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15 tháng 5 2023
a:Sửa đề: \(\dfrac{3}{5x-1}+\dfrac{2}{3-x}=\dfrac{4}{\left(1-5x\right)\left(x-3\right)}\)
=>3x-9-10x+2=-4
=>-7x-7=-4
=>-7x=3
=>x=-3/7
b: =>\(\dfrac{5-x}{4x\left(x-2\right)}+\dfrac{7}{8x}=\dfrac{x-1}{2x\left(x-2\right)}+\dfrac{1}{8\left(x-2\right)}\)
=>\(2\left(5-x\right)+7\left(x-2\right)=4\left(x-1\right)+x\)
=>10-2x+7x-14=4x-4+x
=>5x-4=5x-4
=>0x=0(luôn đúng)
Vậy: S=R\{0;2}
12 tháng 8 2021
1/ x2-3x+2=0
⇒ (x2-2x)-(x-2)=0
⇒ x(x-2)-(x-2)=0
⇒ (x-1)(x-2)=0
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
2) x2-6x+5=0
⇒x2-6x+9-4=0
⇒(x2-6x+9)-22=0
⇒(x-3)2-22=0
⇒(x-3-2)(x-3+2)=0
⇒(x-5)(x-1)=0
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\x-5=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=5\end{matrix}\right.\)
3) 2x2+5x+3=0
⇒ (2x2+2x)+(3x+3)=0
⇒ 2x(x+1)+3(x+1)=0
⇒ (x+1)(2x+3)=0
\(\Rightarrow\left[{}\begin{matrix}x+1=0\\2x+3=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-1\\x=-1,5\end{matrix}\right.\)
4) x2-8x+15=0
⇒ (x2-8x+16)-1=0
⇒ (x-4)2-12=0
⇒ (x-4-1)(x-4+1)=0
⇒ (x-5)(x-3)=0
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\x-5=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=5\end{matrix}\right.\)
5) x2-x-12=0
⇒ (x2-4x)+(3x-12)=0
⇒ x(x-4)+3(x-4)=0
⇒ (x-4)(x+3)=0
\(\Rightarrow\left[{}\begin{matrix}x+3=0\\x-4=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-3\\x=4\end{matrix}\right.\)
12 tháng 8 2021
1: Ta có: \(x^2-3x+2=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
2: Ta có: \(x^2-6x+5=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=5\end{matrix}\right.\)
3: Ta có: \(2x^2+5x+3=0\)
\(\Leftrightarrow\left(x+1\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{3}{2}\end{matrix}\right.\)
4: Ta có: \(x^2-8x+15=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=5\end{matrix}\right.\)
5: Ta có: \(x^2-x-12=0\)
\(\Leftrightarrow\left(x-4\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)
1 tháng 3 2021
`d,(10x+3)/12=1+(6+8x)/9`
`<=>(10x+3)/12=(8x+15)/9`
`<=>30x+9=32x+60`
`<=>2x=-51`
`<=>x=-51/2`
16 tháng 8 2021
a: Ta có: \(8x+11-3=5x+x-3\)
\(\Leftrightarrow8x+8=6x-3\)
\(\Leftrightarrow2x=-11\)
hay \(x=-\dfrac{11}{2}\)
b: Ta có: \(2x\left(x+2\right)^2-8x^2=2\left(x-2\right)\left(x^2+2x+4\right)\)
\(\Leftrightarrow2x\left(x^3+6x^2+12x+8\right)-8x^2=2\left(x^3-8\right)\)
\(\Leftrightarrow2x^4+12x^3+24x^2+16x-8x^2-2x^3+16=0\)
\(\Leftrightarrow2x^4+10x^3+16x^2+16x+16=0\)
\(\Leftrightarrow2x^4+4x^3+6x^3+12x^2+4x^2+8x+8x+16=0\)
\(\Leftrightarrow\left(x+2\right)\left(2x^3+6x^2+4x+8\right)=0\)
\(\Leftrightarrow x+2=0\)
hay x=-2
16 tháng 8 2021
c: Ta có: \(\left(x+1\right)\left(2x-3\right)=\left(2x-1\right)\left(x+5\right)\)
\(\Leftrightarrow2x^2-3x+2x-3-2x^2-10x+x+5=0\)
\(\Leftrightarrow-10x+2=0\)
