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LA
1
17 tháng 7 2022
b: \(\Leftrightarrow\dfrac{2}{\left(x+7\right)\left(x-3\right)}=\dfrac{3x+21}{\left(x-3\right)\left(x+7\right)}\)
=>3x+21=2
=>x=-19/3
d: \(\Leftrightarrow\left(2x+1\right)^2-\left(2x-1\right)^2=8\)
\(\Leftrightarrow4x^2+4x+1-4x^2+4x-1=8\)
=>8x=8
hay x=1
7 tháng 6 2017
giải pt sau
g) 11+8x-3=5x-3+x
\(\Leftrightarrow\) 8x + 8 = 6x - 3
<=> 8x-6x = -3 - 8
<=> 2x = -11
=> x=-\(\dfrac{11}{2}\)
Vậy tập nghiệm của PT là : S={\(-\dfrac{11}{2}\)}
h)4-2x+15=9x+4-2x
<=> 19 - 2x = 7x + 4
<=> -2x - 7x = 4 - 19
<=> -9x = -15
=> x=\(\dfrac{15}{9}=\dfrac{5}{3}\)
Vậy tập nghiệm của pt là : S={\(\dfrac{5}{3}\)}
g)\(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=\dfrac{5}{3}+2x\)
<=> \(\dfrac{3\left(3x+2\right)}{6}-\dfrac{3x+1}{6}=\dfrac{5.2+6.2x}{6}\)
<=> 9x + 6 - 3x + 1 = 10 + 12x
<=> 6x + 7 = 10 + 12x
<=> 6x -12x = 10-7
<=> -6x = 3
=> x= \(-\dfrac{1}{2}\)
Vậy tập nghiệm của PT là : S={\(-\dfrac{1}{2}\)}
\(h,\dfrac{x+4}{5}-x+4=\dfrac{4x+2}{5}-5\)
<=> \(\dfrac{x+4-5\left(x+4\right)}{5}=\dfrac{4x+2-5.5}{5}\)
<=> x + 4 - 5x - 20 = 4x + 2 - 25
<=> x - 5x - 4x = 2-25-4+20
<=> -8x = -7
=> x= \(\dfrac{7}{8}\)
Vậy tập nghiệm của PT là S={\(\dfrac{7}{8}\)}
\(i,\dfrac{4x+3}{5}-\dfrac{6x-2}{7}=\dfrac{5x+4}{3}+3\)
<=> \(\dfrac{21\left(4x+3\right)}{105}\)-\(\dfrac{15\left(6x-2\right)}{105}\)=\(\dfrac{35\left(5x+4\right)+3.105}{105}\)
<=> 84x + 63 - 90x + 30 = 175x + 140 + 315
<=> 84x - 90x - 175x = 140 + 315 - 63 - 30
<=> -181x = 362
=> x = -2
Vậy tập nghiệm của PT là : S={-2}
K) \(\dfrac{5x+2}{6}-\dfrac{8x-1}{3}=\dfrac{4x+2}{5}-5\)
<=> \(\dfrac{5\left(5x+2\right)}{30}-\dfrac{10\left(8x-1\right)}{30}=\dfrac{6\left(4x+2\right)-150}{30}\)
<=> 25x + 10 - 80x - 10 = 24x + 12 - 150
<=> -55x = 24x - 138
<=> -55x - 24x = -138
=> -79x = -138
=> x=\(\dfrac{138}{79}\)
Vậy tập nghiệm của PT là S={\(\dfrac{138}{79}\)}
m) \(\dfrac{2x-1}{5}-\dfrac{x-2}{3}=\dfrac{x+7}{15}\)
<=> \(\dfrac{3\left(2x-1\right)-5\left(x-2\right)}{15}=\dfrac{x+7}{15}\)
<=> 6x - 3 - 5x + 10 = x+7
<=> x + 7 = x+7
<=> 0x = 0
=> PT vô nghiệm
Vậy S=\(\varnothing\)
n)\(\dfrac{1}{4}\left(x+3\right)=3-\dfrac{1}{2}\left(x+1\right)-\dfrac{1}{3}\left(x+2\right)\)
<=> \(\dfrac{1}{4}x+\dfrac{3}{4}=3-\dfrac{1}{2}x-\dfrac{1}{2}-\dfrac{1}{3}x-\dfrac{2}{3}\)
