a) - x - \(\dfrac{2}{3}=-\dfrac{6}{7}\) b) (5x - 1). (2x - \(\dfrac{1}{3}=0\)
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20 tháng 6 2022
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
8 tháng 2 2022
a, \(\Rightarrow10x-4+6x=6+15-9x\Leftrightarrow7x=25\Leftrightarrow x=\dfrac{25}{7}\)
b, \(\Rightarrow2\left(3x^2+5x-2\right)-6x^2-3=33\Leftrightarrow10x-7=33\Leftrightarrow x=4\)
c, \(\Rightarrow12x-10x-4=21-9x\Leftrightarrow11x=25\Leftrightarrow x=\dfrac{25}{11}\)
d, \(\Rightarrow3x-3+2x-2-x+1=12\Leftrightarrow4x=16\Leftrightarrow x=4\)
19 tháng 2 2022
a: \(\Leftrightarrow7\left(7-3x\right)+12\left(5x+2\right)=84\left(x+13\right)\)
\(\Leftrightarrow49-21x+60x+24=84x+1092\)
\(\Leftrightarrow39x-84x=1092-73\)
=>-45x=1019
hay x=-1019/45
b: \(\Leftrightarrow21\left(x+3\right)-14=4\left(5x+9\right)-7\left(7x-9\right)\)
=>21x+63-14=20x+36-49x+63
=>21x+49=-29x+99
=>50x=50
hay x=1
c: \(\Leftrightarrow7\left(2x+1\right)-3\left(5x+2\right)=21x+63\)
=>14x+7-15x-6-21x-63=0
=>-22x-64=0
hay x=-32/11
d: \(\Leftrightarrow35\left(2x-3\right)-15\left(2x+3\right)=21\left(4x+3\right)-17\cdot105\)
=>70x-105-30x-45=84x+63-1785
=>40x-150-84x+1722=0
=>-44x+1572=0
hay x=393/11
25 tháng 7 2023
Bài 3 :
\(\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+...+\dfrac{1}{2023!}\)
\(\dfrac{1}{2!}=\dfrac{1}{2.1}=1-\dfrac{1}{2}< 1\)
\(\dfrac{1}{3!}=\dfrac{1}{3.2.1}=1-\dfrac{1}{2}-\dfrac{1}{3}< 1\)
\(\dfrac{1}{4!}=\dfrac{1}{4.3.2.1}< \dfrac{1}{3!}< \dfrac{1}{2!}< 1\)
.....
\(\)\(\dfrac{1}{2023!}=\dfrac{1}{2023.2022....2.1}< \dfrac{1}{2022!}< ...< \dfrac{1}{2!}< 1\)
\(\Rightarrow\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+...+\dfrac{1}{2023!}< 1\)
TM
24 tháng 4 2022
a) \(\dfrac{3}{x-7}+\dfrac{2}{x+7}=\dfrac{5}{x^2-49}\)
(ĐKXĐ: x khác 7; x khác -7)
<=>\(\dfrac{3.\left(x+7\right)}{\left(x-7\right).\left(x+7\right)}+\dfrac{2.\left(x-7\right)}{\left(x+7\right).\left(x-7\right)}=\dfrac{5}{\left(x+7\right).\left(x-7\right)}\)
=> 3x + 21 + 2x - 14 = 5
<=> 3x + 2x = 5 + 14 - 21
<=> 5x = -2
<=> x = \(\dfrac{-2}{5}\)
Vậy S = { \(\dfrac{-2}{5}\) }
TM
24 tháng 4 2022
b) \(\dfrac{2x-1}{3}-\dfrac{x+3}{2}>1+\dfrac{5x}{6}\)
<=> \(\dfrac{2.\left(2x-1\right)}{3.2}-\dfrac{3.\left(x+3\right)}{3.2}>\dfrac{1.6}{6}+\dfrac{5x}{6}\)
=> 4x - 2 - 3x - 9 > 6 + 5x
<=> 4x - 3x - 5x > 6 + 9 + 2
<=> -4x > 17
<=> \(\dfrac{-17}{4}\)
Vậy S = { \(\dfrac{-17}{4}\) }
TD
13 tháng 2 2018
a:
\(\dfrac{3\left(2x+1\right)}{4}-5-\dfrac{3x+2}{10}=\dfrac{2\left(3x-1\right)}{5}\)
\(\Leftrightarrow\dfrac{15\left(2x+1\right)-100-2\left(3x+2\right)}{20}=\dfrac{8\left(3x-1\right)}{20}\)
\(\Leftrightarrow15\left(2x+1\right)-100-2\left(3x+2\right)=8\left(3x-1\right)\)