\(\Leftrightarrow-10x=-2\)
hay \(x=\dfrac{1}{5}\)
d: Ta có: \(\dfrac{1}{10}-2\cdot\left(\dfrac{1}{2}t-\dfrac{1}{10}\right)=2\left(t-\dfrac{5}{2}\right)-\dfrac{7}{10}\)
\(\Leftrightarrow\dfrac{1}{10}-t+\dfrac{1}{5}=2t-5-\dfrac{7}{10}\)
\(\Leftrightarrow-t-2t=-\dfrac{57}{10}-\dfrac{3}{10}=-6\)
hay t=2
2 tháng 2 2022
a) \(PT\Leftrightarrow3x-2x=2-3\Leftrightarrow x=-1\)
Vậy: \(S=\left\{-1\right\}\)
b) \(PT\Leftrightarrow-2x+3x=-7+22\Leftrightarrow x=15\)
Vậy: \(S=\left\{15\right\}\)
c) \(PT\Leftrightarrow8x-5x=3+12\Leftrightarrow3x=15\Leftrightarrow x=5\)
Vậy: \(S=\left\{5\right\}\)
d) \(PT\Leftrightarrow x+4x-2x=12+25-1\Leftrightarrow3x=36\Leftrightarrow x=12\)
Vậy: \(S=\left\{12\right\}\)
e) \(PT\Leftrightarrow x+2x+3x-3x=19+5\Leftrightarrow3x=24\Leftrightarrow x=8\)
Vậy: \(S=\left\{8\right\}\)
TD
2 tháng 2 2022
a)3x-2=2x-3
=>x=-1
b)7-2x=22-3x
=>x=15
c)8x-3=5x+12
=>3x=15
=>x=5
d)x-12+4x=25+2x-1
=>3x=12
=>x=4
e)x+2x+3x-19=3x+5
=>3x=24
=>x=8
HH
8 tháng 7 2018
1/ \(1+\frac{2}{x-1}+\frac{1}{x+3}=\frac{x^2+2x-7}{x^2+2x-3}\)
ĐKXĐ: \(\hept{\begin{cases}x-1\ne0\\x+3\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne1\\x\ne-3\end{cases}}\)
<=> \(1+\frac{2\left(x+3\right)+x-1}{\left(x-1\right)\left(x+3\right)}=\frac{x^2+2x-3-5}{x^2+2x-3}\)
<=> \(1+\frac{2x+6+x-1}{x^2+2x-3}=1-\frac{5}{x^2+2x-3}\)
<=> \(\frac{3x+5}{x^2+2x-3}+\frac{5}{x^2+2x-3}=1-1\)
<=> \(\frac{3x+5}{x^2+2x-3}+\frac{5}{x^2+2x-3}=0\)
<=> \(\frac{3x+10}{x^2+2x-3}=0\)
<=> \(3x+10=0\)
<=> \(x=-\frac{10}{3}\)
3 tháng 2 2022
a: =>10x-4=15-9x
=>19x=19
hay x=1
b: \(\Leftrightarrow3\left(10x+3\right)=36+4\left(8x+6\right)\)
=>30x+9=36+32x+24
=>30x-32x=60-9
=>-2x=51
hay x=-51/2
c: \(\Leftrightarrow2x+\dfrac{6}{5}=5-\dfrac{13}{5}-x\)
=>3x=6/5
hay x=2/5
d: \(\Leftrightarrow\dfrac{7x}{8}-\dfrac{5\left(x-9\right)}{1}=\dfrac{20x+1.5}{6}\)
\(\Leftrightarrow21x-120\left(x-9\right)=4\left(20x+1.5\right)\)
=>21x-120x+1080=80x+60
=>-179x=-1020
hay x=1020/179
e: \(\Leftrightarrow5\left(7x-1\right)+60x=6\left(16-x\right)\)
=>35x-5+60x=96-6x
=>95x+6x=96+5
=>x=1
f: \(\Leftrightarrow6\left(x+4\right)+30\left(-x+4\right)=10x-15\left(x-2\right)\)
=>6x+24-30x+120=10x-15x+30
=>-24x+96=-5x+30
=>-19x=-66
hay x=66/19
KN
5 tháng 3 2020
a) \(3x-2=2x-3\)
\(\Leftrightarrow x+1=0\)
\(\Leftrightarrow x=-1\)
b) \(3-4y+24+6y=y+27+3y\)
\(\Leftrightarrow-2y=0\Leftrightarrow y=0\)
KN
5 tháng 3 2020
c) \(7-2x=22-3x\)
\(\Leftrightarrow x-15=0\)
\(\Leftrightarrow x=15\)
d) \(8x-3=5x+12\)
\(\Leftrightarrow3x-15=0\Leftrightarrow x=5\)
\(8x-3=5x+12\)
\(\Leftrightarrow8x-5x=12+3\)
\(\Leftrightarrow3x=15\Leftrightarrow x=5\)
vậy tập nghiệm của ptrình là S={5}
\(\dfrac{5}{x+3}=\dfrac{3}{x-1}\)(1)
ĐK \(x\ne-3,x\ne1\)
Từ (1) => 5(x-1)=3(x+3)
<=>5x-5=3x+9
<=>5x-3x=9+5
<=>2x=14<=>x=7(TMĐK)
vậy S={7}