrow x^2-2x+1=0\)
\(\Leftrightarrow\left(x-1\right)^2=0\Leftrightarrow x=1\)(thỏa mãn ĐKXĐ)
Tập nghiệm của pt: \(S=\left\{1\right\}\)
\(\)
Theo BĐT AM-GM,ta có: \(x^2+1\ge2\left|x\right|\ge2x\Rightarrow\frac{x^2+1}{x}\ge2\)
Đặt \(\frac{x^2+t}{x}=t\left(t\ge2\right)\).Bài toán trở thành:
\(t+\frac{1}{t}=\frac{5}{2}\Leftrightarrow\left(\frac{1}{t}+\frac{t}{4}\right)+\frac{3t}{4}=\frac{5}{2}\)
Áp dụng BĐT AM-GM: \(VT\ge1+\frac{3t}{4}\ge1+\frac{6}{4}=\frac{5}{2}\)
Mà \(VT=\frac{5}{2}\) .Dấu "=" xảy ra khi \(\frac{1}{t}=\frac{t}{4}\Leftrightarrow t=2\Leftrightarrow\frac{x^2+1}{x}=2\Leftrightarrow x^2+1=2x\Leftrightarrow x=1\)
Vậy tập hợp nghiệm của phương trình: \(S=\left\{1\right\}\)