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Y
24 tháng 4 2022
\(a,\Leftrightarrow\dfrac{\left(x-3\right)^2-\left(x+3\right)^2-48}{x^2-9}=0\)
\(\Leftrightarrow x^2-6x+9-x^2-6x-9-48=0\)
\(\Leftrightarrow-12x-48=0\)
\(\Leftrightarrow-12x=48\)
\(\Leftrightarrow x=-4\)
\(b,\Leftrightarrow\dfrac{\left(x-5\right)\left(x+1\right)-\left(2x+3\right)-x\left(x-1\right)}{x^2-1}=0\)
\(\Leftrightarrow x^2+x-5x-5-2x-3-x^2+x=0\)
\(\Leftrightarrow-5x-8=0\)
\(\Leftrightarrow-5x=8\)
\(\Leftrightarrow x=-\dfrac{8}{5}\)
NK
1
LT
3 tháng 6 2020
\(\frac{x+3}{x-3}+\frac{48}{9-x^2}=\frac{x-3}{x+3}\)
ĐKXĐ: \(x\ne3;x\ne-3\)
\(\Leftrightarrow\frac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}-\frac{48}{\left(x-3\right)\left(x+3\right)}-\frac{\left(x-3\right)^2}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Rightarrow x^2+6x+9-48-x^2+6x-9=0\)
\(\Leftrightarrow12x-48=0\)
\(\Leftrightarrow12x=48\)
\(\Leftrightarrow x=4\left(TM\right)\)
Vậy phương trình có tập nghiệm là: S ={4}
MH
29 tháng 3 2022
a) \(\dfrac{5x}{2x+2}+1=\dfrac{6}{x+1}\left(đk:x\ne-1\right)\)
\(\dfrac{5x+2x+2}{2x+2}=\dfrac{12}{2x+2}\)
\(7x+2=12\)
\(7x=10\)
\(x=\dfrac{10}{7}\left(TM\right)\)
21 tháng 6 2023
a: ĐKXĐ: x<>2; x<>-2
PT =>(x+2)^2-(x-2)^2=4x^2
=>4x^2=x^2+4x+4-x^2+4x-4=8x
=>4x^2-8x=0
=>4x(x-2)=0
=>x=0(loại) hoặc x=2(loại)
b: ĐKXĐ: x<>1; x<>3
PT =>6x-18-4x+4=8
=>2x-14=8
=>2x=22
=>x=11(nhận)
c: ĐKXĐ: x<>3; x<>-3
PT =>(x+3)^2-48=(x-3)^2
=>x^2+6x+9-48=x^2-6x+9
=>12x=48
=>x=4(nhận)
TT
3
28 tháng 4 2018
a) 5 - 4x = 3x - 9
\(\Leftrightarrow5-4x-3x+9=0\)
\(\Leftrightarrow14-7x=0\)
\(\Leftrightarrow7x=14\Leftrightarrow x=2\)
Vậy \(S=\left\{2\right\}\)
b) \(\left(x-4\right)\left(3x+9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\3x+9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)
Vậy \(S=\left\{-3;4\right\}\)
c) \(\dfrac{x}{x+4}+\dfrac{12}{x-4}=\dfrac{4x+48}{x\cdot x-16}\)(1)
ĐKXĐ: \(x\ne\pm4\)
\(\left(1\right)\Leftrightarrow\dfrac{x\left(x-4\right)+12\left(x+4\right)-4x-48}{\left(x+4\right)\left(x-4\right)}=0\)
\(\Leftrightarrow x^2-4x+12x+48-4x-48=0\)
\(\Leftrightarrow x^2+4x=0\)
\(\Leftrightarrow x\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(TM\right)\\x=-4\left(KTM\right)\end{matrix}\right.\)
Vậy \(S=\left\{0\right\}\)
d) \(4-2x=7-x\)
\(\Leftrightarrow4-2x-7+x=0\)
\(\Leftrightarrow-x-3=0\)
\(\Leftrightarrow-x=3\Leftrightarrow x=-3\)
Vậy \(S=\left\{-3\right\}\)
e) \(\left(x+4\right) \left(8-4x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\8-4x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=2\end{matrix}\right.\)
Vậy \(S=\left\{-4;2\right\}\)
f) \(\dfrac{x}{x+5}+\dfrac{11}{x-5}=\dfrac{x+55}{x\cdot x-25}\left(2\right)\)
ĐKXĐ: \(x\ne\pm5\)
\(\left(2\right)\Leftrightarrow\dfrac{x\left(x-5\right)+11\left(x+5\right)-x-55}{\left(x+5\right)\left(x-5\right)}=0\)
\(\Leftrightarrow x^2-5x+11x+55-x-55=0\)
\(\Leftrightarrow x^2+5x=0\)
\(\Leftrightarrow x\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(TM\right)\\x=-5\left(KTM\right)\end{matrix}\right.\)
Vậy \(S=\left\{0\right\}\)
g) \(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=\dfrac{5}{3}+2x\)
\(\Leftrightarrow\dfrac{3\left(3x+2\right)-3x-1-10-12x}{6}=0\)
\(\Leftrightarrow9x+6-3x-1-10-12x=0\)
\(\Leftrightarrow-6x-5=0\)
\(\Leftrightarrow-6x=5\)
\(\Leftrightarrow x=-\dfrac{5}{6}\)
Vậy \(S=\left\{-\dfrac{5}{6}\right\}\)
h) \(2x-\left(3-5x\right)=4\left(x+3\right)\)
\(\Leftrightarrow2x-3+5x-4x-12=0\)
\(\Leftrightarrow3x-15=0\)
\(\Leftrightarrow x=5\)
Vậy \(S=\left\{5\right\}\)
i) \(3x-6+x=9-x\)
\(\Leftrightarrow3x-6+x-9+x=0\)
\(\Leftrightarrow5x-15=0\)
\(\Leftrightarrow x=3\)
Vậy \(S=\left\{3\right\}\)
k)\(2t-3+5t=4t+12\)
\(\Leftrightarrow2t-3+5t-4t-12=0\)
\(\Leftrightarrow3t-15=0\)
\(\Leftrightarrow t=5\)
Vậy \(S=\left\{5\right\}\)
=>(x+3)^2-48(x-3)=x-3
=>x^2+6x+9-48x+144-x+3=0
=>x^2-43x+156=0
=>x=39 hoặc x=4