<=> \(\dfrac{1}{4}x+\dfrac{1}{2}x+\dfrac{1}{3}x=3-\dfrac{1}{2}-\dfrac{2}{3}-\dfrac{3}{4}\)
<=> \(\dfrac{13}{12}x=\dfrac{13}{12}\)
=> x= 1
Vậy S={1}
p) \(\dfrac{x}{3}-\dfrac{2x+1}{6}=\dfrac{x}{6}-6\)
<=> \(\dfrac{2x-2x+1}{6}=\dfrac{x-36}{6}\)
<=> 2x -2x + 1= x-36
<=> 2x-2x-x = -37
=> x = 37
Vậy S={37}
q) \(\dfrac{2+x}{5}-0,5x=\dfrac{1-2x}{4}+0,25\)
<=> \(\dfrac{4\left(2+x\right)-20.0,5x}{20}=\dfrac{5\left(1-2x\right)+20.0,25}{20}\)
<=> 8 + 4x - 10x = 5 - 10x + 5
<=> 4x-10x + 10x = 5+5-8
<=> 4x = 2
=> x= \(\dfrac{1}{2}\)
Vậy S={\(\dfrac{1}{2}\)}
7 tháng 6 2017
g) \(11+8x-3=5x-3+x\)
\(\Leftrightarrow8+8x=6x-3\)
\(\Leftrightarrow8x-6x=-3-8\)
\(\Leftrightarrow2x=-11\)
\(\Leftrightarrow x=-\dfrac{11}{2}\)
h, \(4-2x+15=9x+4-2x\)
\(\Leftrightarrow-2x-9x+2x=4-4-15\)
\(\Leftrightarrow-9x=-15\)
\(\Leftrightarrow x=\dfrac{-15}{-9}=\dfrac{5}{3}\)
16 tháng 2 2018
điều kiện xác định \(x\ne0\)
ta có : \(\dfrac{x+1}{x^2+2x+4}-\dfrac{x-2}{x^2-2x+4}=\dfrac{6}{x\left(x^4+4x^2+16\right)}\)
\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x^2-2x+4\right)-\left(x-2\right)\left(x^2+2x+4\right)}{\left(x^2+2x+4\right)\left(x^2-2x+4\right)}=\dfrac{6}{x\left(x^4+4x^2+16\right)}\)
\(\Leftrightarrow\dfrac{x^3-2x^2+4x+x^2-2x+4-\left(x^3+2x^2+4x-2x^2-4x-8\right)}{x^4-2x^3+4x^2+2x^3-4x^2+8x+4x^2-8x+16}=\dfrac{6}{x\left(x^4+4x^2+16\right)}\) \(\Leftrightarrow\dfrac{x^3-2x^2+4x+x^2-2x+4-x^3-2x^2-4x+2x^2+4x+8}{x^4-2x^3+4x^2+2x^3-4x^2+8x+4x^2-8x+16}=\dfrac{6}{x\left(x^4+4x^2+16\right)}\) \(\Leftrightarrow\dfrac{-x^2+2x+12}{x^4+4x^2+16}=\dfrac{6}{x\left(x^4+4x^2+16\right)}\)\(\Leftrightarrow-x^2+2x+12=\dfrac{6}{x}\Leftrightarrow x\left(-x^2+2x+12\right)=6\)
\(\Leftrightarrow-x^3+2x^2+12x=6\Leftrightarrow-x^3+2x^2+12x-6=0\)
tới đây bn bấm máy tính nha
21 tháng 1 2019
\(a,\dfrac{2}{2x+1}-\dfrac{3}{2x-1}=\dfrac{4}{4x^2-1}\\ \Leftrightarrow\dfrac{2\left(2x-1\right)-3\left(2x+1\right)}{\left(2x+1\right)\left(2x-1\right)}=\dfrac{4}{4x^2-1}\\ \Leftrightarrow\dfrac{4x-2-6x-3}{4x^2-1}=\dfrac{4}{4x^2-1}\\ \Leftrightarrow\dfrac{-2x-5}{4x^2-1}=\dfrac{4}{4x^2-1}\\ \Leftrightarrow\left(-2x-5\right)\left(4x^2-1\right)=4\left(4x^2-1\right)\\ \Leftrightarrow\left(2x-1\right)\left(2x+1\right)\left(-2x-5-4\right)=0\\ \Leftrightarrow\left(2x-1\right)\left(2x+1\right)\left(-2x-9\right)=0\\ \Rightarrow\left[{}\begin{matrix}2x-1=0\\2x+1=0\\-2x-9=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{1}{2}\\x=-\dfrac{9}{2}\end{matrix}\right.\\ Vậy......\)
