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a: Ta có: \(\dfrac{3}{x^2+x-2}-\dfrac{1}{x-1}=\dfrac{-7}{x+2}\)
\(\Leftrightarrow3-\left(x+2\right)=-7\left(x-1\right)\)
\(\Leftrightarrow3-x-2+7x-7=0\)
\(\Leftrightarrow6x-6=0\)
hay x=1(loại
b: Ta có: \(\dfrac{2}{-x^2+6x-8}-\dfrac{x-1}{x-2}=\dfrac{x+3}{x-4}\)
\(\Leftrightarrow\dfrac{-2}{\left(x-2\right)\left(x-4\right)}-\dfrac{\left(x-1\right)\left(x-4\right)}{\left(x-2\right)\left(x-4\right)}=\dfrac{\left(x+3\right)\left(x-2\right)}{\left(x-4\right)\left(x-2\right)}\)
Suy ra: \(-2-x^2+5x-4=x^2+x-6\)
\(\Leftrightarrow-x^2+5x-6-x^2-x+6=0\)
\(\Leftrightarrow-2x^2+4x=0\)
\(\Leftrightarrow-2x\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=2\left(loại\right)\end{matrix}\right.\)
\(\dfrac{3}{x^2+x-2}-\dfrac{1}{x-1}=-\dfrac{7}{x+2}\)
\(\Rightarrow\dfrac{3}{\left(x^2-x\right)+\left(2x-2\right)}-\dfrac{1}{x-1}=-\dfrac{7}{x+2}\)
\(\Rightarrow\dfrac{3}{x\left(x-1\right)+2\left(x-1\right)}-\dfrac{1}{x-1}=-\dfrac{7}{x+2}\)
\(\Rightarrow\dfrac{3}{\left(x+2\right)\left(x-1\right)}-\dfrac{1}{x-1}+\dfrac{7}{x+2}=0\)
\(\Rightarrow\dfrac{3}{\left(x+2\right)\left(x-1\right)}-\dfrac{x+2}{\left(x+2\right)\left(x-1\right)}+\dfrac{7\left(x-1\right)}{\left(x+2\right)\left(x-1\right)}=0\)
\(\Rightarrow\dfrac{3-\left(x+2\right)+7\left(x-1\right)}{\left(x+2\right)\left(x-1\right)}=0\)
\(\Rightarrow3-x-2+7x-7=0\)
\(\Rightarrow6x-6=0\)
\(\Rightarrow x=1\)
a, ĐK: \(x\ge2\)
\(\sqrt{2x+1}-\sqrt{x-2}=x+3\)
\(\Leftrightarrow\dfrac{x+3}{\sqrt{2x+1}+\sqrt{x-2}}=x+3\)
\(\Leftrightarrow\left(x+3\right)\left(\dfrac{1}{\sqrt{2x+1}+\sqrt{x-2}}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\left(l\right)\\\sqrt{2x+1}+\sqrt{x-2}=1\left(vn\right)\end{matrix}\right.\)
Phương trình vô nghiệm.
b, ĐK: \(x\ge-1\)
\(\sqrt{x+3}+2x\sqrt{x+1}=2x+\sqrt{x^2+4x+3}\)
\(\Leftrightarrow\sqrt{x+3}+2x\sqrt{x+1}=2x+\sqrt{\left(x+3\right)\left(x+1\right)}\)
\(\Leftrightarrow-\sqrt{x+3}\left(\sqrt{x+1}-1\right)+2x\left(\sqrt{x+1}-1\right)=0\)
\(\Leftrightarrow\left(2x-\sqrt{x+3}\right)\left(\sqrt{x+1}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+3}=2x\\\sqrt{x+1}=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge0\\x+3=4x^2\end{matrix}\right.\\x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=0\left(tm\right)\end{matrix}\right.\)
Ta có: \(\dfrac{3}{1-x^2}-\dfrac{1}{x+1}=\dfrac{2}{x^3-x^2-x+1}\)
\(\Leftrightarrow\dfrac{-3}{\left(x-1\right)\left(x+1\right)}-\dfrac{x-1}{\left(x+1\right)\left(x-1\right)}=\dfrac{2}{\left(x-1\right)^2\cdot\left(x+1\right)}\)
\(\Leftrightarrow\dfrac{-\left(x+2\right)\left(x-1\right)}{\left(x-1\right)^2\cdot\left(x+1\right)}=\dfrac{2}{\left(x-1\right)^2\cdot\left(x+1\right)}\)
\(\Leftrightarrow-\left(x^2-x+2x-2\right)=2\)
\(\Leftrightarrow x^2+x-2=-2\)
\(\Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=-1\left(loại\right)\end{matrix}\right.\)
Vậy: S={0}
\(x^2+1+3x=x\sqrt{x^2+1}+3\sqrt{x^2+1}\)
<=> \(\sqrt{x^2+1}\left(\sqrt{x^2+1}-x\right)-3\left(\sqrt{x^2+1}-x\right)=0\)
