giai phuong trinh : \(\dfrac{4x^2+14}{x^2+6}-\dfrac{5}{x^2+1}=\dfrac{7}{x^2+3}+\dfrac{9}{x^2+5}\)
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b) \(\left(x-3\right)^2+3x-22=\sqrt{x^2-3x+7}\)
\(\Leftrightarrow x^2-6x+9+3x-22=\sqrt{x^2-3x+7}\)
\(\Leftrightarrow\left(x^2-3x+7\right)-\sqrt{x^2-3x+7}-20=0\)
Đặt \(\sqrt{x^2-3x+7}=t\left(t\ge0\right)\left(1\right)\)
\(\Rightarrow t^2-t-20=0\)
\(\Rightarrow x_1=5\left(TM\right);x_2=-4\left(KTM\right)\)
Thay t=5 vào (1), ta có :
\(\sqrt{x^2-3x+7}=5\)
\(\Leftrightarrow x^2-3x+7=25\)
\(\Leftrightarrow x^2-3x-18=0\)
\(\Rightarrow x_1=6;x_2=-3\)
vậy...
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\(1,\dfrac{4x-4}{3}=\dfrac{7-x}{5}\\ \Leftrightarrow5\left(4x-4\right)=3\left(7-x\right)\\ \Leftrightarrow20x-20=21-3x\\ \Leftrightarrow17x=41\Leftrightarrow x=\dfrac{41}{17}\)
\(2,\dfrac{3x-9}{5}=\dfrac{3-x}{2}\\ \Leftrightarrow6x-18=15-5x\\ \Leftrightarrow11x=33\\ \Leftrightarrow x=3\)
\(3,\dfrac{2x-1}{5}-\dfrac{3-x}{3}=1\\ \Leftrightarrow\dfrac{6x-3-15+5x}{15}=1\\ \Leftrightarrow11x-18=1\\ \Leftrightarrow x=\dfrac{19}{11}\)
\(4,\dfrac{x-5}{3}+\dfrac{3x+4}{2}=\dfrac{5x+2}{6}\\ \Leftrightarrow2x-10+9x+12=5x+2\\ \Leftrightarrow6x=0\Leftrightarrow x=0\)
\(5,\dfrac{x-3}{2}+\dfrac{2x+3}{5}=\dfrac{2x+5}{10}\\ \Leftrightarrow5x-15+4x+6=2x+5\\ \Leftrightarrow7x=14\\ \Leftrightarrow x=2\)
Tick nha
2: Ta có: \(\dfrac{3x-9}{5}=\dfrac{3-x}{2}\)
\(\Leftrightarrow6x-18=15-5x\)
\(\Leftrightarrow11x=33\)
hay x=3
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\(\left\{{}\begin{matrix}4x+3x=-6\\\dfrac{x+3y}{3}-\dfrac{y-2}{5}=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}7x=-6\\\dfrac{5\left(x+3y\right)-3\left(y-2\right)}{15}=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=-\dfrac{6}{7}\\5x+15y-3y+6=15\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=-\dfrac{6}{7}\\12y=9-5x=9+5\cdot\dfrac{6}{7}=9+\dfrac{30}{7}=\dfrac{93}{7}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=-\dfrac{6}{7}\\y=\dfrac{93}{7\cdot12}=\dfrac{93}{84}=\dfrac{31}{28}\end{matrix}\right.\)
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=>12/(x+y-1)-15/(2x-y+3)=15/2 và 12/(x+y-1)-4/(2x-y+3)=28/5
=>x+y-1=22/9; 2x-y+3=-110/19
=>x+y=31/9; 2x-y=-167/19
=>x=-914/513; y=2681/513
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left\{{}\begin{matrix}\dfrac{4}{x+y-1}-\dfrac{5}{2x-y+3}=\dfrac{5}{2}\\\dfrac{3}{x+y-1}-\dfrac{1}{2x-y+3}=\dfrac{7}{5}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
b , \(\sqrt{1-4x+4x^2}-3=0\)
\(\Leftrightarrow\sqrt{\left(1-2x\right)^2}=3\)
\(\Leftrightarrow\left|1-2x\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}1-2x=3\\1-2x=-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-2x=2\\-2x=-4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)
Vậy nghiệm của phương trình là \(S=\left\{-1,2\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
5x-2>2(x+3)\(\Leftrightarrow\)5x-2>2x+6
\(\Leftrightarrow\) 5x-2x>6+2
\(\Leftrightarrow\)3x>8
\(\Leftrightarrow\)x>\(\dfrac{8}{3}\)
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