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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: 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 \(x\ge\dfrac{1}{5}\)
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: \(x\ge-1\)
\(\sqrt{x+3}+2x\sqrt{x+1}=2x+\sqrt{x^2+4x+3}\)
\(\Leftrightarrow\sqrt{x+3}+2x\sqrt{x+1}=2x+\sqrt{\left(x+3\right)\left(x+1\right)}\)
\(\Leftrightarrow-\sqrt{x+3}\left(\sqrt{x+1}-1\right)+2x\left(\sqrt{x+1}-1\right)=0\)
\(\Leftrightarrow\left(2x-\sqrt{x+3}\right)\left(\sqrt{x+1}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+3}=2x\\\sqrt{x+1}=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge0\\x+3=4x^2\end{matrix}\right.\\x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=0\left(tm\right)\end{matrix}\right.\)
\(\Leftrightarrow\sqrt{x^2-\frac{1}{4}+\sqrt{\left(x+\frac{1}{2}\right)^2}}=\frac{1}{2}\left(2x^3+x^2+2x+1\right)\)\(\Leftrightarrow\sqrt{x^2+x+\frac{1}{4}}=\frac{1}{2}\left(2x^3+x^2+2x+1\right)\)\(\Leftrightarrow x+\frac{1}{2}=\frac{1}{2}\left(2x^3+x^2+2x+1\right)\Leftrightarrow2x+1=2x^3+x^2+2x+1\)\(\Leftrightarrow2x^3+x^2=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{1}{2}\end{cases}}\)
\(\sqrt{x^2-\frac{1}{4}+\sqrt{x^2+x+\frac{1}{4}}}=\frac{1}{2}\left(2x^3+x^2+2x+1\right)\left(1\right)\)
\(\left(1\right)\Leftrightarrow\sqrt{x^2-\frac{1}{4}+\sqrt{\left(x+\frac{1}{2}\right)^2}}=\frac{1}{2}\left(2x+1\right)\left(x^2+1\right)\)
\(x^2+1\ge1\forall x\Rightarrow2x+1\ge0!2x+1!=2x+1\)
\(\left(1\right)\Leftrightarrow\sqrt{x^2+x+\frac{1}{4}}=\frac{1}{2}\left(2x+1\right)\left(x^2+1\right)\)
\(\left(1\right)\Leftrightarrow x+\frac{1}{2}=\frac{1}{2}\left(2x+1\right)\left(x^2+1\right)\)
\(\left(1\right)\Leftrightarrow2x+1=\left(2x+1\right)\left(x^2+1\right)\Leftrightarrow\left(2x+1\right).\left(1-\left(x^2+1\right)\right)=0\)
\(\hept{\begin{cases}2x+1=0\\-x^2=0\end{cases}\Rightarrow\hept{\begin{cases}x=-\frac{1}{2}\\x=0\end{cases}}}\)
Chúc bạn học tốt !!!
d: Ta có: \(\dfrac{2x+1}{3}-\dfrac{1-x}{2}\ge1-\dfrac{x}{4}\)
\(\Leftrightarrow8x+4-6+6x\ge12-3x\)
\(\Leftrightarrow14x+3x\ge12+2=14\)
\(\Leftrightarrow x\ge\dfrac{14}{17}\)
e: Ta có: \(\dfrac{x+1}{2}-\dfrac{2-x}{3}< \dfrac{2x-3}{4}\)
\(\Leftrightarrow6x+12+4x-8< 6x-9\)
\(\Leftrightarrow4x< -9+8-12=-13\)
hay \(x< -\dfrac{13}{4}\)
f: Ta có: \(\left(x+1\right)\left(x-2\right)-\left(2-x\right)\left(3-x\right)>0\)
\(\Leftrightarrow x^2-2x+x-2-\left(x-2\right)\left(x-3\right)>0\)
\(\Leftrightarrow x^2-x-2-x^2+5x-6>0\)
\(\Leftrightarrow4x>8\)
hay x>2
g: Ta có: \(\left(2x-1\right)^2\le2\left(x-1\right)^2\)
\(\Leftrightarrow4x^2-4x+1-2x^2+4x-2\le0\)
\(\Leftrightarrow2x^2\le1\)
\(\Leftrightarrow x^2\le\dfrac{1}{2}\)
\(\Leftrightarrow-\dfrac{\sqrt{2}}{2}\le x\le\dfrac{\sqrt{2}}{2}\)
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 \(x=-\dfrac{1}{13}\)
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}