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LT
4
LT
11
NB
4 tháng 3 2023
a) 6/-x=x/-24
=> -x.x=6.-24
=>-x.x=-144
=>x=12 hay x=-12
b)9/x=-35/105
=>9/x=-1/3
=>x=9.3/-1=-27
=>x=-27
c)x-1/8=5/8
=>x=5/8+1/8
=>x=3/4
d)x-1/2-(3/2+x)=-2
=>-x+4/2=-2
=>-x/2=0
=>x=0
e)x+1/3=-12/5.10/6
x+1/3=-4
x=-4-1/3
x= -13/3
#YM
31 tháng 3 2021
Bài 1:
a) Ta có: \(\dfrac{2}{5}\cdot x+\dfrac{1}{3}=\dfrac{1}{5}\)
\(\Leftrightarrow\dfrac{2}{5}\cdot x=\dfrac{1}{5}-\dfrac{1}{3}=\dfrac{-2}{15}\)
\(\Leftrightarrow x=\dfrac{-2}{15}:\dfrac{2}{5}=\dfrac{-2}{15}\cdot\dfrac{5}{2}\)
hay \(x=-\dfrac{1}{3}\)
Vậy: \(x=-\dfrac{1}{3}\)
b) Ta có: \(\dfrac{1}{5}+\dfrac{5}{3}:x=\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{5}{3}:x=\dfrac{1}{2}-\dfrac{1}{5}=\dfrac{3}{10}\)
\(\Leftrightarrow x=\dfrac{5}{3}:\dfrac{3}{10}=\dfrac{5}{3}\cdot\dfrac{10}{3}\)
hay \(x=\dfrac{50}{9}\)
Vậy: \(x=\dfrac{50}{9}\)
c) Ta có: \(\dfrac{4}{9}-\dfrac{5}{3}\cdot x=-2\)
\(\Leftrightarrow\dfrac{5}{3}x=\dfrac{4}{9}+2=\dfrac{22}{9}\)
\(\Leftrightarrow x=\dfrac{22}{9}:\dfrac{5}{3}=\dfrac{22}{9}\cdot\dfrac{3}{5}\)
hay \(x=\dfrac{22}{15}\)
Vậy: \(x=\dfrac{22}{15}\)
d) Ta có: \(\dfrac{5}{7}:x-3=\dfrac{-2}{7}\)
\(\Leftrightarrow\dfrac{5}{7}:x=\dfrac{-2}{7}+3=\dfrac{19}{21}\)
\(\Leftrightarrow x=\dfrac{5}{7}:\dfrac{19}{21}=\dfrac{5}{7}\cdot\dfrac{21}{19}\)
hay \(x=\dfrac{15}{19}\)
Vậy:\(x=\dfrac{15}{19}\)
24 tháng 1 2018
Mik chỉ làm 1 câu chung cho bài 1 thôi nha , mấy câu sau giống .
Tìm x , biết :
a) ( x + 1) 2 . ( x - 2 )2 = 0
=> \(\left\{{}\begin{matrix}\left(x+1\right)^2=0\\\left(x-2\right)^2=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x+1=0\\x-2=0\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)
Vậy x = -1 hoặc x = 2 .
24 tháng 1 2018
Bài 2 , rút gọn biểu thức :
A = a.(b -c) - b.(a+c)
= ab - ac - ( ab + bc )
= ab - ac - ab - bc
= ac - bc
= c .(a-b)
C = (a+3b).c - d - (3a-d).(b+c) - 2c.(b - a) + 2b.(a+d)
= ac + 3bc - d - (3a - d).(b+c) - 2cb - 2ca + 2ba + 2bd
= ac + ( 3bc - 2bc ) - d - ( 3a - d) . ( b+c) +(-2ca + 2ba ) +2db
= ac + bc - d - ( 3a -d) . ( b+c) -2a + cb + 2db
= (a+b).c - d - (3a-d) . ( b+c) - 2a + (2d+c).b
= .........(mik chịu )..........
N
1
20 tháng 1 2022
a: \(\Leftrightarrow\left|x\cdot\dfrac{1}{3}-\dfrac{1}{4}\right|=\dfrac{1}{2}+\dfrac{1}{6}=\dfrac{3}{6}+\dfrac{1}{6}=\dfrac{2}{3}\)
\(\Leftrightarrow\left[{}\begin{matrix}x\cdot\dfrac{1}{3}-\dfrac{1}{4}=\dfrac{2}{3}\\x\cdot\dfrac{1}{3}-\dfrac{1}{4}=-\dfrac{2}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{11}{4}\\x=-\dfrac{5}{4}\end{matrix}\right.\)
b: \(\Leftrightarrow\left|x\cdot\dfrac{4}{7}+\dfrac{3}{4}\right|=\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\)
\(\Leftrightarrow\left[{}\begin{matrix}x\cdot\dfrac{4}{7}+\dfrac{3}{4}=\dfrac{1}{6}\\x\cdot\dfrac{4}{7}+\dfrac{3}{4}=-\dfrac{1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-49}{48}\\x=-\dfrac{77}{48}\end{matrix}\right.\)
18 tháng 1
b: \(\dfrac{5}{7}-\dfrac{2}{3}\cdot x=\dfrac{4}{5}\)
=>\(\dfrac{2}{3}x=\dfrac{5}{7}-\dfrac{4}{5}=\dfrac{25-28}{35}=\dfrac{-3}{35}\)
=>\(x=-\dfrac{3}{35}:\dfrac{2}{3}=\dfrac{-3}{35}\cdot\dfrac{3}{2}=-\dfrac{9}{70}\)
c: \(\dfrac{1}{2}x+\dfrac{3}{5}x=-\dfrac{2}{3}\)
=>\(x\left(\dfrac{1}{2}+\dfrac{3}{5}\right)=-\dfrac{2}{3}\)
=>\(x\cdot\dfrac{5+6}{10}=\dfrac{-2}{3}\)
=>\(x\cdot\dfrac{11}{10}=-\dfrac{2}{3}\)
=>\(x=-\dfrac{2}{3}:\dfrac{11}{10}=-\dfrac{2}{3}\cdot\dfrac{10}{11}=\dfrac{-20}{33}\)
d: \(\dfrac{4}{7}\cdot x-x=-\dfrac{9}{14}\)
=>\(\dfrac{-3}{7}\cdot x=\dfrac{-9}{14}\)
=>\(\dfrac{3}{7}\cdot x=\dfrac{9}{14}\)
=>\(x=\dfrac{9}{14}:\dfrac{3}{7}=\dfrac{9}{14}\cdot\dfrac{7}{3}=\dfrac{3}{2}\)
L
3
22 tháng 6 2021
a) \(\dfrac{5}{24}+x=\dfrac{7}{12}\)
<=> \(x=\dfrac{7}{12}-\dfrac{5}{24}=\dfrac{14}{24}-\dfrac{5}{24}=\dfrac{9}{24}=\dfrac{3}{8}\)
b) \(x-\dfrac{3}{4}=\dfrac{1}{2}\)
<=> \(x=\dfrac{1}{2}+\dfrac{3}{4}=\dfrac{2}{4}+\dfrac{3}{4}=\dfrac{5}{4}\)
c) bn ghi rõ đề chút