\(b,\dfrac{2x}{x+1}+\dfrac{18}{x^2+2x-3}=\dfrac{2x-5}{x+3}\\ \Leftrightarrow\dfrac{2x}{x+1}+\dfrac{18}{x^2+3x-\left(x+3\right)}=\dfrac{2x-5}{x+3}\\ \Leftrightarrow\dfrac{2x\left(x^2+2x-3\right)+18\left(x+1\right)}{\left(x+1\right)\left(x-1\right)\left(x+3\right)}=\dfrac{2x-5}{x+3}\\ \Leftrightarrow\dfrac{2x^3+4x^2-6x+18x+18}{\left(x^2-1\right)\left(x+3\right)}=\dfrac{2x-5}{x+3}\\ \Leftrightarrow\dfrac{2x^3+4x^2+12x+18}{\left(x^2-1\right)\left(x+3\right)}=\dfrac{\left(2x-5\right)}{x+3}\\ \Leftrightarrow2\left(x^3+2x^2+6x+9\right)\left(x+3\right)=\left(2x-5\right)\left(x^2-1\right)\left(x+3\right)\\ \Leftrightarrow\left(x+3\right)\left(2x^3+4x^2+12x+18+2x^3-5x^2-2x+5\right)=0\\ \Leftrightarrow\left(x+3\right)\left(4x^3-x^2+10x+23\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+3=0\\4x^3-x^2+10x+23=0\end{matrix}\right.\)
\(\dfrac{1}{x-1}+\dfrac{2x^2-5}{x^3-1}=\dfrac{4}{x^2+x+1}\\ \Leftrightarrow\dfrac{x^2+x+1+2x^2-5}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{4}{x^2+x+1}\\ \Leftrightarrow\dfrac{3x^2+x-4}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{4}{x^2+x+1}\\ \Leftrightarrow\dfrac{3x^2+4x-3x-4}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{4}{x^2+x+1}\\ \Leftrightarrow\dfrac{\left(3x+4\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{4}{x^2+x+1}\\ \Leftrightarrow\left(3x+4\right)\left(x-1\right)\left(x^2+x+1\right)=4\left(x^2+x+1\right)\left(x-1\right)\\ \Leftrightarrow\left(x^2+x+1\right)\left(x-1\right)3x=0\\\Rightarrow\left[{}\begin{matrix}x^2+x+1=0\\x-1=0\\3x=0\end{matrix}\right.\\ Vìx^2+x+1=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\\ \Rightarrow\left[{}\begin{matrix}x-1=0\\3x=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=0\end{matrix}\right.\\ Vậy.....\)
CW
26 tháng 1 2017
a)\(\frac{3+2x}{2+x}-1=\frac{2-x}{2+x}\) (x khác -2)
\(\Leftrightarrow\frac{3+2x}{2+x}-\frac{2-x}{2+x}=1\)
\(\Leftrightarrow\frac{1+3x}{2+x}=1\)
\(\Leftrightarrow1+3x=2+x\)
\(\Leftrightarrow2x=1\Leftrightarrow x=\frac{1}{2}\)