\(\Leftrightarrow30x+15-100-6x+4=24x-8\)\(\Leftrightarrow30x-6x-24x=100-4-8\)
\(\Leftrightarrow0x=88\)
Vậy pt vô nghiệm
b:
\(\dfrac{x-15}{23}+\dfrac{x-23}{15}-2=0\)
\(\Leftrightarrow\dfrac{x-15}{23}+\dfrac{x-23}{15}=2\)
\(\Leftrightarrow\dfrac{x-15}{23}-1+\dfrac{x-23}{15}-1=2-2\)
\(\Leftrightarrow\dfrac{x-15-23}{23}+\dfrac{x-23-15}{15}=0\)
\(\Leftrightarrow\dfrac{x-38}{23}+\dfrac{x-23}{15}=0\)
\(\Leftrightarrow\left(x+38\right)\left(\dfrac{1}{23}+\dfrac{1}{15}\right)=0\)
Vì \(\dfrac{1}{23}+\dfrac{1}{15}\ne0\) nên x + 38 =0 \(\Leftrightarrow x=-38\)
Vậy tập nghiện của pt S= {-38}
c:
\(\dfrac{3\left(2x+1\right)}{4}-\dfrac{5x+3}{6}+\dfrac{x+1}{3}=x+\dfrac{7}{12}\)
\(\Leftrightarrow\dfrac{9\left(2x+1\right)-2\left(5x+3\right)+4\left(x+1\right)}{12}=\dfrac{12x+7}{12}\)
\(\Leftrightarrow9\left(2x+1\right)-2\left(5x+3\right)+4\left(x+1\right)=12x+7\)
\(\Leftrightarrow18x+9-10x-6+4x+4=12x+7\)
\(\Leftrightarrow18x-10x+4x-12x=7-9+6-4\)
\(\Leftrightarrow0x=0\)
Vậy pt vô số nghiệm
MH
18 tháng 3 2022
\(1\text{)}\left(x-5\right)^2+3\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-5+3\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)
\(2\text{)}\dfrac{2x-1}{3}-\dfrac{5x+2}{7}=x+13\)
\(\Leftrightarrow\dfrac{7\left(2x-1\right)-3\left(5x+2\right)}{21}=\dfrac{21\left(x+13\right)}{21}\)
\(\Leftrightarrow14x-7-15x-6=21x+273\)
\(\Leftrightarrow-x-13=21x+273\)
\(\Leftrightarrow-22x=286\)
\(\Rightarrow x=-\dfrac{286}{22}=-\dfrac{143}{11}\)
\(3\text{)}\dfrac{x-1}{x+2}-\dfrac{x}{x-2}=\dfrac{7x-6}{4-x^2}\left(đk:x\ne\pm2\right)\)
\(\Leftrightarrow\dfrac{x-1}{2+x}+\dfrac{x}{2-x}=\dfrac{7x-6}{4-x^2}\)
\(\Leftrightarrow\dfrac{\left(x-1\right)\left(2-x\right)+x\left(2+x\right)}{4-x^2}=\dfrac{7x-6}{4-x^2}\)
\(\Leftrightarrow2x-x^2-2+x+2x+x^2=7x-6\)
\(\Leftrightarrow x-2=7x-6\)
\(\Leftrightarrow-6x=-4\)
\(\Rightarrow x=\dfrac{2}{3}\)
OP
29 tháng 1 2022
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
21 tháng 2 2021
a) Ta có: \(\left(x+1\right)\left(2x-3\right)-3\left(x-2\right)=2\left(x-1\right)^2\)
\(\Leftrightarrow2x^2-3x+2x-3-3x+6=2\left(x^2-2x+1\right)\)
\(\Leftrightarrow2x^2-4x+3-2x^2+4x-2=0\)
\(\Leftrightarrow1=0\)(vô lý)
Vậy: \(S=\varnothing\)
a/ \(-x-\dfrac{2}{3}=-\dfrac{6}{7}\)
\(\Leftrightarrow-x=-\dfrac{6}{7}+\dfrac{2}{3}\)
\(\Leftrightarrow-x=-\dfrac{18}{21}+\dfrac{14}{21}\)
\(\Leftrightarrow-x=-\dfrac{4}{21}\)
\(\Leftrightarrow x=\dfrac{4}{21}\)
Vậy ....
b/ \(\left(5x-1\right)\left(2x-\dfrac{1}{3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-1=0\\2x-\dfrac{1}{3}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}5x=1\\2x=\dfrac{1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=\dfrac{1}{6}\end{matrix}\right.\)
Vậy ..