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\(\dfrac{x-1}{65}+\dfrac{x-3}{63}=\dfrac{x-5}{61}+\dfrac{x-7}{59}\)
\(\Leftrightarrow\dfrac{x-1}{65}-1+\dfrac{x-3}{63}-1=\dfrac{x-5}{61}-1+\dfrac{x-7}{59}-1\)
\(\Leftrightarrow\dfrac{x-66}{65}+\dfrac{x-66}{63}=\dfrac{x-66}{61}+\dfrac{x-66}{59}\)
\(\Leftrightarrow\left(x-66\right)\left(\dfrac{1}{65}+\dfrac{1}{63}-\dfrac{1}{61}-\dfrac{1}{59}\right)=0\)
\(\Leftrightarrow x-66=0\)
\(\Leftrightarrow x=66\)
Vậy x=66.
a, Theo đề ta có:
\(2.3^x-405=3^{x-1}\)
=> \(2.3^x-405=3^x:3\)
=> \(405=(2.3^x)-(3^x:3)\)
=>\(405=(2.3^x)-(3^x.\dfrac{1}{3})\)
=> \(405=3^x(2-\dfrac{1}{3})\)
=>\(405=3^x(\dfrac{6}{3}-\dfrac{1}{3})\)
=> \(405=3^x.\dfrac{5}{3}\)
=> \(3^x=405:\dfrac{5}{3}\)
=>\(3^x=405.\dfrac{3}{5}\)
=> \(3^x=81.3\)
=> \(3^x=243\)
=> \(3^x=3^5\)
=> x=5
Vậy:..............................
2, \(\Rightarrow\left\{{}\begin{matrix}\\\dfrac{5}{4}x-\dfrac{7}{2}=0\\\dfrac{5}{8}x+\dfrac{3}{5}=0\\\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{14}{5}\\\\x=\dfrac{-24}{25}\\\end{matrix}\right.\)
a: |x-1/2|=7/2
=>x-1/2=7/2 hoặc x-1/2=-7/2
=>x=4 hoặc x=-3
b: \(x:\dfrac{3}{8}+\dfrac{5}{8}=x\)
=>8/3x-x=-5/8
=>5/3x=-5/8
hay x=-5/8:5/3=-5/8x3/5=-15/40=-3/8
c: \(\dfrac{5}{6}-\left|x-\dfrac{1}{2}\right|=\dfrac{15}{18}=\dfrac{5}{6}\)
=>|x-1/2|=0
=>x-1/2=0
hay x=1/2
e: \(\left(5x-3\right)^2-\dfrac{1}{64}=0\)
=>(5x-3)2=1/64
=>5x-3=1/8 hoặc 5x-3=-1/8
=>5x=25/8 hoặc 5x=23/8
=>x=5/8 hoặc x=23/40
1: \(\Leftrightarrow3x+4=2\)
=>3x=-2
=>x=-2/3
2: \(\Leftrightarrow7x-7=6x-30\)
=>x=-23
3: =>\(5x-5=3x+9\)
=>2x=14
=>x=7
4: =>9x+15=14x+7
=>-5x=-8
=>x=8/5
a. \(\dfrac{-39}{7}:x=26\)
x = \(\dfrac{-39}{7}:26\)
x = \(\dfrac{-3}{14}\)
b. \(x:\dfrac{13}{5}=\dfrac{7}{4}\)
x = \(\dfrac{7}{4}.\dfrac{13}{5}\)
x = \(\dfrac{91}{20}\)
c. x = \(\dfrac{-3}{5}-\dfrac{1}{2}\)
x = \(\dfrac{-11}{10}\)
d. \(x-\dfrac{3}{4}=\dfrac{9}{4}\)
x = \(\dfrac{9}{4}+\dfrac{3}{4}\)
x = 3
e. \(\dfrac{7}{8}:x=\dfrac{14}{3}\)
x = \(\dfrac{7}{8}:\dfrac{14}{3}\)
x = \(\dfrac{3}{16}\)
f. \(x:\dfrac{8}{3}=\dfrac{13}{3}\)
x = \(\dfrac{13}{3}.\dfrac{8}{3}\)
x = \(\dfrac{104}{9}\)
g. x = \(\dfrac{4}{10}-\dfrac{2}{5}\)
x = 0
chúc bạn học tốt 
Giải:
a) \(\dfrac{1}{3}x+\dfrac{1}{5}-\dfrac{1}{2}x=1\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{1}{5}-\dfrac{1}{6}x=\dfrac{5}{4}\)
\(\Leftrightarrow\dfrac{1}{6}x=\dfrac{-21}{20}\)
\(\Leftrightarrow x=\dfrac{-63}{10}\)
Vậy ...