b) \(\frac{5-2x}{3}+\frac{x^2-1}{3}x-1=\frac{\left(x-2\right)\left(1-3x\right)}{9x-3}\) (x khác 1/3)
\(\Leftrightarrow\frac{x^3-3x+5}{3}+\frac{\left(x-2\right)\left(3x-1\right)}{3\left(3x-1\right)}=1\)
\(\Leftrightarrow\frac{x^2-2x+3}{3}=1\)
\(\Leftrightarrow x\left(x-2\right)=0\Leftrightarrow\left[\begin{matrix}x=0\\x=2\end{matrix}\right.\)
c) \(\frac{1}{\left(3-2x\right)^2}-\frac{4}{\left(3+2x\right)^2}=\frac{3}{9-4x^2}\) (x khác +- 3/2)
\(\Leftrightarrow\frac{\left(3+2x\right)^2}{\left(3+2x\right)^2\left(3-2x\right)^2}-\frac{4\left(3-2x\right)^2}{\left(3+2x\right)^2\left(3-2x\right)^2}=\frac{9}{\left(3+2x\right)^2\left(3-2x\right)^2}\)
\(\Leftrightarrow9+12x+4x^2-4\left(9-12x+4x^2\right)-9=0\)
\(\Leftrightarrow-12x^2+60x-36=0\)
\(\Leftrightarrow-12\left(x^2-5x+3\right)=0\Leftrightarrow x^2-5x+3=0\)
\(\Rightarrow\Delta=b^2-4ac=25-12=13>0\)
\(x_1=\frac{-b+\sqrt{\Delta}}{2ac}=\frac{5+\sqrt{13}}{6}\)
\(x_2=\frac{5-\sqrt{13}}{6}\)
d) \(\frac{1}{x^2+2x+1}=\frac{4}{x+2x^2+x^3}=\frac{5}{2x+2x^2}\)
\(\Leftrightarrow\frac{x^2+2x+1}{1}=\frac{x+2x^2+x^3}{4}=\frac{2x+2x^2}{5}\)
Áp dụng tính chất của dãy tỉ số bằng nhau:
\(\frac{x^2+2x+1}{1}=\frac{x+2x^2+x^3}{4}=\frac{2x+2x^2}{5}=\frac{x^2+2x+1-\left(x+2x^2+x^3\right)+2x+2x^2}{1-4+5}\)
(dấu bằng thứ nhất của câu d là dấu cộng à???)
8 tháng 5 2017
Đặt t=x2-2x+3(t\(\ge\)2)
PTTT: \(\dfrac{1}{t-1}+\dfrac{1}{t}=\dfrac{9}{2\left(t+1\right)}\)
<=>2t2+2t+2t2-2=9t2-9
<=>5t2-2t-7=0
<=>(t+1)(5t-7)=0
Do t\(\ge\)2
=>t+1>0 5t-7>0
Vậy pt vô nghiệm
HN
9 tháng 5 2017
\(\dfrac{1}{x^2-2x+2}+\dfrac{1}{x^2-2x+3}=\dfrac{9}{2\left(x^2-2x+4\right)}\)
Đặt \(t=x^2-2x+2=\left(x-1\right)^2+1\ge1\)
Thì ta có:
\(PT\Leftrightarrow\dfrac{1}{t}+\dfrac{1}{t+1}=\dfrac{9}{2\left(t+2\right)}\)
\(\Leftrightarrow5t^2-t-4=0\)
\(\Leftrightarrow\left(5t^2-5t\right)+\left(4t-4\right)=0\)
\(\Leftrightarrow\left(t-1\right)\left(5t+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5t+4=0\\t-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}t=-\dfrac{4}{5}\left(l\right)\\t=1\end{matrix}\right.\)
\(\Rightarrow x^2-2x+2=1\)
\(\Leftrightarrow x=1\)
Vậy PT có 1 nghiệm là x = 1
\(\dfrac{2x-1}{2x+1}+\dfrac{2x+1}{2x-1}=\dfrac{4}{1-4x^2}\)
\(\Leftrightarrow4x^2-4x+1+4x^2+4x+1=-4\)
\(\Leftrightarrow8x^2+2+4=0\)
\(\Leftrightarrow8x^2+6=0\)(vô lý)