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\(5,\dfrac{4}{x-2}+\dfrac{x}{x+1}-\dfrac{x^2-2}{\left(x-2\right)\left(x+1\right)}=0\left(dkxd:x\ne2;-1\right)\)
\(\Rightarrow4\left(x+1\right)+x\left(x-2\right)-x^2-2=0\)
\(\Rightarrow4x+4+x^2-2x-x^2-2=0\)
\(\Rightarrow2x+2=0\)
\(\Rightarrow x=-1\left(loai\right)\)
Vậy \(S=\varnothing\)
1.
<=> \(\left[{}\begin{matrix}4-3x=0\\10-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=2\end{matrix}\right.\)
2.
<=>\(\left[{}\begin{matrix}7-2x=0\\4+8x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
3.
<=>\(\left[{}\begin{matrix}9-7x=0\\11-3x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{7}\\x=\dfrac{11}{3}\end{matrix}\right.\)
4.
<=>\(\left[{}\begin{matrix}7-14x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=2\end{matrix}\right.\)
5.
<=>\(\left[{}\begin{matrix}\dfrac{7}{8}-2x=0\\3x+\dfrac{1}{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{16}\\x=-\dfrac{1}{9}\end{matrix}\right.\)
6,7. ko đủ điều kiện tìm
Bài 1:
a.
$(4x^2+4x+1)-x^2=0$
$\Leftrightarrow (2x+1)^2-x^2=0$
$\Leftrightarrow (2x+1-x)(2x+1+x)=0$
$\Leftrightarrow (x+1)(3x+1)=0$
$\Rightarrow x+1=0$ hoặc $3x+1=0$
$\Rightarrow x=-1$ hoặc $x=-\frac{1}{3}$
b.
$x^2-2x+1=4$
$\Leftrightarrow (x-1)^2=2^2$
$\Leftrightarrow (x-1)^2-2^2=0$
$\Leftrightarrow (x-1-2)(x-1+2)=0$
$\Leftrightarrow (x-3)(x+1)=0$
$\Leftrightarrow x-3=0$ hoặc $x+1=0$
$\Leftrightarrow x=3$ hoặc $x=-1$
c.
$x^2-5x+6=0$
$\Leftrightarrow (x^2-2x)-(3x-6)=0$
$\Leftrightarrow x(x-2)-3(x-2)=0$
$\Leftrightarrow (x-2)(x-3)=0$
$\Leftrightarrow x-2=0$ hoặc $x-3=0$
$\Leftrightarrow x=2$ hoặc $x=3$
2c.
ĐKXĐ: $x\neq 0$
PT $\Leftrightarrow x-\frac{6}{x}=x+\frac{3}{2}$
$\Leftrightarrow -\frac{6}{x}=\frac{3}{2}$
$\Leftrightarrow x=-4$ (tm)
2d.
ĐKXĐ: $x\neq 2$
PT $\Leftrightarrow \frac{1+3(x-2)}{x-2}=\frac{3-x}{x-2}$
$\Leftrightarrow \frac{3x-5}{x-2}=\frac{3-x}{x-2}$
$\Rightarrow 3x-5=3-x$
$\Leftrightarrow 4x=8$
$\Leftrightarrow x=2$ (không tm)
Vậy pt vô nghiệm.