b) \(\dfrac{3}{2}\left(x+\dfrac{1}{2}\right)-\dfrac{1}{8}x=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{3}{2}x+\dfrac{3}{4}-\dfrac{1}{8}x=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{11}{8}x=\dfrac{-1}{2}\)
\(\Leftrightarrow x=\dfrac{-4}{11}\)
Vậy ...
Các câu sau làm tương tự câu b)
a: x>-3/5 nên x+3/5>0
x<1/7 nên x-1/7<0
A=1/7-x-x-3/5+4/5=-2x+12/35
b: B=|x-1/7|+|x+3/5|-1/3
x>-3/5 nên x+3/5>0
x<1/7 nên x-1/7<0
B=1/7-x+3/5+x-1/3=43/105
a) \(\dfrac{x}{48}=-\dfrac{4}{7}\Rightarrow x=-\dfrac{192}{7}\)
b) \(\left(x+\dfrac{4}{5}\right)-\dfrac{2}{5}=\dfrac{3}{5}\Rightarrow x+\dfrac{4}{5}=1\)
\(\Rightarrow x=\dfrac{1}{5}\)
c) \(2\left|x-1\right|^2=72\Rightarrow\left|x-1\right|^2=36\)
\(\Rightarrow\left|x-1\right|=6\)
TH1: x - 1 = -6 => x = -5
TH2: x - 1 = 6 => x = 7
e) \(\dfrac{x}{2,5}=\dfrac{4}{5}\Rightarrow x=2\)
f) | x - 2 | = 1 + 4 = 5
TH1: x - 2 = -5 => x = -3
TH2: x - 2 = 5 => x = 7
a) \(\dfrac{x}{48}=\dfrac{-4}{7}\)
⇒ x.7=48.(-4)
7x = -192
x=\(\dfrac{-192}{7}\) Vậy x=\(\dfrac{-192}{7}\)
b) \(\left(x+\dfrac{4}{5}\right)-\dfrac{2}{5}=\dfrac{3}{5}\)
\(\left(x+\dfrac{4}{5}\right)=\dfrac{3}{5}+\dfrac{2}{5}\)
\(x+\dfrac{4}{5}=1\)
\(x=1-\dfrac{4}{5}\)
\(x=\dfrac{1}{5}\)
c) chưa từng gặp dạng với giá trị tuyệt đối sory
d) \(\dfrac{1}{6}x-\dfrac{2}{3}=2\)
\(\dfrac{1}{6}x=2+\dfrac{2}{3}\)
\(\dfrac{1}{6}x=\dfrac{8}{3}\)
\(x=\dfrac{8}{3}:\dfrac{1}{6}\)
\(x=16\)
e) \(\dfrac{x}{2,5}=\dfrac{4}{5}\)
=> x.5 = 4.2,5
5x=10
x=10:5
x=2
f) |x-2|-4=1
|x-2|=1+4
|x-2|=5
=>\(\left[{}\begin{matrix}x-2=5\\x-2=-5\end{matrix}\right.\) =>\(\left[{}\begin{matrix}x=5+2\\x=-5+2\end{matrix}\right.\) =>\(\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\)
đôi khi cũng có sai sót , hãy xem lại thật kĩ
\(\dfrac{x-1}{65}+\dfrac{x-3}{63}=\dfrac{x-5}{61}+\dfrac{x-7}{59}\)
=>\(\left(\dfrac{x-1}{65}-1\right)+\left(\dfrac{x-3}{63}-1\right)=\left(\dfrac{x-5}{61}-1\right)+\left(\dfrac{x-7}{59}-1\right)\)
=>\(\dfrac{x-66}{65}+\dfrac{x-66}{63}-\dfrac{x-66}{61}-\dfrac{x-66}{59}=0\)
=>x-66=0
=>x=66