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a: =>4(2x-1)-12x=3(x+3)+24
=>8x-4-12x=3x+9+24
=>-4x-4=3x+33
=>-7x=37
=>x=-37/7
b: =>(x-2)(x+2+x-9)=0
=>(2x-7)(x-2)=0
=>x=2 hoặc x=7/2
c: =>(x-1)(x+3)-x+3=3x+3
=>x^2+2x-3-x+3=3x+3
=>x^2+x-3x-3=0
=>x^2-2x-3=0
=>(x-3)(x+1)=0
=>x=-1
a. 3x-1=x-5 <=> 2x=-4 <=> x=-2
Vậy tập no của phương trình là S={-2}
b.\(\dfrac{2x-1}{3}\)+\(\dfrac{3x-5}{4}\)=\(\dfrac{x-1}{5}\)
<=>40x-20+45x-75=12x-12
<=>73x=83 <=> x= \(\dfrac{83}{73}\)
Vậy tập no của phương trình là S={\(\dfrac{83}{73}\)}
c.(2x-6)(x+20)=0
<=> 2x-6=0 hoặc x+20=0
1) 2x-6=0 <=> x= 3
2) x+20=0 <=> x=-20
Vậy tập no của phương trình là S={-20 ; 3}
d. \(\dfrac{x-3}{x+3}\)+\(\dfrac{x+3}{x-3}\)=\(\dfrac{2x\left(x+1\right)}{x^2-9}\)
ĐKXĐ: x ≠ 3 và x ≠ -3
Ta có \(\dfrac{x-3}{x+3}\)+\(\dfrac{x+3}{x-3}\)=\(\dfrac{2x\left(x+1\right)}{x^2-9}\)
<=> (x-3)2 + (x+3)2 = 2x2+2x
<=> x2 -6x +9 +x2 +6x +9=2x2+2x
<=> 2x=18 <=> x=9
Vậy tập no của phương trình là S={9}
giải luôn ko chép đề nhé
a,
<=>(3x-5)(x-1)=(3x+1)(x-2)-3(x-1)
<=>3x^2-8x+5=3x^2-5x-2-3x+3
<=>3x^2-8x-3x^2+5x+3x=-5+3
<=>0x=-2
vậy s=\(\varnothing\)
b: \(\Leftrightarrow\dfrac{7x+10}{x+1}\left(x^2-x-2-2x^2+3x+5\right)=0\)
\(\Leftrightarrow\left(7x+10\right)\left(-x^2+2x+3\right)=0\)
\(\Leftrightarrow\left(7x+10\right)\left(x^2-2x-3\right)=0\)
=>(7x+10)(x-3)=0
hay \(x\in\left\{-\dfrac{10}{7};3\right\}\)
d: \(\Leftrightarrow\dfrac{13}{2x^2+7x-6x-21}+\dfrac{1}{2x+7}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\dfrac{13}{\left(2x+7\right)\left(x-3\right)}+\dfrac{1}{\left(2x+7\right)}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow26x+91+x^2-9-12x-14=0\)
\(\Leftrightarrow x^2+14x+68=0\)
hay \(x\in\varnothing\)
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 \(S=\left\{-1\right\}\)
\(a,3x^2+2x-1\Leftrightarrow3x^2-x+3x-1\Leftrightarrow x\left(3x-1\right)+\left(3x-1\right)\Leftrightarrow\left(3x-1\right)\left(x-1\right)\)
ĐKXĐ:\(x\ne\pm1\)
\(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}+\dfrac{x^2+3x-2}{1-x^2}=0\\ \Leftrightarrow\dfrac{\left(x+1\right)^2}{\left(x+1\right)\left(x-1\right)}-\dfrac{\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)}-\dfrac{x^2+3x-2}{\left(x+1\right)\left(x-1\right)}=0\\ \Leftrightarrow\dfrac{x^2+2x+1-x^2+2x-1-x^2-3x+2}{\left(x+1\right)\left(x-1\right)}=0\\ \Rightarrow-x^2+x+2=0\\ \Leftrightarrow x^2-x-2=0\\ \Leftrightarrow\left(x^2-2x\right)+\left(x-2\right)=0\\ \Leftrightarrow x\left(x-2\right)+\left(x-2\right)=0\\ \Leftrightarrow\left(x+1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\left(ktm\right)\\x=2\left(tm\right)\end{matrix}\right.\)
\(ĐK:x\ne\pm1\)
\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x+1\right)-\left[\left(x-1\right)\left(x-1\right)\right]-\left(x^2+3x-2\right)}{\left(x-1\right)\left(x+1\right)}=0\)
\(\Leftrightarrow\left(x+1\right)^2-\left(x-1\right)^2-\left(x^2+3x-2\right)=0\)
\(\Leftrightarrow x^2+2x+1-x^2+2x-1-x^2-3x+2=0\)
\(\Leftrightarrow-x^2-x+2=0\)
\(\Leftrightarrow-x^2+x-2x+2=0\)
\(\Leftrightarrow-x\left(x-1\right)-2\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(ktm\right)\\x=-2\left(tm\right)\end{matrix}\right.\)