Câu 2:
ĐKXĐ: \(\left[{}\begin{matrix}1-9x^2\ne0\\1+3x\ne0\\1-3x\ne0\end{matrix}\right.\Rightarrow \left[{}\begin{matrix}x\ne\dfrac{-1}{3}\\x\ne\dfrac{1}{3}\end{matrix}\right.\)
\(\dfrac{12}{1-9x^2}=\dfrac{1-3x}{1+3x}-\dfrac{1+3x}{1-3x}\left(1\right)\)
\(\left(1\right):\dfrac{12}{\left(1-3x\right)\left(1+3x\right)}-\dfrac{\left(1-3x\right)\left(1-3x\right)}{\left(1-3x\right)\left(1+3x\right)}+\dfrac{\left(1+3x\right)\left(1+3x\right)}{\left(1-3x\right)\left(1+3x\right)}=0\)
\(\Leftrightarrow 12-\left(1-3x-3x+9x^2\right)+\left(1+3x+3x+9x^2\right)=0\)
\(\Leftrightarrow 12-1+3x+3x-9x^2+1+3x+3x+9x^2=0\)
\(\Leftrightarrow12x+12=0\\ \Leftrightarrow12x=-12\\ \Leftrightarrow x=-1\left(TM\right)\)
Vậy \(S=\left\{-1\right\}\)
\(1,\dfrac{4x-4}{3}=\dfrac{7-x}{5}\\ \Leftrightarrow5\left(4x-4\right)=3\left(7-x\right)\\ \Leftrightarrow20x-20=21-3x\\ \Leftrightarrow17x=41\Leftrightarrow x=\dfrac{41}{17}\)
\(2,\dfrac{3x-9}{5}=\dfrac{3-x}{2}\\ \Leftrightarrow6x-18=15-5x\\ \Leftrightarrow11x=33\\ \Leftrightarrow x=3\)
\(3,\dfrac{2x-1}{5}-\dfrac{3-x}{3}=1\\ \Leftrightarrow\dfrac{6x-3-15+5x}{15}=1\\ \Leftrightarrow11x-18=1\\ \Leftrightarrow x=\dfrac{19}{11}\)
\(4,\dfrac{x-5}{3}+\dfrac{3x+4}{2}=\dfrac{5x+2}{6}\\ \Leftrightarrow2x-10+9x+12=5x+2\\ \Leftrightarrow6x=0\Leftrightarrow x=0\)
\(5,\dfrac{x-3}{2}+\dfrac{2x+3}{5}=\dfrac{2x+5}{10}\\ \Leftrightarrow5x-15+4x+6=2x+5\\ \Leftrightarrow7x=14\\ \Leftrightarrow x=2\)
Tick nha
2: Ta có: \(\dfrac{3x-9}{5}=\dfrac{3-x}{2}\)
\(\Leftrightarrow6x-18=15-5x\)
\(\Leftrightarrow11x=33\)
hay x=3
1,\(3x-1=0\Leftrightarrow3x=-1\Leftrightarrow x=-\dfrac{1}{3}\)
2,\(2-x=3x+1\Leftrightarrow2-1=3x+x\rightarrow1=4x\Rightarrow x=-\dfrac{1}{4}\)
3,\(2\left(x-2\right)-1=5x\Leftrightarrow2x-4-1=5x\Leftrightarrow2x-5x=4+1\Rightarrow3x=5\Rightarrow x=\dfrac{5}{3}\)
4,\(\dfrac{x}{3}-\dfrac{x}{5}=4\Leftrightarrow\dfrac{5x}{15}-\dfrac{3x}{15}=\dfrac{60}{15}\Rightarrow5x-3x=60\Rightarrow2x=60\Rightarrow x=\dfrac{60}{2}=30\)
a: =>5-x+6=12-8x
=>-x+11=12-8x
=>7x=1
hay x=1/7
b: \(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=2x+\dfrac{5}{3}\)
\(\Leftrightarrow9x+6-3x-1=12x+10\)
=>12x+10=6x+5
=>6x=-5
hay x=-5/6
d: =>(x-2)(x-3)=0
=>x=2 hoặc x=3
\(1,\left(dk:x\ne0,-1,4\right)\)
\(\Leftrightarrow\dfrac{9}{x+1}+\dfrac{2}{x-4}-\dfrac{11}{x}=0\)
\(\Leftrightarrow\dfrac{9x\left(x-4\right)+2x\left(x+1\right)-11\left(x+1\right)\left(x-4\right)}{x\left(x+1\right)\left(x-4\right)}=0\)
\(\Leftrightarrow9x^2-36x+2x^2+2x-11x^2+44x-11x+44=0\)
\(\Leftrightarrow-x=-44\)
\(\Leftrightarrow x=44\left(tm\right)\)
\(2,\left(đk:x\ne4\right)\)
\(\Leftrightarrow\dfrac{14}{3\left(x-4\right)}-\dfrac{2+x}{x-4}-\dfrac{3}{2\left(x-4\right)}+\dfrac{5}{6}=0\)
\(\Leftrightarrow\dfrac{14.2-6\left(2+x\right)-3.3+5\left(x-4\right)}{6\left(x-4\right)}=0\)
\(\Leftrightarrow28-12-6x-9+5x-20=0\)
\(\Leftrightarrow-x=13\)
\(\Leftrightarrow x=-13\left(tm\right)\)
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