\(\Leftrightarrow\left(\sqrt{x^2+1}-x\right)\left(\sqrt{x^2+1}-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+1}=x\\\sqrt{x^2+1}=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2+1=x^2\\x^2=8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}∃x̸\\x=\pm\sqrt{8}\end{matrix}\right.\)
`x^2 + 3x + 1 = (x + 3) \sqrt{x^2 + 1}`
Nghiệm của pt là `x = +- 2 \sqrt{2}`
a: Ta có: \(2x+3>1-x\)
\(\Leftrightarrow3x>-2\)
hay \(x>-\dfrac{2}{3}\)
b: Ta có: \(15-2\left(x-3\right)< -2x+5\)
\(\Leftrightarrow15-2x+6+2x-5< 0\)
\(\Leftrightarrow16< 0\left(vôlý\right)\)
c: Ta có: \(\left(x+1\right)\left(x-3\right)\le\left(x+4\right)\left(x-1\right)\)
\(\Leftrightarrow x^2-3x+x-3-x^2+x-4x+4\le0\)
\(\Leftrightarrow-5x\le-1\)
hay \(x\ge\dfrac{1}{5}\)
Ta có: \(\dfrac{4}{x^2+2x-3}=\dfrac{2x-5}{x+3}-\dfrac{2x}{x-1}\)
\(\Leftrightarrow\dfrac{\left(2x-5\right)\left(x-1\right)-2x\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}=\dfrac{4}{\left(x+3\right)\left(x-1\right)}\)
Suy ra: \(2x^2-2x-5x+5-2x^2-6x=4\)
\(\Leftrightarrow13x=-1\)
hay \(x=-\dfrac{1}{13}\)
Điều kiện xác định của pt : \(\hept{\begin{cases}\frac{x^3+1}{x+3}\ge0\\x+1\ge0\\x+3\ge0\end{cases}}\) \(\Leftrightarrow x\ge-1\)
Ta có : \(\sqrt{\frac{x^3+1}{x+3}}+\sqrt{x+1}=\sqrt{x^2-x+1}+\sqrt{x+3}\)
\(\Leftrightarrow\sqrt{\left(x+1\right)\left(x^2-x+1\right)}+\sqrt{x+1}.\sqrt{x+3}=\sqrt{x^2-x+1}.\sqrt{x+3}+\left(x+3\right)\)
\(\Leftrightarrow\sqrt{x^2-x+1}\left(\sqrt{x+1}-\sqrt{x+3}\right)+\sqrt{x+3}\left(\sqrt{x+1}-\sqrt{x+3}\right)=0\)
\(\Leftrightarrow\left(\sqrt{x+1}-\sqrt{x+3}\right)\left(\sqrt{x^2-x+1}+\sqrt{x+3}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x+1}-\sqrt{x+3}=0\\\sqrt{x^2-x+1}+\sqrt{x+3}=0\end{cases}}\)
- Nếu \(\sqrt{x+1}-\sqrt{x+3}=0\Rightarrow x+1=x+3\Leftrightarrow1=3\)(vô lí - loại)
- Nếu \(\sqrt{x^2-x+1}+\sqrt{x+3}=0\)(1).
Từ điều kiện : Với \(x\ge-1\)thì \(\sqrt{x+3}\ge\sqrt{2}>0\);
\(\sqrt{x^2-x+1}=\sqrt{\left(x-\frac{1}{2}\right)^2+\frac{3}{4}}\ge\frac{\sqrt{3}}{2}>0\)
Do đó pt (1) vô nghiệm.
Vậy pt ban đầu vô nghiệm.
Điều kiện xác định của pt : \(\hept{\begin{cases}\frac{x^3+1}{x+3}\ge0\\x+1\ge0\\x+3\ge0\end{cases}}\) \(\Leftrightarrow x\ge-1\)
Ta có : \(\sqrt{\frac{x^3+1}{x+3}}+\sqrt{x+1}=\sqrt{x^2-x+1}+\sqrt{x+3}\)
\(\Leftrightarrow\sqrt{\left(x+1\right)\left(x^2-x+1\right)}+\sqrt{x+1}.\sqrt{x+3}=\sqrt{x^2-x+1}.\sqrt{x+3}+\left(x+3\right)\)
\(\Leftrightarrow\sqrt{x^2-x+1}\left(\sqrt{x+1}-\sqrt{x+3}\right)+\sqrt{x+3}\left(\sqrt{x+1}-\sqrt{x+3}\right)=0\)
\(\Leftrightarrow\left(\sqrt{x+1}-\sqrt{x+3}\right)\left(\sqrt{x^2-x+1}+\sqrt{x+3}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x+1}-\sqrt{x+3}=0\\\sqrt{x^2-x+1}+\sqrt{x+3}=0\end{cases}}\)
- Nếu \(\sqrt{x+1}-\sqrt{x+3}=0\Rightarrow x+1=x+3\Leftrightarrow1=3\)(vô lí - loại)
- Nếu \(\sqrt{x^2-x+1}+\sqrt{x+3}=0\)(1). So sánh từ điều kiện : Với mọi \(x\ge-1\)thì \(\sqrt{x+3}\ge\sqrt{2}>0\), \(\sqrt{x^2-x+1}=\sqrt{\left(x-\frac{1}{2}\right)^2+\frac{3}{4}}\ge\frac{\sqrt{3}}{2}>\)với mọi x
Do đó pt (1) vô nghiệm.
Vậy pt ban đầu vô nghiệm.